358 multiple-choice questions and 266 exam-style questions with mark schemes, organised by chapter, with answers you can check as you go. Free, no sign-up.
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Q1The equation 2x² + 3x + 5 = 0 has:
Correct answer: C — No real roots
Δ = b² − 4ac = 9 − 40 = −31. A negative discriminant means no real roots. Since a > 0 the parabola opens upward, and its minimum lies entirely above the x-axis.
Q2For x² − 6x + 9 = 0, the discriminant is:
Correct answer: A — 0
Δ = (−6)² − 4(1)(9) = 36 − 36 = 0, so the roots are equal. The expression is (x − 3)², and the graph touches the x-axis at x = 3 instead of cutting it.
Q3If α and β are the roots of x² − 7x + 12 = 0, then α + β equals:
Correct answer: B — 7
α + β = −b/a = −(−7)/1 = 7. The two negative signs cancel, which is where most errors occur. Check: the roots are 3 and 4, and 3 + 4 = 7, 3 × 4 = 12 = c/a.
Q4Solving x² = 4x by dividing both sides by x gives x = 4. The error is that:
Correct answer: B — The root x = 0 has been lost
Dividing by x assumes x ≠ 0 and discards that root. Correct method: x² − 4x = 0, so x(x − 4) = 0, giving x = 0 or x = 4.
Q5The graph of y = −2x² + 3x + 1 has:
Correct answer: B — A maximum point
a = −2 is negative, so the parabola opens downward and its turning point is a maximum. Every quadratic has exactly one turning point.
Q6The quadratic equation whose roots are 2 and −5 is:
Correct answer: A — x² + 3x − 10 = 0
Sum = −3 and product = −10, so x² − (−3)x + (−10) = x² + 3x − 10 = 0. Expanding (x − 2)(x + 5) confirms it. Option B comes from dropping the negative sign in the sum formula.
Q7The domain of f(x) = 1/(x − 7) is:
Correct answer: C — all real x except 7
The only forbidden value is the one making the denominator zero, namely x = 7. Everything else, including negatives, is allowed.
Q8The range of f(x) = |x| is:
Correct answer: B — y ≥ 0
A modulus is never negative, and it reaches 0 at x = 0. So the range includes zero, which rules out option C.
Q9If f(x) = 5x − 2, then f⁻¹(x) is:
Correct answer: A — (x + 2)/5
From y = 5x − 2 we get x = (y + 2)/5. Check: f((x+2)/5) = 5(x+2)/5 − 2 = x ✓. Option D confuses the inverse function with the reciprocal.
Q10If f(x) = x + 3 and g(x) = 2x, then fg(4) equals:
Correct answer: A — 11
The inner function acts first: g(4) = 8, then f(8) = 11. Option B is gf(4), which shows why the order matters.
Q11Which function has no inverse over all of ℝ?
Correct answer: C — f(x) = x²
x² is not one-one, since x and −x share an image. A cubic is one-one over all of ℝ because it is always increasing, so it does have an inverse.
Q12The graph of y = 3ˣ passes through:
Correct answer: B — (0, 1)
Any positive base to the power 0 equals 1, so every exponential graph of this form passes through (0, 1) and never touches the x-axis.
Q13The range of f(x) = −x² + 5 is:
Correct answer: B — y ≤ 5
The negative coefficient turns the parabola downward, so the vertex (0, 5) is a maximum and every output is 5 or less.
Q14For a function to have an inverse it must be:
Correct answer: C — bijective
One-one guarantees each output identifies its input; onto guarantees every element of the codomain has one. Both together — bijective — are exactly what an inverse needs.
Q15The gradient of 6x + 3y − 9 = 0 is:
Correct answer: B — −2
m = −a/b = −6/3 = −2. Rearranging gives y = −2x + 3, which agrees.
Q16The distance from (0, 0) to the line 3x + 4y − 10 = 0 is:
Correct answer: B — 2
Substitute the point into the general form and divide by √(a² + b²): |3(0) + 4(0) − 10| / √(9 + 16) = 10/5 = 2. The 3–4–5 pattern in the coefficients is why the square root comes out exactly, and examiners choose such coefficients deliberately.
Q17Two lines have gradients 4 and −1/4. They are:
Correct answer: B — perpendicular
4 × (−1/4) = −1, which is the perpendicularity condition. It also makes the denominator of the angle formula zero, so tan θ is undefined and θ = 90°.
Q18The line with x-intercept 2 and y-intercept 5 is:
Correct answer: A — x/2 + y/5 = 1
The intercept form puts each intercept under its own variable. Checking: at y = 0 we get x = 2 ✓, and at x = 0 we get y = 5 ✓.
Q19The centroid of the triangle with vertices (0,0), (6,0) and (0,9) is:
Correct answer: B — (2, 3)
Average each coordinate: x = (0+6+0)/3 = 2 and y = (0+0+9)/3 = 3. Option A is the midpoint of the hypotenuse, a different point.
Q20The point dividing the segment from (0,0) to (10,0) in the ratio 3 : 2 is at:
Correct answer: B — (6, 0)
(3×10 + 2×0)/(3+2) = 30/5 = 6. Three parts of the five lie behind it, so it sits closer to the far end — as the answer shows.
Q21The perpendicular distance formula gives a negative value when:
Correct answer: B — never — the modulus makes it positive
The numerator is taken in modulus, so the result is always positive. A distance cannot be negative, whichever side of the line the point lies on.
Q22A line parallel to 2x − y + 3 = 0 through (1, 1) is:
Correct answer: A — 2x − y − 1 = 0
Parallel means the same gradient of 2, so only the constant changes. Substituting (1,1) into 2x − y + k = 0 gives k = −1.
Q23One radian is the angle for which:
Correct answer: B — the arc equals the radius
The definition compares the arc with the radius. The last option describes a different angle — an arc equal to the radius is 1 rad, but a chord equal to the radius is π/3.
Q24135° in radians is:
Correct answer: A — 3π/4
135 × π/180 = 135π/180. Dividing top and bottom by 45 gives 3π/4.
Q25An arc of length 15 cm subtends 1.25 rad at the centre. The radius is:
Correct answer: B — 12 cm
Rearranging s = rθ gives r = s/θ = 15 / 1.25 = 12 cm. Multiplying instead of dividing gives 18.75, the tempting wrong answer.
Q26A sector has radius 6 cm and angle 0.5 rad. Its perimeter is:
Correct answer: C — 15 cm
The arc is rθ = 3 cm, but the perimeter also includes both radii: 3 + 6 + 6 = 15 cm. Answering 3 cm is the most common error in the topic.
Q27The area of a sector of radius 4 cm with angle 1.5 rad is:
Correct answer: B — 12 cm²
A = ½r²θ = ½ × 16 × 1.5 = 12 cm². Using ½rθ instead of ½r²θ gives 3, and forgetting the ½ gives 24.
Q28Which expression gives the area of a segment?
Correct answer: B — ½r²(θ − sin θ)
A segment is the sector minus the triangle: ½r²θ − ½r² sin θ, which factorises to ½r²(θ − sin θ). The order matters — for θ in this range θ > sin θ, so the reversed version would be negative.
Q29Your calculator gives sin 1.2 = 0.0209. This tells you:
Correct answer: B — the calculator is in degree mode
In radians sin 1.2 ≈ 0.932. Getting 0.0209 means the calculator read 1.2 as 1.2 degrees, so it is in degree mode and every result in the question will be wrong.
Q30Why does s = rθ produce a length even though θ has no unit?
Correct answer: B — because a length times a pure number is a length
θ is a ratio of two lengths, so it is dimensionless. Multiplying the radius by a pure number leaves the unit of the radius untouched, which is exactly what an arc length should be.
Q31On the unit circle, what does cos θ represent?
Correct answer: B — The horizontal distance of the point from the centre
The point sits at (cos θ, sin θ). Cosine is the x-coordinate — how far across — and sine is the y-coordinate — how far up. This is why cos starts at 1 (fully right) while sin starts at 0.
Q32Why is tan(90°) undefined?
Correct answer: B — Because cos(90°) = 0 and you cannot divide by zero
tan θ = sin θ / cos θ. At 90° the point is straight up at (0, 1), so cos(90°) = 0 and the fraction has a zero denominator. On a graph tan shoots off to infinity there — a vertical asymptote.
Q33What is sin(210°)?
Correct answer: B — −0.5
210° is 30° past the 180° mark, putting the point in the third quadrant — left and below centre. The height has the same magnitude as sin(30°) = 0.5 but is now below the axis, so sin(210°) = −0.5.
Q34Convert 60° to radians.
Correct answer: B — π/3
Since 180° = π rad, one degree is π/180 rad. So 60° = 60π/180 = π/3. Quick check: π/3 ≈ 1.047 rad, and 1.047 × 57.3 ≈ 60°.
Q35sin²θ + cos²θ equals:
Correct answer: B — 1
It is Pythagoras applied to the radius. The point (cos θ, sin θ) is 1 unit from the origin, so cos²θ + sin²θ = 1². It holds for every angle without exception, which is why it is the workhorse identity of trigonometry.
Q36The graph of y = sin x is shifted so it starts at its maximum. What function is that?
Correct answer: A — cos x
Cosine is sine shifted left by 90°: cos x = sin(x + 90°). At x = 0 cosine is at its peak of 1, while sine is at 0 and climbing. Both are the same wave viewed from a different starting angle.
Q37The 15th term of the AP 3, 7, 11, … is:
Correct answer: A — 59
a = 3 and d = 4, so a₁₅ = 3 + 14(4) = 59. Using 15 × 4 instead of 14 × 4 gives 63, the standard off-by-one.
Q38The common ratio of 81, 27, 9, 3, … is:
Correct answer: C — 1/3
Divide any term by the one before: 27/81 = 1/3. The terms are shrinking, so r must be less than 1 in size.
Q39The sum to infinity of 6 + 3 + 1.5 + … is:
Correct answer: B — 12
a = 6 and r = 0.5, and |r| < 1 so the sum exists: 6/(1 − 0.5) = 12.
Q40The series 2 + 6 + 18 + 54 + … has a sum to infinity of:
Correct answer: D — none — it diverges
r = 3, so |r| ≥ 1 and the terms grow rather than shrink. Applying the formula anyway would give 2/(1−3) = −1, an obviously impossible negative total for a series of positive terms.
Q41Inserting three arithmetic means between 4 and 20 gives a sequence with how many terms?
Correct answer: C — 5
The two given numbers plus the three inserted make five terms in all, so 20 = 4 + 4d and d = 4.
Q42The geometric mean of 4 and 25 is:
Correct answer: B — 10
√(4 × 25) = √100 = 10. Option A is the arithmetic mean, which is larger — as AM ≥ GM guarantees.
Q43Σr from r = 1 to 20 equals:
Correct answer: A — 210
Use the standard result Σr = n(n+1)/2 with n = 20: (20 × 21)/2 = 210. Adding the twenty numbers by hand gives the same answer, but the formula is the point — it turns a twenty-step addition into one multiplication and one division, and the same idea scales to Σr² and Σr³.
Q44The sum of the first n terms of an AP is quadratic in n because:
Correct answer: B — Sₙ = (n/2)[2a + (n−1)d] expands to include n²
Multiplying out gives an + n(n−1)d/2, which contains an n² term. That is why the running-total line on the diagram curves while the terms themselves lie on a straight line.
Q45What does f′(3) = 5 tell you about the graph of f?
Correct answer: B — At x = 3 the curve is rising with slope 5
A derivative is a slope, not a height. f′(3) = 5 says that at the instant x = 3, the curve climbs 5 units of y for every 1 unit of x. The actual height at x = 3 is f(3), which is a completely different number.
Q46Differentiate f(x) = 4x³ − 7x + 2.
Correct answer: A — 12x² − 7
Power rule on each term: 4x³ → 3·4x² = 12x². Then −7x → −7 (the x disappears). The constant +2 → 0, because a constant never changes, so its rate of change is zero.
Q47In the limit definition, why must h approach 0 rather than simply equal 0?
Correct answer: B — Because the formula would give 0/0, which is undefined
Substituting h = 0 directly gives [f(x) − f(x)]/0 = 0/0 — meaningless. The limit lets us ask what the expression heads toward as h shrinks, without ever dividing by zero.
Q48What is the derivative of sin(5x)?
Correct answer: B — 5 cos(5x)
Chain rule. Outer function sin → cos(5x). Inner function 5x has derivative 5. Multiply them: 5 cos(5x). Forgetting the inner 5 is the single most common slip in calculus.
Q49A Riemann sum with n = 10 left rectangles underestimates the area under an increasing curve. What happens as n increases?
Correct answer: B — The estimate converges toward the true area
Each rectangle misses a small triangular sliver above it. Narrower rectangles mean smaller slivers, so the total gap shrinks toward zero. That convergence is exactly what the integral is defined as.
Q50Evaluate ∫₁³ 2x dx.
Correct answer: B — 8
The antiderivative of 2x is x². Apply the Fundamental Theorem: F(3) − F(1) = 9 − 1 = 8. Sanity check: the region is a trapezium with parallel sides 2 and 6 and width 2, giving ½(2+6)(2) = 8.
Q51∫ x³ dx equals:
Correct answer: C — x⁴/4 + C
Raise the power by one and divide by the new power: x⁴/4 + C. Answer A is the derivative rather than the integral — the classic direction error. Check any integration by differentiating your answer: d/dx (x⁴/4) = x³. ✓
Q52Why does the power rule for integration exclude n = −1?
Correct answer: B — It would require dividing by zero
The rule divides by n + 1, which is zero when n = −1. The integral certainly exists — it is ln|x| + C. The modulus signs matter, since 1/x is defined for negative x but the logarithm is not.
Q53∫₀² 3x² dx equals:
Correct answer: A — 8
The antiderivative is x³, so the value is 2³ − 0³ = 8. A common slip is forgetting to subtract the lower limit, or integrating to x³/3 by dividing by the new power while ignoring the 3 already in front.
Q54The "+C" appears in indefinite integrals because:
Correct answer: B — Differentiation destroys constants, so many functions share one derivative
x², x² + 5 and x² − 100 all differentiate to 2x, so reversing the process cannot recover which one you started from. C stands for that whole family. It disappears in definite integrals because it cancels in F(b) − F(a).
Q55Which technique is the reverse of the chain rule?
Correct answer: B — Integration by substitution
Substitution undoes the chain rule; integration by parts undoes the product rule. Recognising which rule created the expression is usually the fastest route to choosing the right technique.
Q56A velocity–time graph is integrated with respect to time. The result represents:
Correct answer: B — Displacement
Integrating velocity accumulates distance travelled, giving displacement — the same thing as the area under the graph. Differentiating instead would give acceleration, which is the mirror-image operation.
The discriminant is b² − 4ac. It gives the number of real roots: two if positive, one repeated if zero, none if negative.
x² + kx + 9 = 0 has exactly one real root.b² − 4ac = 0 → k² − 36 = 0 → k = ±6
y = (x − 3)² + 5 and state whether it is a maximum or a minimum.(3, 5), a minimum, because the coefficient of the squared term is positive.
2x² − 7x + 3 = 0 by two different methods and show that they agree.(2x − 1)(x − 3) = 0[1]x = 1/2 or x = 3[1]a = 2, b = −7, c = 3[1]x = [7 ± √(49 − 24)] / 4[1]= (7 ± 5)/4[1]x = 3 or x = 1/2, the same pair[1]x = 1/2 and x = 3
h = 20t − 5t².h = −5(t² − 4t)[1]t² − 4t = (t − 2)² − 4[1]h = −5[(t − 2)² − 4] = 20 − 5(t − 2)²[1]= 20 mthe squared term is zero there[1]t = 2 s[1]h = 0: t(20 − 5t) = 0[1]t = 4 st = 0 is the moment of throwing[1](a) 20 − 5(t − 2)² (b) 20 m at t = 2 s (c) 4 s
y = x + k is a tangent to the curve y = x² + 3x + 4.x² + 2x + (4 − k) = 0. [2]x + k = x² + 3x + 4[1]x² + 2x + 4 − k = 0[1]b² − 4ac = 0[1]4 − 4(4 − k) = 0[1]4 − 16 + 4k = 0 → k = 3[1]x² + 2x + 1 = 0 → x = −1, so the point is (−1, 2)[1]k = 3, touching at (−1, 2)
f(x) = 1 / (x² − 9).The denominator is zero when x² = 9, that is at x = 3 and x = −3. The domain is therefore all real numbers except 3 and −3.
f(x) = x² defined on all of ℝ has no inverse function.It is not one-one: f(3) = f(−3) = 9, so the output 9 does not identify a unique input. An inverse would have to assign two values to 9, which no function may do. Restricting the domain to x ≥ 0 makes it one-one and the inverse √x then exists.
f(x) = x² + 3.Since x² ≥ 0 for every real x, the smallest value of f is 3, reached at x = 0. The range is f(x) ≥ 3.
f(x) = 2x + 1 and g(x) = x² − 3, find fg(x), gf(x) and the value of x for which fg(x) = gf(x).fg(x) = f(x² − 3) = 2(x² − 3) + 1 = 2x² − 5substitute the whole of g into f[1]gf(x) = g(2x + 1) = (2x + 1)² − 3 = 4x² + 4x − 2expand the bracket fully[1]2x² − 5 = 4x² + 4x − 2, so 2x² + 4x + 3 = 0[1]= 16 − 24 = −8 < 0, so there is no real value of x for which they are equala reasoned "no solution" is the answer, not an omission[1]fg(x) = 2x² − 5; gf(x) = 4x² + 4x − 2; no real x makes them equal
f(x) = (2x − 3)/(x + 1) is defined for x ≠ −1. Find f⁻¹(x).y = (2x − 3)/(x + 1) and multiply up: y(x + 1) = 2x − 3clear the fraction before rearranging[1]xy + y = 2x − 3, so xy − 2x = −3 − ygather every term containing x on one side[1]x(y − 2) = −(3 + y), so x = −(y + 3)/(y − 2)factorising out x is the key step[1]f⁻¹(x) = −(x + 3)/(x − 2), or equivalently (x + 3)/(2 − x)either form accepted[1]f⁻¹(x) = (x + 3)/(2 − x), x ≠ 2
f is defined by f(x) = √(x − 4).x − 4 ≥ 0, so the domain is x ≥ 4a square root requires a non-negative argument[1]f(x) ≥ 0[1]y = √(x − 4) gives y² = x − 4, so x = y² + 4squaring is safe here because y ≥ 0[1]f⁻¹(x) = x² + 4[1]x ≥ 0not all of ℝ — it must match the range of f[1](a) domain x ≥ 4, range f(x) ≥ 0 (b) f⁻¹(x) = x² + 4 with domain x ≥ 0 (c) inverting swaps domain and range
5x − 2y + 7 = 0.Using m = −a/b = −5/(−2) = 5/2. Rearranging confirms it: 2y = 5x + 7, so y = (5/2)x + 3.5.
Intercept form: x/4 + y/(−3) = 1, which rearranges to 3x − 4y = 12.
Their gradients multiply to −1, so each is the negative reciprocal of the other. Example: y = 2x and y = −x/2, since 2 × (−1/2) = −1.
(−1, 4) to the line 3x + 4y − 10 = 0.a = 3, b = 4, c = −10, with the line already in general form[1]3(−1) + 4(4) − 10 = −3 + 16 − 10 = 3[1]√(9 + 16) = 5[1]d = |3|/5 = 0.6 unitsthe modulus makes the distance positive[1]0.6 units
y = 3x + 1 and y = −2x + 5.m₁ = 3 and m₂ = −2[1]tan θ = |(−2 − 3)/(1 + (3)(−2))| = |−5/−5|both numerator and denominator are negative[1]= 1[1]θ = 45°[1]45°
A(1, 2), B(7, 4) and C(3, 8).= ((7+3)/2, (4+8)/2) = (5, 6)a median joins a vertex to the midpoint of the opposite side[1]= (6 − 2)/(5 − 1) = 1, so y − 2 = 1(x − 1), giving y = x + 1[1]= ((1+7+3)/3, (2+4+8)/3) = (11/3, 14/3)average of the three vertices[1]= (4 − 2)/(7 − 1) = 1/3, so AB is x − 3y + 5 = 0from y − 2 = (1/3)(x − 1), cleared of fractions[1]|3 − 24 + 5| = 16[1]d = 16/√(1 + 9) = 16/√10 = 5.06 unitsaccept 16√10/10[1](a) y = x + 1 (b) (11/3, 14/3) (c) 16/√10 ≈ 5.06 units
A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. It has no units because it is a length divided by a length, so the units cancel.
Arc 7.2 cm; perimeter 31.2 cm
20.7 cm²
θ = 5/3 rad; sector 67.5 cm²; triangle 40.3 cm²; difference 27.2 cm² is the segment
Substituting θ = 2π gives A = ½r²(2π) = πr², the standard area of a circle. This works because 2π radians is a complete turn, so the sector becomes the entire disc.
a/sin A = b/sin B = c/sin C. Use it when you have a matched pair — a side and the angle opposite it — plus one more piece of information.
sin 30°, cos 60° and tan 45°.sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1
(1 − cos²θ)/(sin θ cos θ) = tan θ.1 − cos²θ = sin²θ, so the expression is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ.
AB = 7.0 cm, AC = 9.0 cm and angle BAC = 52°.a² = b² + c² − 2bc cos Atwo sides and the included angle[1]BC² = 49 + 81 − 2(7)(9)cos 52°[1]BC = 7.29 cm[1]½ab sin C = ½(7)(9)sin 52°[1]= 24.8 cm²[1]sin B = 9 sin 52° / 7.29 → B = 76.6°[1](a) 7.29 cm (b) 24.8 cm² (c) 76.6°
h = d tan 58°[1]h = (d + 45) tan 34°[1]d tan 58° = (d + 45) tan 34°[1]d(1.600 − 0.6745) = 30.35 → d = 32.8 m[1]h = 32.8 × tan 58° = 52.5 m[1]height ≈ 52.5 m, Q is ≈ 32.8 m from the foot
2 sin θ = 1 for 0° ≤ θ ≤ 360°.sin θ = 1/2[1]θ = 30°[1]θ = 180° − 30° = 150°[1]θ = 30° and 150°[1]θ = 30° and 150°
A sequence is an ordered list of terms, such as 2, 5, 8, 11. A series is the sum of the terms of a sequence, such as 2 + 5 + 8 + 11 = 26.
a = 7 and d = 4, so a₂₀ = 7 + 19(4) = 7 + 76 = 83.
The sum to infinity exists only when |r| < 1, and then S∞ = a / (1 − r).
S₁₀ = 5[2a + 9d] = 155, so 2a + 9d = 31divide through by 5 immediately to keep the numbers small[1]S₂₀ = 10[2a + 19d] = 610, so 2a + 19d = 61[1]10d = 30, so d = 3the 2a terms cancel[1]2a + 27 = 31, so a = 2check: S₁₀ = 5(4 + 27) = 155 ✓[1]a = 2, d = 3
0.4747… as a fraction by summing an infinite geometric series.0.47 + 0.0047 + 0.000047 + …each block of two digits is one term[1]a = 0.47 and r = 0.01each term is the previous one divided by 100[1]|r| < 1, so S∞ = 0.47 / (1 − 0.01) = 0.47 / 0.99the condition should be stated[1]= 47/99[1]47/99
a = 8 and r = 0.75; after the 4th bounce the height is 8(0.75)⁴the drop height is the term before the first bounce[1]= 2.53 m (3 s.f.)[1]8(0.75) = 6 and r = 0.75, so their sum is 6/(1 − 0.75) = 24[1]= 8 + 2(24) = 56 m[1]|r| < 1, the heights shrink geometrically towards zero fast enough for the infinite sum to converge to a finite limitthe explanation must refer to |r| < 1[1](a) 2.53 m (b) 56 m (c) the heights form a convergent GP since |r| < 1
y = 5x³ − 2/x with respect to x.Write it as 5x³ − 2x⁻¹. Then dy/dx = 15x² + 2x⁻², that is 15x² + 2/x².
y = x² − 4x + 1 at the point where x = 3.dy/dx = 2x − 4, so at x = 3 the gradient is 2(3) − 4 = 2.
At a stationary point, if d²y/dx² < 0 it is a maximum; if d²y/dx² > 0 it is a minimum.
y = x³ − 3x² − 9x + 5.dy/dx. [1]dy/dx = 3x² − 6x − 9[1]3(x² − 2x − 3) = 0[1](x − 3)(x + 1) = 0, so x = 3 or x = −1[1]x = 3: y = 27 − 27 − 27 + 5 = −22[1]x = −1: y = −1 − 3 + 9 + 5 = 10[1]d²y/dx² = 6x − 6; at x = 3 it is +12, a minimum[1]x = −1 it is −12, a maximum[1]Minimum at (3, −22), maximum at (−1, 10)
V = x(24 − 2x)². [2]24 − 2xa square is removed from both ends[1]V = x(24 − 2x)²[1]V = 576x − 96x² + 4x³[1]dV/dx = 576 − 192x + 12x²[1]12(x² − 16x + 48) = 0 → (x − 4)(x − 12) = 0[1]x = 4; x = 12 is rejected because it leaves no base[1]V = 4 × 16² = 4 × 256[1]V = 1024 cm³[1]x = 4 cm giving V = 1024 cm³
s = 2t³ − 9t² + 12t metres after t seconds.v = ds/dt = 6t² − 18t + 12[1]a = dv/dt = 12t − 18[1]v = 0: 6(t² − 3t + 2) = 0 → (t − 1)(t − 2) = 0[1]t = 1 s and t = 2 s[1]t = 2: a = 24 − 18 = 6 m s⁻²[1]v = 6t² − 18t + 12, a = 12t − 18; at rest at t = 1 s and 2 s; a = 6 m s⁻² at t = 2 s
∫ (6x² − 4x + 3) dx.2x³ − 2x² + 3x + c
∫₁³ (2x + 1) dx.[x² + x]₁³ = (9 + 3) − (1 + 1) = 10
The area between the curve and the x axis, between the two limits. Area below the axis counts as negative.
y = x² − 4x crosses the x axis at two points.y = 0: x(x − 4) = 0[1]x = 0 and x = 4[1]∫(x² − 4x) dx = x³/3 − 2x²[1](64/3 − 32) − 0 = −32/3[1]= 32/3 ≈ 10.7 square unitsthe magnitude is taken[1](a) x = 0 and x = 4 (b) 32/3 square units
(2, 9) and has gradient dy/dx = 3x² − 2x.x = 0 and x = 2. [3]y = x³ − x² + c[1]9 = 8 − 4 + c[1]c = 5[1]y = x³ − x² + 5[1]∫₀² (x³ − x² + 5) dx = [x⁴/4 − x³/3 + 5x]₀²[1]= 4 − 8/3 + 10[1]= 34/3 ≈ 11.3 square units[1](a) y = x³ − x² + 5 (b) 34/3 square units
y = x + 2 and y = x² is to be found.x + 2 = x² → x² − x − 2 = 0[1](x − 2)(x + 1) = 0, so x = −1 and x = 2[1]= ∫₋₁² [(x + 2) − x²] dx[1]= [x²/2 + 2x − x³/3]₋₁²[1]= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2[1]x = −1 and 2; area = 9/2 square units
Q1|−7| equals:
Correct answer: B — 7
The modulus discards the sign and returns the size, which is 7.
Q2The corner of y = |5x − 15| is at:
Correct answer: B — (3, 0)
The corner is where the bracket is zero: 5x − 15 = 0 gives x = 3, and the y-value there is 0.
Q3How many solutions does |x + 1| = 6 have?
Correct answer: C — two
Both x + 1 = 6 and x + 1 = −6 are possible, giving x = 5 and x = −7. A horizontal line above the corner cuts the V twice.
Q4|x − 2| < 5 is equivalent to:
Correct answer: B — −3 < x < 7
x is within 5 of 2, so it lies between 2 − 5 = −3 and 2 + 5 = 7. Less-than always gives one interval.
Q5|x + 4| > 2 is equivalent to:
Correct answer: B — x < −6 or x > −2
x is further than 2 from −4, so it lies outside the interval, giving two separate regions.
Q6√(x²) is equal to:
Correct answer: B — |x|
The square root sign denotes the non-negative root, so for x = −3 it gives 3, not −3. That is exactly what |x| does.
Q7To solve |2x − 1| = |x + 4|, the safest first step is:
Correct answer: A — square both sides
With moduli on both sides, |a| = |b| ⟺ a² = b², so squaring removes both bars without introducing false solutions.
Q8A student solves |x − 3| = 2x by squaring and gets x = 1 and x = −3. They should:
Correct answer: B — check both, and reject x = −3
Only one side carried a modulus, so squaring can invent solutions. At x = −3 the right side is −6, and a modulus cannot be negative, so that root must be discarded.
Q9The remainder when x³ − 2x + 4 is divided by (x − 1) is:
Correct answer: B — 3
f(1) = 1 − 2 + 4 = 3. The remainder theorem replaces the whole long division with one substitution.
Q10(x − 5) is a factor of f(x) if:
Correct answer: A — f(5) = 0
The root of x − 5 is +5, so the factor theorem requires f(5) = 0.
Q11Before dividing x⁴ − 1 by (x − 1), you should write it as:
Correct answer: B — x⁴ + 0x³ + 0x² + 0x − 1
Long division aligns terms by degree, so every missing power must appear with a zero coefficient or the columns go out of step.
Q12The remainder when f(x) is divided by (2x − 1) is:
Correct answer: C — f(1/2)
Substitute the root of the divisor. 2x − 1 = 0 gives x = 1/2.
Q13If f(x) = x³ + kx − 10 and (x − 2) is a factor, then k is:
Correct answer: A — 1
f(2) = 8 + 2k − 10 = 0 gives 2k = 2, so k = 1.
Q14Dividing a cubic by a linear factor gives a quotient that is:
Correct answer: B — quadratic
The degrees subtract: 3 − 1 = 2. That is why factorising a cubic reduces the problem to a quadratic you can already handle.
Q15A whole-number root of x³ + 2x² − 5x − 6 must divide:
Correct answer: C — 6
Any integer root divides the constant term, so the candidates are ±1, ±2, ±3, ±6. Testing x = −1 gives 0, so (x + 1) is a factor.
Q16If dividing f(x) by (x − a) leaves remainder 0, then:
Correct answer: B — (x − a) is a factor of f(x)
A zero remainder means the division is exact, which is precisely what it means for (x − a) to be a factor.
Q17log₄ 64 equals:
Correct answer: A — 3
Ask "4 to what power gives 64". 4¹ = 4, 4² = 16, 4³ = 64, so the answer is 3. Option B confuses the base with the answer.
Q18log 2 + log 50 equals:
Correct answer: B — 2
The product law gives log(2 × 50) = log 100 = 2, since 10² = 100. Option A applies a law for sums that does not exist.
Q19The characteristic of log 0.0072 is:
Correct answer: C — −3
In standard form 0.0072 = 7.2 × 10⁻³, so the characteristic is the power of 10, namely −3, written 3̄. Shortcut for a number below 1: count the zeros between the decimal point and the first significant figure (here 2) and add one.
Q20If log x = 2.4771, and log 3 = 0.4771, then x is:
Correct answer: A — 300
Same mantissa 0.4771 means the same digits as 3. Characteristic 2 means the power of 10 is 2, so x = 3 × 10² = 300. This is exactly how log tables were used for multiplication.
Q21Which expression equals 3 log x − log y?
Correct answer: B — log(x³/y)
The multiplier 3 goes back up as a power giving log(x³), and a difference of logs is the log of a quotient. So the answer is log(x³/y).
Q22Solve 2ˣ = 1/8.
Correct answer: B — x = −3
1/8 = 8⁻¹ = (2³)⁻¹ = 2⁻³, so x = −3. A number smaller than 1 always gives a negative logarithm when the base is greater than 1.
Q23log₃ 5 written in common logs is:
Correct answer: B — log 5 / log 3
Change of base: logₐ b = log b / log a. A quotient of two logarithms is not the same as the logarithm of a quotient — that would be option A or D.
Q24For log₁₀ x to exist, x must be:
Correct answer: B — greater than 0
10 raised to any real power is strictly positive, so no exponent produces 0 or a negative number. The domain of the log function is x > 0, with zero excluded as well as the negatives.
Q25sec θ is equal to:
Correct answer: B — 1/cos θ
The third letter of "sec" is c, which points to cosine. Matching the first letters instead gives the standard wrong answer.
Q26The identity 1 + cot²θ = cosec²θ comes from dividing sin²θ + cos²θ = 1 by:
Correct answer: B — sin²θ
Dividing by sin²θ turns the first term into 1, the second into cot²θ, and the right side into cosec²θ.
Q27cos(A + B) equals:
Correct answer: B — cos A cos B − sin A sin B
The cosine formula reverses the sign: a plus inside gives a minus outside. The last option is sin(A + B).
Q28Which is NOT a valid form of cos 2A?
Correct answer: D — 2 sin A cos A
2 sin A cos A is sin 2A, not cos 2A. The other three are equivalent forms of cos 2A.
Q29For 8 sin θ + 6 cos θ = R sin(θ + α), the value of R is:
Correct answer: B — 10
R = √(8² + 6²) = √100 = 10. Adding the coefficients instead of using Pythagoras gives 14, the tempting wrong answer.
Q30The maximum value of 7 sin θ − 24 cos θ is:
Correct answer: B — 25
The maximum is R = √(49 + 576) = √625 = 25, since the combined wave has amplitude R.
Q31Solving sin 2θ = 0.5 for 0° ⩽ θ ⩽ 360°, the interval you should search is:
Correct answer: B — 0° to 720°
If θ reaches 360° then 2θ reaches 720°, so all solutions must be found across two revolutions before halving.
Q32To solve 2 sin θ cos θ = sin θ, you should:
Correct answer: B — factorise as sin θ(2cos θ − 1) = 0
Dividing by sin θ throws away every solution where sin θ = 0. Factorising keeps both families of solutions.
Q33What does f′(3) = 5 tell you about the graph of f?
Correct answer: B — At x = 3 the curve is rising with slope 5
A derivative is a slope, not a height. f′(3) = 5 says that at the instant x = 3, the curve climbs 5 units of y for every 1 unit of x. The actual height at x = 3 is f(3), which is a completely different number.
Q34Differentiate f(x) = 4x³ − 7x + 2.
Correct answer: A — 12x² − 7
Power rule on each term: 4x³ → 3·4x² = 12x². Then −7x → −7 (the x disappears). The constant +2 → 0, because a constant never changes, so its rate of change is zero.
Q35In the limit definition, why must h approach 0 rather than simply equal 0?
Correct answer: B — Because the formula would give 0/0, which is undefined
Substituting h = 0 directly gives [f(x) − f(x)]/0 = 0/0 — meaningless. The limit lets us ask what the expression heads toward as h shrinks, without ever dividing by zero.
Q36What is the derivative of sin(5x)?
Correct answer: B — 5 cos(5x)
Chain rule. Outer function sin → cos(5x). Inner function 5x has derivative 5. Multiply them: 5 cos(5x). Forgetting the inner 5 is the single most common slip in calculus.
Q37A Riemann sum with n = 10 left rectangles underestimates the area under an increasing curve. What happens as n increases?
Correct answer: B — The estimate converges toward the true area
Each rectangle misses a small triangular sliver above it. Narrower rectangles mean smaller slivers, so the total gap shrinks toward zero. That convergence is exactly what the integral is defined as.
Q38Evaluate ∫₁³ 2x dx.
Correct answer: B — 8
The antiderivative of 2x is x². Apply the Fundamental Theorem: F(3) − F(1) = 9 − 1 = 8. Sanity check: the region is a trapezium with parallel sides 2 and 6 and width 2, giving ½(2+6)(2) = 8.
Q39∫ x³ dx equals:
Correct answer: C — x⁴/4 + C
Raise the power by one and divide by the new power: x⁴/4 + C. Answer A is the derivative rather than the integral — the classic direction error. Check any integration by differentiating your answer: d/dx (x⁴/4) = x³. ✓
Q40Why does the power rule for integration exclude n = −1?
Correct answer: B — It would require dividing by zero
The rule divides by n + 1, which is zero when n = −1. The integral certainly exists — it is ln|x| + C. The modulus signs matter, since 1/x is defined for negative x but the logarithm is not.
Q41∫₀² 3x² dx equals:
Correct answer: A — 8
The antiderivative is x³, so the value is 2³ − 0³ = 8. A common slip is forgetting to subtract the lower limit, or integrating to x³/3 by dividing by the new power while ignoring the 3 already in front.
Q42The "+C" appears in indefinite integrals because:
Correct answer: B — Differentiation destroys constants, so many functions share one derivative
x², x² + 5 and x² − 100 all differentiate to 2x, so reversing the process cannot recover which one you started from. C stands for that whole family. It disappears in definite integrals because it cancels in F(b) − F(a).
Q43Which technique is the reverse of the chain rule?
Correct answer: B — Integration by substitution
Substitution undoes the chain rule; integration by parts undoes the product rule. Recognising which rule created the expression is usually the fastest route to choosing the right technique.
Q44A velocity–time graph is integrated with respect to time. The result represents:
Correct answer: B — Displacement
Integrating velocity accumulates distance travelled, giving displacement — the same thing as the area under the graph. Differentiating instead would give acceleration, which is the mirror-image operation.
Q45f is continuous, f(2) = −3 and f(3) = 4. This shows that:
Correct answer: B — f has a root between 2 and 3
A sign change guarantees at least one root, not exactly one — there could be three. The word "exactly" makes the third option a stronger claim than the evidence supports.
Q46The sign-change test can fail to indicate a root when:
Correct answer: B — the function is not continuous
A discontinuity such as an asymptote lets the function change sign without ever taking the value zero. 1/(x − 2) is the standard counter-example.
Q47For the iteration x₍ₙ₊₁₎ = F(xₙ), the value it converges to satisfies:
Correct answer: B — x = F(x)
If the sequence settles, the input and output stop differing, so the limit satisfies x = F(x). That is why the rearrangement had to be into this form.
Q48An iteration produces 2, 5, 122, 1815848. This means:
Correct answer: B — it has diverged
The values are growing without limit, so this rearrangement diverges. Continuing will not help — a different rearrangement is needed.
Q49To show x³ − 5x + 1 = 0 can be written as x = ∛(5x − 1), you should:
Correct answer: B — start from the given equation and rearrange forwards
Examiners want the given equation transformed into the target. Working backwards from the answer proves the same thing but can lose the mark for direction.
Q50Iterating gives x₄ = 2.4471 and x₅ = 2.4468. To 3 decimal places the root is:
Correct answer: A — 2.447
Both values round to 2.447, so they agree at 3 d.p. and the iteration can stop. Quoting 2.4470 gives four decimal places, which is more than was asked for.
Q51Why keep more decimal places than the final answer needs?
Correct answer: B — because rounding each step lets the error build up
Each rounded value is fed back into the formula, so an early rounding error is carried into every subsequent term and can shift the final digit.
Q52Near the root, an iteration converges when the graph of y = F(x) is:
Correct answer: B — shallower than y = x
A shallower curve means each bounce between the curve and the line lands nearer the root. A steeper one magnifies the gap each time, which is the diverging case.
A modulus measures a distance, so |2x + 1| is never negative for any real x. It can therefore never equal −3, and the equation has no solutions.
x = 11 or x = −3
1 < x < 4
(a) corners (−2, 0) and (2, 0); (b) x = 1 or x = 4; (c) 1 < x < 4
The bracket is zero when 4x + 10 = 0, so x = −2.5 and the corner is at (−2.5, 0). The arms have gradients 4 and −4.
When a polynomial f(x) is divided by (x − a), the remainder is f(a) — the value obtained by substituting x = a into the polynomial.
x³ + 2x² − 5x + 1 is divided by (x − 2).f(2) = 8 + 8 − 10 + 1 = 7.
(x + 1) is a factor of x³ + 3x² + 3x + 1.f(−1) = −1 + 3 − 3 + 1 = 0. Since the remainder is zero, by the factor theorem (x + 1) is a factor. (In fact the polynomial is (x + 1)³.)
f(x) = 2x³ + ax² + bx − 6 is divided by (x − 1) the remainder is −6, and (x + 2) is a factor. Find a and b.f(1) = 2 + a + b − 6 = −6, so a + b = −2remainder theorem[1]f(−2) = −16 + 4a − 2b − 6 = 0, so 4a − 2b = 22 and 2a − b = 11factor theorem: the remainder is zero[1]3a = 9, so a = 3the b terms cancel because their coefficients are +1 and −1[1]b = −2 − 3 = −5check: f(1) = 2 + 3 − 5 − 6 = −6 ✓ and f(−2) = −16 + 12 + 10 − 6 = 0 ✓[1]a = 3, b = −5
x³ − 4x² + x + 6 by (x − 3) and hence factorise the polynomial completely.f(3) = 27 − 36 + 3 + 6 = 0, so (x − 3) is a factorconfirming the division will be exact[1]x² − x − 2by long or synthetic division[1]x² − x − 2 = (x − 2)(x + 1)[1]f(x) = (x − 3)(x − 2)(x + 1)[1](x − 3)(x − 2)(x + 1)
p(x) = x³ + px² + qx + 12 has (x − 2) and (x + 3) as factors.p(2) = 8 + 4p + 2q + 12 = 0, so 4p + 2q = −20 and 2p + q = −10factor theorem[1]p(−3) = −27 + 9p − 3q + 12 = 0, so 9p − 3q = 15 and 3p − q = 5[1]5p = −5, so p = −1the q terms cancel[1]q = −10 − 2(−1) = −8so p(x) = x³ − x² − 8x + 12[1]−12 (from −constant/leading coefficient), and two roots are 2 and −3or divide out both known factors[1]= −12 / (2 × −3) = 2, so the third factor is (x − 2) and p(x) = (x − 2)²(x + 3)check: p(x) = (x−2)²(x+3) expands to x³ − x² − 8x + 12 ✓[1](a) 2p + q = −10 and 3p − q = 5 (b) p = −1, q = −8 (c) (x − 2), giving p(x) = (x − 2)²(x + 3)
If ax = y, where a > 0 and a ≠ 1, then x is called the logarithm of y to the base a, written loga y = x. It is the power to which the base must be raised to give the number.
Because 1x = 1 for every value of x. No power of 1 can produce any number other than 1, so log₁ y would have no value for y ≠ 1 and infinitely many for y = 1.
In standard form 0.00456 = 4.56 × 10⁻³, so the characteristic is −3, written 3̄. The characteristic equals the power of 10 in standard form.
log₂ 64 + log₃ 81 − log₅ 125 without tables.log₂ 64 = 6 because 2⁶ = 64[1]log₃ 81 = 4 because 3⁴ = 81[1]log₅ 125 = 3, so the value is 6 + 4 − 3 = 7final answer required[1]7
log₂(x + 3) + log₂(x − 3) = 4.log₂[(x+3)(x−3)] = 4sum of logs with the same base becomes the log of a product[1](x+3)(x−3) = 2⁴ = 16[1]x² − 9 = 16, so x² = 25 and x = ±5difference of two squares[1]x = −5 because it makes x − 3 negative and the log undefined; x = 5the rejection must be stated with a reason[1]x = 5
N = 500 × 2t.N = 500 × 2⁶ = 500 × 64[1]N = 32 000 bacteria[1]32 000 = 500 × 2ᵗ so 2ᵗ = 64divide by the initial amount first[1]t log 2 = log 64power law brings t down[1]t = log 64 / log 2 = 1.8062 / 0.3010 = 6.00 hoursaccept t = 6 from inspection with working shown[1](a) 32 000 (b) t = 6.00 hours (c) in (b) the unknown is the exponent
Dividing every term of sin²θ + cos²θ = 1 by cos²θ gives sin²θ/cos²θ + 1 = 1/cos²θ. Since sin θ/cos θ = tan θ and 1/cos θ = sec θ, this is tan²θ + 1 = sec²θ.
sec θ = 1/cos θ, and cos 90° = 0. Division by zero is undefined, so sec 90° does not exist. The graph has a vertical asymptote at θ = 90°, with the curve tending to +∞ on one side and −∞ on the other.
θ = 0°, 60°, 300°, 360°
(a) 13 sin(θ − 67.4°); (b) θ = 97.4°, 217.4°; (c) maximum 13 at θ = 157.4°
Using cos 2θ = 1 − 2sin²θ, the numerator becomes 1 − (1 − 2sin²θ) = 2sin²θ. Using sin 2θ = 2 sin θ cos θ, the denominator is 2 sin θ cos θ. So the fraction is 2sin²θ / (2 sin θ cos θ) = sin θ / cos θ = tan θ.
y = 5x³ − 2/x with respect to x.Write it as 5x³ − 2x⁻¹. Then dy/dx = 15x² + 2x⁻², that is 15x² + 2/x².
y = x² − 4x + 1 at the point where x = 3.dy/dx = 2x − 4, so at x = 3 the gradient is 2(3) − 4 = 2.
At a stationary point, if d²y/dx² < 0 it is a maximum; if d²y/dx² > 0 it is a minimum.
y = x³ − 3x² − 9x + 5.dy/dx. [1]dy/dx = 3x² − 6x − 9[1]3(x² − 2x − 3) = 0[1](x − 3)(x + 1) = 0, so x = 3 or x = −1[1]x = 3: y = 27 − 27 − 27 + 5 = −22[1]x = −1: y = −1 − 3 + 9 + 5 = 10[1]d²y/dx² = 6x − 6; at x = 3 it is +12, a minimum[1]x = −1 it is −12, a maximum[1]Minimum at (3, −22), maximum at (−1, 10)
V = x(24 − 2x)². [2]24 − 2xa square is removed from both ends[1]V = x(24 − 2x)²[1]V = 576x − 96x² + 4x³[1]dV/dx = 576 − 192x + 12x²[1]12(x² − 16x + 48) = 0 → (x − 4)(x − 12) = 0[1]x = 4; x = 12 is rejected because it leaves no base[1]V = 4 × 16² = 4 × 256[1]V = 1024 cm³[1]x = 4 cm giving V = 1024 cm³
s = 2t³ − 9t² + 12t metres after t seconds.v = ds/dt = 6t² − 18t + 12[1]a = dv/dt = 12t − 18[1]v = 0: 6(t² − 3t + 2) = 0 → (t − 1)(t − 2) = 0[1]t = 1 s and t = 2 s[1]t = 2: a = 24 − 18 = 6 m s⁻²[1]v = 6t² − 18t + 12, a = 12t − 18; at rest at t = 1 s and 2 s; a = 6 m s⁻² at t = 2 s
∫ (6x² − 4x + 3) dx.2x³ − 2x² + 3x + c
∫₁³ (2x + 1) dx.[x² + x]₁³ = (9 + 3) − (1 + 1) = 10
The area between the curve and the x axis, between the two limits. Area below the axis counts as negative.
y = x² − 4x crosses the x axis at two points.y = 0: x(x − 4) = 0[1]x = 0 and x = 4[1]∫(x² − 4x) dx = x³/3 − 2x²[1](64/3 − 32) − 0 = −32/3[1]= 32/3 ≈ 10.7 square unitsthe magnitude is taken[1](a) x = 0 and x = 4 (b) 32/3 square units
(2, 9) and has gradient dy/dx = 3x² − 2x.x = 0 and x = 2. [3]y = x³ − x² + c[1]9 = 8 − 4 + c[1]c = 5[1]y = x³ − x² + 5[1]∫₀² (x³ − x² + 5) dx = [x⁴/4 − x³/3 + 5x]₀²[1]= 4 − 8/3 + 10[1]= 34/3 ≈ 11.3 square units[1](a) y = x³ − x² + 5 (b) 34/3 square units
y = x + 2 and y = x² is to be found.x + 2 = x² → x² − x − 2 = 0[1](x − 2)(x + 1) = 0, so x = −1 and x = 2[1]= ∫₋₁² [(x + 2) − x²] dx[1]= [x²/2 + 2x − x³/3]₋₁²[1]= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2[1]x = −1 and 2; area = 9/2 square units
If f(a) and f(b) have opposite signs then the curve must cross zero somewhere between a and b, so a root lies in that interval. The argument requires f to be continuous on the interval — otherwise the sign change could be caused by an asymptote rather than a root.
f(1) = −4, f(2) = 5; f continuous with a sign change, so a root lies in (1, 2)
1.57 (2 d.p.)
Starting from 3x³ − 4x + 2 = 0, move the terms across to get 3x³ = 4x − 2. Dividing by 3 gives x³ = (4x − 2)/3, and taking the cube root of both sides gives x = ∛((4x − 2)/3).
(a) sign change between 0.7 and 0.8; (b) α = 0.739; (c) the iteration diverges — use x = cos x instead
Q1|−7| equals:
Correct answer: B — 7
The modulus discards the sign and returns the size, which is 7.
Q2The corner of y = |5x − 15| is at:
Correct answer: B — (3, 0)
The corner is where the bracket is zero: 5x − 15 = 0 gives x = 3, and the y-value there is 0.
Q3How many solutions does |x + 1| = 6 have?
Correct answer: C — two
Both x + 1 = 6 and x + 1 = −6 are possible, giving x = 5 and x = −7. A horizontal line above the corner cuts the V twice.
Q4|x − 2| < 5 is equivalent to:
Correct answer: B — −3 < x < 7
x is within 5 of 2, so it lies between 2 − 5 = −3 and 2 + 5 = 7. Less-than always gives one interval.
Q5|x + 4| > 2 is equivalent to:
Correct answer: B — x < −6 or x > −2
x is further than 2 from −4, so it lies outside the interval, giving two separate regions.
Q6√(x²) is equal to:
Correct answer: B — |x|
The square root sign denotes the non-negative root, so for x = −3 it gives 3, not −3. That is exactly what |x| does.
Q7To solve |2x − 1| = |x + 4|, the safest first step is:
Correct answer: A — square both sides
With moduli on both sides, |a| = |b| ⟺ a² = b², so squaring removes both bars without introducing false solutions.
Q8A student solves |x − 3| = 2x by squaring and gets x = 1 and x = −3. They should:
Correct answer: B — check both, and reject x = −3
Only one side carried a modulus, so squaring can invent solutions. At x = −3 the right side is −6, and a modulus cannot be negative, so that root must be discarded.
Q9The remainder when x³ − 2x + 4 is divided by (x − 1) is:
Correct answer: B — 3
f(1) = 1 − 2 + 4 = 3. The remainder theorem replaces the whole long division with one substitution.
Q10(x − 5) is a factor of f(x) if:
Correct answer: A — f(5) = 0
The root of x − 5 is +5, so the factor theorem requires f(5) = 0.
Q11Before dividing x⁴ − 1 by (x − 1), you should write it as:
Correct answer: B — x⁴ + 0x³ + 0x² + 0x − 1
Long division aligns terms by degree, so every missing power must appear with a zero coefficient or the columns go out of step.
Q12The remainder when f(x) is divided by (2x − 1) is:
Correct answer: C — f(1/2)
Substitute the root of the divisor. 2x − 1 = 0 gives x = 1/2.
Q13If f(x) = x³ + kx − 10 and (x − 2) is a factor, then k is:
Correct answer: A — 1
f(2) = 8 + 2k − 10 = 0 gives 2k = 2, so k = 1.
Q14Dividing a cubic by a linear factor gives a quotient that is:
Correct answer: B — quadratic
The degrees subtract: 3 − 1 = 2. That is why factorising a cubic reduces the problem to a quadratic you can already handle.
Q15A whole-number root of x³ + 2x² − 5x − 6 must divide:
Correct answer: C — 6
Any integer root divides the constant term, so the candidates are ±1, ±2, ±3, ±6. Testing x = −1 gives 0, so (x + 1) is a factor.
Q16If dividing f(x) by (x − a) leaves remainder 0, then:
Correct answer: B — (x − a) is a factor of f(x)
A zero remainder means the division is exact, which is precisely what it means for (x − a) to be a factor.
Q17Which fraction is proper?
Correct answer: B — (x + 5)/(x² − 9)
Degree 1 over degree 2 is proper. Option D has equal degrees, which counts as improper and must be divided first.
Q18The partial fraction form of 1/[(x−1)(x+4)] is:
Correct answer: A — A/(x−1) + B/(x+4)
Two distinct linear factors give one constant over each. Option B is just the original fraction rewritten and decomposes nothing.
Q19The form for 1/[(x+2)(x+2)] is:
Correct answer: B — A/(x+2) + B/(x+2)²
A repeated factor needs a term for each power up to its multiplicity. Option C writes the same denominator twice, which just gives one constant A + B.
Q20For (2x+3)/[(x−1)(x²+1)], the numerator over x²+1 should be:
Correct answer: C — Ax + B
x² + 1 has no real factors, so it takes a numerator one degree lower — a general linear expression Ax + B.
Q21To find A in (3x−2)/[(x−4)(x+1)] = A/(x−4) + B/(x+1), substitute:
Correct answer: B — x = 4
x = 4 makes (x − 4) zero, killing the B term and leaving A alone. Substituting x = −1 would isolate B instead.
Q22(x² + 1)/(x² − 1) should first be:
Correct answer: B — divided out
The degrees are equal, so the fraction is improper. Division gives 1 + 2/(x² − 1), and only that remainder is decomposed.
Q23How many unknown constants does (x+1)/[(x−2)³] require?
Correct answer: C — 3
A cubed factor contributes A/(x−2) + B/(x−2)² + C/(x−2)³ — one term for each power from 1 to 3.
Q24If your equations for the constants are inconsistent, the likely cause is:
Correct answer: B — the wrong decomposition shape
An inconsistent system usually means too few unknowns were allowed for — a repeated factor given only one term, or a quadratic given only a constant on top.
Q25In a proof by induction, the basis step establishes:
Correct answer: B — that the statement is true for the first value of n
The basis anchors the chain at its start, usually n = 1. Establishing P(k) ⟹ P(k+1) is the separate inductive step.
Q26The expansion of (x + 2)⁵ has how many terms?
Correct answer: B — 6
n + 1 = 6. The powers of x run from 5 down to 0, which is six values.
Q27The fourth term of (a + b)⁸ uses which coefficient?
Correct answer: B — ⁸C₃
T_(r+1) uses ⁿCᵣ, so the fourth term has r = 3 and uses ⁸C₃ = 56. The off-by-one here is examined deliberately.
Q28In (a + b)⁹, the powers of a and b in any term sum to:
Correct answer: A — 9
Every term is aⁿ⁻ʳbʳ, and (n − r) + r = n = 9 whatever r is.
Q29The sum of all the entries in row 6 of Pascal's triangle is:
Correct answer: C — 64
Setting a = b = 1 in the binomial theorem gives Σ⁶Cᵣ = 2⁶ = 64. Checking: 1+6+15+20+15+6+1 = 64 ✓
Q30To find the coefficient of x³ in (1 + 2x)⁶, you would use:
Correct answer: B — ⁶C₃ × 2³
The general term is ⁶Cᵣ(2x)ʳ, so at r = 3 the coefficient is ⁶C₃ × 2³ = 20 × 8 = 160. The 2 must be cubed along with the x.
Q31An induction proof with a valid inductive step but no basis proves:
Correct answer: C — nothing about whether the statement is ever true
The step only says truth propagates forwards. Without a starting case that is actually true, the chain never begins — a false statement can have a perfectly valid inductive step.
Q32The term independent of x exists in an expansion when:
Correct answer: B — the equation for r has a whole-number solution in range
Setting the collected power of x to zero gives an equation for r. If its solution is not a whole number between 0 and n, no term has power zero and there is no constant term.
Q33log₄ 64 equals:
Correct answer: A — 3
Ask "4 to what power gives 64". 4¹ = 4, 4² = 16, 4³ = 64, so the answer is 3. Option B confuses the base with the answer.
Q34log 2 + log 50 equals:
Correct answer: B — 2
The product law gives log(2 × 50) = log 100 = 2, since 10² = 100. Option A applies a law for sums that does not exist.
Q35The characteristic of log 0.0072 is:
Correct answer: C — −3
In standard form 0.0072 = 7.2 × 10⁻³, so the characteristic is the power of 10, namely −3, written 3̄. Shortcut for a number below 1: count the zeros between the decimal point and the first significant figure (here 2) and add one.
Q36If log x = 2.4771, and log 3 = 0.4771, then x is:
Correct answer: A — 300
Same mantissa 0.4771 means the same digits as 3. Characteristic 2 means the power of 10 is 2, so x = 3 × 10² = 300. This is exactly how log tables were used for multiplication.
Q37Which expression equals 3 log x − log y?
Correct answer: B — log(x³/y)
The multiplier 3 goes back up as a power giving log(x³), and a difference of logs is the log of a quotient. So the answer is log(x³/y).
Q38Solve 2ˣ = 1/8.
Correct answer: B — x = −3
1/8 = 8⁻¹ = (2³)⁻¹ = 2⁻³, so x = −3. A number smaller than 1 always gives a negative logarithm when the base is greater than 1.
Q39log₃ 5 written in common logs is:
Correct answer: B — log 5 / log 3
Change of base: logₐ b = log b / log a. A quotient of two logarithms is not the same as the logarithm of a quotient — that would be option A or D.
Q40For log₁₀ x to exist, x must be:
Correct answer: B — greater than 0
10 raised to any real power is strictly positive, so no exponent produces 0 or a negative number. The domain of the log function is x > 0, with zero excluded as well as the negatives.
Q41sec θ is equal to:
Correct answer: B — 1/cos θ
The third letter of "sec" is c, which points to cosine. Matching the first letters instead gives the standard wrong answer.
Q42The identity 1 + cot²θ = cosec²θ comes from dividing sin²θ + cos²θ = 1 by:
Correct answer: B — sin²θ
Dividing by sin²θ turns the first term into 1, the second into cot²θ, and the right side into cosec²θ.
Q43cos(A + B) equals:
Correct answer: B — cos A cos B − sin A sin B
The cosine formula reverses the sign: a plus inside gives a minus outside. The last option is sin(A + B).
Q44Which is NOT a valid form of cos 2A?
Correct answer: D — 2 sin A cos A
2 sin A cos A is sin 2A, not cos 2A. The other three are equivalent forms of cos 2A.
Q45For 8 sin θ + 6 cos θ = R sin(θ + α), the value of R is:
Correct answer: B — 10
R = √(8² + 6²) = √100 = 10. Adding the coefficients instead of using Pythagoras gives 14, the tempting wrong answer.
Q46The maximum value of 7 sin θ − 24 cos θ is:
Correct answer: B — 25
The maximum is R = √(49 + 576) = √625 = 25, since the combined wave has amplitude R.
Q47Solving sin 2θ = 0.5 for 0° ⩽ θ ⩽ 360°, the interval you should search is:
Correct answer: B — 0° to 720°
If θ reaches 360° then 2θ reaches 720°, so all solutions must be found across two revolutions before halving.
Q48To solve 2 sin θ cos θ = sin θ, you should:
Correct answer: B — factorise as sin θ(2cos θ − 1) = 0
Dividing by sin θ throws away every solution where sin θ = 0. Factorising keeps both families of solutions.
Q49What does f′(3) = 5 tell you about the graph of f?
Correct answer: B — At x = 3 the curve is rising with slope 5
A derivative is a slope, not a height. f′(3) = 5 says that at the instant x = 3, the curve climbs 5 units of y for every 1 unit of x. The actual height at x = 3 is f(3), which is a completely different number.
Q50Differentiate f(x) = 4x³ − 7x + 2.
Correct answer: A — 12x² − 7
Power rule on each term: 4x³ → 3·4x² = 12x². Then −7x → −7 (the x disappears). The constant +2 → 0, because a constant never changes, so its rate of change is zero.
Q51In the limit definition, why must h approach 0 rather than simply equal 0?
Correct answer: B — Because the formula would give 0/0, which is undefined
Substituting h = 0 directly gives [f(x) − f(x)]/0 = 0/0 — meaningless. The limit lets us ask what the expression heads toward as h shrinks, without ever dividing by zero.
Q52What is the derivative of sin(5x)?
Correct answer: B — 5 cos(5x)
Chain rule. Outer function sin → cos(5x). Inner function 5x has derivative 5. Multiply them: 5 cos(5x). Forgetting the inner 5 is the single most common slip in calculus.
Q53A Riemann sum with n = 10 left rectangles underestimates the area under an increasing curve. What happens as n increases?
Correct answer: B — The estimate converges toward the true area
Each rectangle misses a small triangular sliver above it. Narrower rectangles mean smaller slivers, so the total gap shrinks toward zero. That convergence is exactly what the integral is defined as.
Q54Evaluate ∫₁³ 2x dx.
Correct answer: B — 8
The antiderivative of 2x is x². Apply the Fundamental Theorem: F(3) − F(1) = 9 − 1 = 8. Sanity check: the region is a trapezium with parallel sides 2 and 6 and width 2, giving ½(2+6)(2) = 8.
Q55∫ x³ dx equals:
Correct answer: C — x⁴/4 + C
Raise the power by one and divide by the new power: x⁴/4 + C. Answer A is the derivative rather than the integral — the classic direction error. Check any integration by differentiating your answer: d/dx (x⁴/4) = x³. ✓
Q56Why does the power rule for integration exclude n = −1?
Correct answer: B — It would require dividing by zero
The rule divides by n + 1, which is zero when n = −1. The integral certainly exists — it is ln|x| + C. The modulus signs matter, since 1/x is defined for negative x but the logarithm is not.
Q57∫₀² 3x² dx equals:
Correct answer: A — 8
The antiderivative is x³, so the value is 2³ − 0³ = 8. A common slip is forgetting to subtract the lower limit, or integrating to x³/3 by dividing by the new power while ignoring the 3 already in front.
Q58The "+C" appears in indefinite integrals because:
Correct answer: B — Differentiation destroys constants, so many functions share one derivative
x², x² + 5 and x² − 100 all differentiate to 2x, so reversing the process cannot recover which one you started from. C stands for that whole family. It disappears in definite integrals because it cancels in F(b) − F(a).
Q59Which technique is the reverse of the chain rule?
Correct answer: B — Integration by substitution
Substitution undoes the chain rule; integration by parts undoes the product rule. Recognising which rule created the expression is usually the fastest route to choosing the right technique.
Q60A velocity–time graph is integrated with respect to time. The result represents:
Correct answer: B — Displacement
Integrating velocity accumulates distance travelled, giving displacement — the same thing as the area under the graph. Differentiating instead would give acceleration, which is the mirror-image operation.
Q61f is continuous, f(2) = −3 and f(3) = 4. This shows that:
Correct answer: B — f has a root between 2 and 3
A sign change guarantees at least one root, not exactly one — there could be three. The word "exactly" makes the third option a stronger claim than the evidence supports.
Q62The sign-change test can fail to indicate a root when:
Correct answer: B — the function is not continuous
A discontinuity such as an asymptote lets the function change sign without ever taking the value zero. 1/(x − 2) is the standard counter-example.
Q63For the iteration x₍ₙ₊₁₎ = F(xₙ), the value it converges to satisfies:
Correct answer: B — x = F(x)
If the sequence settles, the input and output stop differing, so the limit satisfies x = F(x). That is why the rearrangement had to be into this form.
Q64An iteration produces 2, 5, 122, 1815848. This means:
Correct answer: B — it has diverged
The values are growing without limit, so this rearrangement diverges. Continuing will not help — a different rearrangement is needed.
Q65To show x³ − 5x + 1 = 0 can be written as x = ∛(5x − 1), you should:
Correct answer: B — start from the given equation and rearrange forwards
Examiners want the given equation transformed into the target. Working backwards from the answer proves the same thing but can lose the mark for direction.
Q66Iterating gives x₄ = 2.4471 and x₅ = 2.4468. To 3 decimal places the root is:
Correct answer: A — 2.447
Both values round to 2.447, so they agree at 3 d.p. and the iteration can stop. Quoting 2.4470 gives four decimal places, which is more than was asked for.
Q67Why keep more decimal places than the final answer needs?
Correct answer: B — because rounding each step lets the error build up
Each rounded value is fed back into the formula, so an early rounding error is carried into every subsequent term and can shift the final digit.
Q68Near the root, an iteration converges when the graph of y = F(x) is:
Correct answer: B — shallower than y = x
A shallower curve means each bounce between the curve and the line lands nearer the root. A steeper one magnifies the gap each time, which is the diverging case.
Q69The magnitude of 2i − 3j + 6k is:
Correct answer: B — 7
Square each component, add, then take the root: √(2² + (−3)² + 6²) = √(4 + 9 + 36) = √49 = 7. The minus sign on the j component disappears when it is squared, so a negative component never reduces the magnitude.
Q70The dot product of two perpendicular vectors is:
Correct answer: B — 0
a · b = |a||b| cos 90° = 0. This is the standard test for perpendicularity.
Q71a × b is:
Correct answer: B — a vector perpendicular to both
The cross product returns a vector at right angles to the plane containing a and b, with direction given by the right-hand rule.
Q72If a × b = 0 for non-zero vectors, then a and b are:
Correct answer: B — parallel
|a × b| = |a||b| sin θ, which vanishes only when sin θ = 0, so θ is 0° or 180° — the vectors are parallel. Perpendicular is the condition for the dot product.
Q73i · j equals:
Correct answer: A — 0
i and j are perpendicular unit vectors, so their dot product is 1 × 1 × cos 90° = 0. Note that i × j = k, which is a different product entirely.
Q74The area of the triangle formed by vectors a and b is:
Correct answer: B — ½|a × b|
The cross product magnitude gives the parallelogram area, and a triangle is half of that parallelogram.
Q75If a · (b × c) = 0, the three vectors are:
Correct answer: B — coplanar
The triple product is the volume of the solid they span. Zero volume means they all lie in one plane.
Q76Which is true?
Correct answer: B — a · b = b · a but a × b = −(b × a)
The dot product is commutative because multiplication of components is. The cross product reverses direction when the order is swapped, which is why the order must be preserved exactly as the question gives it.
Q77The solution of a differential equation is:
Correct answer: B — a function
The equation describes how a quantity changes; solving it recovers the quantity itself as a function of the independent variable.
Q78"A population grows at a rate proportional to its size" is:
Correct answer: B — dP/dt = kP
Proportional to its size means the right-hand side is k times P. The first option would be constant growth, and the third would be growth that slows as the population rises.
Q79A body cools in surroundings at 15 °C. The correct model is:
Correct answer: B — dθ/dt = −k(θ − 15)
Cooling is proportional to the excess over the surroundings. The second option correctly gives dθ/dt = 0 when θ = 15, so the body stops cooling at room temperature.
Q80Separating the variables in dy/dx = xy² gives:
Correct answer: B — y⁻² dy = x dx
Dividing both sides by y² moves it to the left as 1/y², which is y⁻². The first option multiplies instead of divides.
Q81Integrating (1/y) dy gives:
Correct answer: B — ln|y|
The integral of 1/y is ln|y| + c. The modulus can usually be dropped once a condition shows the sign of y is fixed.
Q82How many arbitrary constants does a first order differential equation need?
Correct answer: B — one
Integrating once introduces one constant. Writing one on each side is not wrong but they immediately combine, so a single c is standard.
Q83For V = 400e^(−0.05t), as t becomes large V approaches:
Correct answer: B — 0
e^(−0.05t) tends to 0 as t grows, so the whole product tends to 0. The value 400 is the starting volume at t = 0, not the limit.
Q84A student solves a cooling problem and gets θ → −40 °C in a room at 18 °C. The most likely error is:
Correct answer: B — the model used θ rather than the excess θ − 18
Using dθ/dt = −kθ makes the temperature decay towards zero and beyond rather than towards room temperature. Checking the long-run value against common sense catches this immediately.
Q85Simplify i⁵⁰.
Correct answer: C — −1
50 ÷ 4 leaves remainder 2, so i⁵⁰ = i² = −1. Only the remainder matters, because i⁴ = 1.
Q86(2 + 3i)(2 − 3i) equals:
Correct answer: B — 13
This is a² − (bi)² = 4 − 9i² = 4 + 9 = 13. A complex number times its conjugate is always the real number a² + b².
Q87The modulus of 3 − 4i is:
Correct answer: C — 5
√(9 + 16) = √25 = 5. Squaring removes the sign of each part, so 3 − 4i and 3 + 4i have the same modulus.
Q88√(−16) equals:
Correct answer: B — 4i
√(−16) = √16 × √(−1) = 4i. It is defined perfectly well in ℂ, which is the entire reason complex numbers were introduced.
Q89(5 + 2i) − (3 − 4i) equals:
Correct answer: B — 2 + 6i
Real parts: 5 − 3 = 2. Imaginary parts: 2 − (−4) = 6. So the answer is 2 + 6i; the double negative is where marks are lost.
Q90The conjugate of −7i is:
Correct answer: A — 7i
Write it as 0 − 7i. Flipping the sign of the imaginary part gives 0 + 7i = 7i.
Q91If x² + 9 = 0, then x equals:
Correct answer: B — ±3i
x² = −9, so x = ±√(−9) = ±3i. In ℝ there is no solution, but in ℂ there are exactly two.
Q92To divide by (3 + 2i), multiply top and bottom by:
Correct answer: B — (3 − 2i)
The conjugate 3 − 2i gives a denominator of 9 + 4 = 13, which is real. Multiplying by 3 + 2i itself would leave an imaginary part behind.
Q93The modulus of −3 + 4i is:
Correct answer: B — 5
√(9 + 16) = √25 = 5. The signs disappear because both parts are squared.
Q94The argument of the complex number i is:
Correct answer: C — 90°
i = 0 + 1i sits on the positive imaginary axis, a quarter turn anticlockwise from the positive real axis.
Q95If z₁ = 2 cis 40° and z₂ = 3 cis 50°, then z₁z₂ is:
Correct answer: B — 6 cis 90°
Multiply the moduli (2 × 3 = 6) and add the arguments (40 + 50 = 90). Adding the moduli, as option A does, is the standard error.
Q96(cis 30°)⁶ equals:
Correct answer: A — cis 180°
De Moivre multiplies the argument by the power: 6 × 30° = 180°. The modulus is 1⁶ = 1, so the answer is cis 180° = −1.
Q97The argument of −1 − i is:
Correct answer: D — −135°
Both parts negative puts it in the third quadrant, so θ = α − 180° = 45 − 180 = −135°. The calculator would return +45°, which points the opposite way.
Q98How many distinct fourth roots does 16 have in ℂ?
Correct answer: C — 4
An n-th root has exactly n values in ℂ. Here they are 2, 2i, −2 and −2i — the vertices of a square of radius 2.
Q99Dividing z₁ = 12 cis 100° by z₂ = 4 cis 30° gives:
Correct answer: A — 3 cis 70°
Divide the moduli (12 ÷ 4 = 3) and subtract the arguments (100 − 30 = 70).
Q100The n distinct n-th roots of a complex number lie:
Correct answer: B — evenly spaced on a circle
They all share the modulus r^(1/n), so they are equidistant from the origin, and their arguments differ by 360°/n, so they are evenly spaced around that circle.
A modulus measures a distance, so |2x + 1| is never negative for any real x. It can therefore never equal −3, and the equation has no solutions.
x = 11 or x = −3
1 < x < 4
(a) corners (−2, 0) and (2, 0); (b) x = 1 or x = 4; (c) 1 < x < 4
The bracket is zero when 4x + 10 = 0, so x = −2.5 and the corner is at (−2.5, 0). The arms have gradients 4 and −4.
When a polynomial f(x) is divided by (x − a), the remainder is f(a) — the value obtained by substituting x = a into the polynomial.
x³ + 2x² − 5x + 1 is divided by (x − 2).f(2) = 8 + 8 − 10 + 1 = 7.
(x + 1) is a factor of x³ + 3x² + 3x + 1.f(−1) = −1 + 3 − 3 + 1 = 0. Since the remainder is zero, by the factor theorem (x + 1) is a factor. (In fact the polynomial is (x + 1)³.)
f(x) = 2x³ + ax² + bx − 6 is divided by (x − 1) the remainder is −6, and (x + 2) is a factor. Find a and b.f(1) = 2 + a + b − 6 = −6, so a + b = −2remainder theorem[1]f(−2) = −16 + 4a − 2b − 6 = 0, so 4a − 2b = 22 and 2a − b = 11factor theorem: the remainder is zero[1]3a = 9, so a = 3the b terms cancel because their coefficients are +1 and −1[1]b = −2 − 3 = −5check: f(1) = 2 + 3 − 5 − 6 = −6 ✓ and f(−2) = −16 + 12 + 10 − 6 = 0 ✓[1]a = 3, b = −5
x³ − 4x² + x + 6 by (x − 3) and hence factorise the polynomial completely.f(3) = 27 − 36 + 3 + 6 = 0, so (x − 3) is a factorconfirming the division will be exact[1]x² − x − 2by long or synthetic division[1]x² − x − 2 = (x − 2)(x + 1)[1]f(x) = (x − 3)(x − 2)(x + 1)[1](x − 3)(x − 2)(x + 1)
p(x) = x³ + px² + qx + 12 has (x − 2) and (x + 3) as factors.p(2) = 8 + 4p + 2q + 12 = 0, so 4p + 2q = −20 and 2p + q = −10factor theorem[1]p(−3) = −27 + 9p − 3q + 12 = 0, so 9p − 3q = 15 and 3p − q = 5[1]5p = −5, so p = −1the q terms cancel[1]q = −10 − 2(−1) = −8so p(x) = x³ − x² − 8x + 12[1]−12 (from −constant/leading coefficient), and two roots are 2 and −3or divide out both known factors[1]= −12 / (2 × −3) = 2, so the third factor is (x − 2) and p(x) = (x − 2)²(x + 3)check: p(x) = (x−2)²(x+3) expands to x³ − x² − 8x + 12 ✓[1](a) 2p + q = −10 and 3p − q = 5 (b) p = −1, q = −8 (c) (x − 2), giving p(x) = (x − 2)²(x + 3)
A proper rational fraction has a numerator of lower degree than its denominator, such as (x + 1)/(x² − 4). (x² + 3)/(x − 1) is improper, because the numerator has the higher degree.
1 / [(x + 3)(x − 2)²].A/(x + 3) + B/(x − 2) + C/(x − 2)².
The numerator over any factor must be permitted to be one degree lower than that factor. A quadratic denominator therefore takes a linear numerator, Ax + B. A single constant would give too few unknowns and the resulting equations would be inconsistent.
(7x − 1) / [(x − 3)(x + 1)] into partial fractions.A/(x − 3) + B/(x + 1) and multiplies up to 7x − 1 = A(x + 1) + B(x − 3)correct shape[1]x = 3: 20 = 4A, so A = 5x = 3 kills the B term[1]x = −1: −8 = −4B, so B = 2[1]5/(x − 3) + 2/(x + 1)accept a verification by recombining[1]5/(x − 3) + 2/(x + 1)
(x + 4) / (x − 2)² into partial fractions.A/(x − 2) + B/(x − 2)², giving x + 4 = A(x − 2) + Ba repeated factor needs both powers[1]x = 2: 6 = B[1]1 = Ano substitution can isolate A here[1]1/(x − 2) + 6/(x − 2)²[1]1/(x − 2) + 6/(x − 2)²
f(x) = (2x² + 5x + 3) / (x² + 2x + 1).2x² + 5x + 3 = 2(x² + 2x + 1) + (x + 1), so f(x) = 2 + (x + 1)/(x + 1)²x² + 2x + 1 = (x + 1)²[1](x + 1)/(x + 1)² = 1/(x + 1)cancelling one factor, valid for x ≠ −1[1]1/(x + 1)accept working through A/(x+1) + B/(x+1)² and finding B = 0[1]f(x) = 2 + 1/(x + 1)[1]x ≠ −1the excluded value must be stated[1]f(x) = 2 + 1/(x + 1), for x ≠ −1
Basis: show the statement is true for the first value, usually n = 1. Inductive step: assume it is true for n = k and prove it must then be true for n = k + 1. Together these prove it for every natural number.
(x + y)¹², and what is the coefficient of the third term?There are 12 + 1 = 13 terms. The third term uses r = 2, so its coefficient is ¹²C₂ = 66.
(3x − 2)ⁿ.T_(r+1) = ⁿCᵣ (3x)ⁿ⁻ʳ (−2)ʳ.
2ⁿ > n for all natural numbers n.2¹ = 2 > 1, so P(1) is trueboth sides evaluated[1]2ᵏ > k for some k ≥ 1the hypothesis stated as an assumption[1]2ᵏ⁺¹ = 2 × 2ᵏ > 2k, using the hypothesisthe hypothesis is used here and must be visible[1]k ≥ 1, 2k = k + k ≥ k + 1, so 2ᵏ⁺¹ > k + 1; hence by induction the result holds for all nthe k ≥ 1 is what makes the last inequality work[1]Proved by induction for all n ≥ 1.
x⁴ in the expansion of (2 + x)⁷.T_(r+1) = ⁷Cᵣ (2)⁷⁻ʳ xʳ[1]r = 4[1]⁷C₄ = 35 and 2³ = 87 − 4 = 3, so the 2 is cubed[1]= 35 × 8 = 280[1]280
(x² − 2/x)⁶.T_(r+1) = ⁶Cᵣ (x²)⁶⁻ʳ (−2/x)ʳthe minus stays with the 2[1]x¹²⁻²ʳ × x⁻ʳ = x¹²⁻³ʳ[1]12 − 3r = 0 gives r = 4[1]= ⁶C₄ (−2)⁴ = 15 × 16 = 240the power 4 is even, so the result is positive[1]6 + 1 = 7 terms[1](b) 240 (c) 7 terms; a constant term requires an integer solution for r
If ax = y, where a > 0 and a ≠ 1, then x is called the logarithm of y to the base a, written loga y = x. It is the power to which the base must be raised to give the number.
Because 1x = 1 for every value of x. No power of 1 can produce any number other than 1, so log₁ y would have no value for y ≠ 1 and infinitely many for y = 1.
In standard form 0.00456 = 4.56 × 10⁻³, so the characteristic is −3, written 3̄. The characteristic equals the power of 10 in standard form.
log₂ 64 + log₃ 81 − log₅ 125 without tables.log₂ 64 = 6 because 2⁶ = 64[1]log₃ 81 = 4 because 3⁴ = 81[1]log₅ 125 = 3, so the value is 6 + 4 − 3 = 7final answer required[1]7
log₂(x + 3) + log₂(x − 3) = 4.log₂[(x+3)(x−3)] = 4sum of logs with the same base becomes the log of a product[1](x+3)(x−3) = 2⁴ = 16[1]x² − 9 = 16, so x² = 25 and x = ±5difference of two squares[1]x = −5 because it makes x − 3 negative and the log undefined; x = 5the rejection must be stated with a reason[1]x = 5
N = 500 × 2t.N = 500 × 2⁶ = 500 × 64[1]N = 32 000 bacteria[1]32 000 = 500 × 2ᵗ so 2ᵗ = 64divide by the initial amount first[1]t log 2 = log 64power law brings t down[1]t = log 64 / log 2 = 1.8062 / 0.3010 = 6.00 hoursaccept t = 6 from inspection with working shown[1](a) 32 000 (b) t = 6.00 hours (c) in (b) the unknown is the exponent
Dividing every term of sin²θ + cos²θ = 1 by cos²θ gives sin²θ/cos²θ + 1 = 1/cos²θ. Since sin θ/cos θ = tan θ and 1/cos θ = sec θ, this is tan²θ + 1 = sec²θ.
sec θ = 1/cos θ, and cos 90° = 0. Division by zero is undefined, so sec 90° does not exist. The graph has a vertical asymptote at θ = 90°, with the curve tending to +∞ on one side and −∞ on the other.
θ = 0°, 60°, 300°, 360°
(a) 13 sin(θ − 67.4°); (b) θ = 97.4°, 217.4°; (c) maximum 13 at θ = 157.4°
Using cos 2θ = 1 − 2sin²θ, the numerator becomes 1 − (1 − 2sin²θ) = 2sin²θ. Using sin 2θ = 2 sin θ cos θ, the denominator is 2 sin θ cos θ. So the fraction is 2sin²θ / (2 sin θ cos θ) = sin θ / cos θ = tan θ.
y = 5x³ − 2/x with respect to x.Write it as 5x³ − 2x⁻¹. Then dy/dx = 15x² + 2x⁻², that is 15x² + 2/x².
y = x² − 4x + 1 at the point where x = 3.dy/dx = 2x − 4, so at x = 3 the gradient is 2(3) − 4 = 2.
At a stationary point, if d²y/dx² < 0 it is a maximum; if d²y/dx² > 0 it is a minimum.
y = x³ − 3x² − 9x + 5.dy/dx. [1]dy/dx = 3x² − 6x − 9[1]3(x² − 2x − 3) = 0[1](x − 3)(x + 1) = 0, so x = 3 or x = −1[1]x = 3: y = 27 − 27 − 27 + 5 = −22[1]x = −1: y = −1 − 3 + 9 + 5 = 10[1]d²y/dx² = 6x − 6; at x = 3 it is +12, a minimum[1]x = −1 it is −12, a maximum[1]Minimum at (3, −22), maximum at (−1, 10)
V = x(24 − 2x)². [2]24 − 2xa square is removed from both ends[1]V = x(24 − 2x)²[1]V = 576x − 96x² + 4x³[1]dV/dx = 576 − 192x + 12x²[1]12(x² − 16x + 48) = 0 → (x − 4)(x − 12) = 0[1]x = 4; x = 12 is rejected because it leaves no base[1]V = 4 × 16² = 4 × 256[1]V = 1024 cm³[1]x = 4 cm giving V = 1024 cm³
s = 2t³ − 9t² + 12t metres after t seconds.v = ds/dt = 6t² − 18t + 12[1]a = dv/dt = 12t − 18[1]v = 0: 6(t² − 3t + 2) = 0 → (t − 1)(t − 2) = 0[1]t = 1 s and t = 2 s[1]t = 2: a = 24 − 18 = 6 m s⁻²[1]v = 6t² − 18t + 12, a = 12t − 18; at rest at t = 1 s and 2 s; a = 6 m s⁻² at t = 2 s
∫ (6x² − 4x + 3) dx.2x³ − 2x² + 3x + c
∫₁³ (2x + 1) dx.[x² + x]₁³ = (9 + 3) − (1 + 1) = 10
The area between the curve and the x axis, between the two limits. Area below the axis counts as negative.
y = x² − 4x crosses the x axis at two points.y = 0: x(x − 4) = 0[1]x = 0 and x = 4[1]∫(x² − 4x) dx = x³/3 − 2x²[1](64/3 − 32) − 0 = −32/3[1]= 32/3 ≈ 10.7 square unitsthe magnitude is taken[1](a) x = 0 and x = 4 (b) 32/3 square units
(2, 9) and has gradient dy/dx = 3x² − 2x.x = 0 and x = 2. [3]y = x³ − x² + c[1]9 = 8 − 4 + c[1]c = 5[1]y = x³ − x² + 5[1]∫₀² (x³ − x² + 5) dx = [x⁴/4 − x³/3 + 5x]₀²[1]= 4 − 8/3 + 10[1]= 34/3 ≈ 11.3 square units[1](a) y = x³ − x² + 5 (b) 34/3 square units
y = x + 2 and y = x² is to be found.x + 2 = x² → x² − x − 2 = 0[1](x − 2)(x + 1) = 0, so x = −1 and x = 2[1]= ∫₋₁² [(x + 2) − x²] dx[1]= [x²/2 + 2x − x³/3]₋₁²[1]= (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2[1]x = −1 and 2; area = 9/2 square units
If f(a) and f(b) have opposite signs then the curve must cross zero somewhere between a and b, so a root lies in that interval. The argument requires f to be continuous on the interval — otherwise the sign change could be caused by an asymptote rather than a root.
f(1) = −4, f(2) = 5; f continuous with a sign change, so a root lies in (1, 2)
1.57 (2 d.p.)
Starting from 3x³ − 4x + 2 = 0, move the terms across to get 3x³ = 4x − 2. Dividing by 3 gives x³ = (4x − 2)/3, and taking the cube root of both sides gives x = ∛((4x − 2)/3).
(a) sign change between 0.7 and 0.8; (b) α = 0.739; (c) the iteration diverges — use x = cos x instead
v = 3i − 4j + 12k.|v| = √(9 + 16 + 144) = √169 = 13.
The dot product gives a scalar and is commutative; the cross product gives a vector perpendicular to both, and reversing the order reverses its direction. The dot product is zero for perpendicular vectors, the cross product for parallel ones.
a · b = 0 and neither vector is zero, what can you conclude?Since a · b = |a||b| cos θ and neither magnitude is zero, cos θ = 0, so θ = 90° — the vectors are perpendicular.
a = i + 2j − k and b = 3i − j + 2k, find a · b and the angle between them.a · b = (1)(3) + (2)(−1) + (−1)(2) = 3 − 2 − 2 = −1a negative dot product means an obtuse angle[1]|a| = √(1 + 4 + 1) = √6[1]|b| = √(9 + 1 + 4) = √14[1]cos θ = −1/(√6 √14) = −1/9.165 = −0.109, so θ = 96.3°obtuse, as the negative dot product predicted[1]a · b = −1; θ ≈ 96.3°
a × b for a = 2i + j + k and b = i − j + 3k, and hence the area of the triangle they form.(1)(3) − (1)(−1) = 3 + 1 = 4[1]−[(2)(3) − (1)(1)] = −5the minus sign on j is where marks are lost[1](2)(−1) − (1)(1) = −3, so a × b = 4i − 5j − 3k[1]= ½|a × b| = ½√(16 + 25 + 9) = ½√50 = 3.54accept (5√2)/2[1]a × b = 4i − 5j − 3k; area ≈ 3.54 square units
a = i + j, b = j + k and c = i + k.b × c.a · (b × c).= (1)(1) − (1)(0) = 1[1]= −[(0)(1) − (1)(1)] = 1; k component = (0)(0) − (1)(1) = −1so b × c = i + j − k[1]a · (b × c) = (1)(1) + (1)(1) + (0)(−1)a = ⟨1, 1, 0⟩[1]= 2[1]= |2| = 2 cubic units[1](a) i + j − k (b) 2 (c) not coplanar; volume 2 cubic units
The general solution contains an arbitrary constant and therefore describes a whole family of curves. A particular solution is the single member of that family obtained by using a given condition to find the value of the constant.
dN/dt = −kN, where k is a positive constant. The minus sign is needed because the colony is decreasing, so the rate of change is negative while N and k are both positive.
y = 2e^(x³)
(a) V = 500e^(−0.02t); (b) 274 litres; (c) 34.7 min; (d) V → 0 but never reaches it
y = tan(ln x + π/4)
A complex number is any number of the form z = a + bi, where a and b are real and i = √(−1). The real part is a and the imaginary part is b.
i¹⁰³.The powers of i repeat every four. 103 = 4 × 25 + 3, so i¹⁰³ = i³ = −i.
z = −5 + 12i.|z| = √((−5)² + 12²) = √(25 + 144) = √169 = 13.
(3 + 2i) / (1 − 4i) in the form a + bi.1 + 4i[1](1 − 4i)(1 + 4i) = 1 + 16 = 17a² + b², because −16i² = +16[1](3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = −5 + 14i[1]= −5/17 + (14/17)imust be split into a + bi form[1]−5/17 + (14/17)i
z₁ = 2 + 3i and z₂ = 4 − i, find z₁z₂ and verify that |z₁z₂| = |z₁||z₂|.z₁z₂ = (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 11 + 10i−3i² = +3, which combines with the 8[1]|z₁z₂| = √(121 + 100) = √221[1]|z₁| = √13 and |z₂| = √17[1]|z₁||z₂| = √13 × √17 = √221, equal to |z₁z₂|, so the result is verifiedthe concluding comparison is required[1]z₁z₂ = 11 + 10i, and both sides equal √221.
x² − 6x + 25 = 0.= (−6)² − 4(1)(25) = 36 − 100 = −64[1]x = [6 ± √(−64)] / 2 with √(−64) = 8i[1]x = 3 ± 4i[1]= (3 + 4i) + (3 − 4i) = 6 ✓the imaginary parts cancel[1]= (3 + 4i)(3 − 4i) = 9 − 16i² = 9 + 16 = 25 ✓difference of two squares, with i² = −1[1](a) discriminant = −64 < 0 (b) x = 3 ± 4i (c) sum 6, product 25
The modulus |z| = √(a² + b²) is the distance of the point from the origin on the Argand diagram. The argument is the angle the line from the origin to the point makes with the positive real axis, measured anticlockwise.
z = 1 − i.r = √(1 + 1) = √2. The point is in the fourth quadrant with α = tan⁻¹(1) = 45°, so θ = −45°.
i?i has modulus 1 and argument 90°, so multiplying leaves the modulus unchanged and adds 90° to the argument: the point is rotated a quarter turn anticlockwise about the origin.
z = −2 − 2i in polar form.r = √(4 + 4) = √8 = 2√2[1]α = tan⁻¹(2/2) = 45°use the absolute values to get the acute angle[1]θ = 45° − 180° = −135°accept 225° if the range is not restricted[1]z = 2√2 (cos(−135°) + i sin(−135°))[1]z = 2√2 cis(−135°)
(1 + i)⁸.r = √2 and θ = 45°, so 1 + i = √2 cis 45°first quadrant, so no adjustment needed[1](√2)⁸ cis(8 × 45°)[1](√2)⁸ = 2⁴ = 16 and 8 × 45° = 360°[1]16(cos 360° + i sin 360°) = 16a real answer, since the argument came back to a full turn[1]16
z³ = 8.8 = 8(cos 0° + i sin 0°)8 lies on the positive real axis, so its argument is 0°[1]8^(1/3) = 2[1](0 + 360k)/3 for k = 0, 1, 2, giving 0°, 120°, 240°adding 360° each time before dividing is what produces the extra roots[1]2 cis 0°, 2 cis 120°, 2 cis 240°the first is the obvious real root 2[1](b) 2 cis 0°, 2 cis 120°, 2 cis 240° (c) equally spaced 120° apart on a circle of radius 2
Q1Two forces of 3 N and 4 N act at a point. Which resultant is IMPOSSIBLE?
Correct answer: D — 8 N
The resultant of two vectors ranges from |4 − 3| = 1 N when they are antiparallel to 4 + 3 = 7 N when they are parallel. 8 N lies outside that range and cannot be produced at any angle. 5 N is the perpendicular case.
Q2A force of 20 N acts at 60° above the horizontal. Its horizontal component is:
Correct answer: A — 10 N
20 cos 60° = 20 × 0.5 = 10 N. The 17.3 N answer is 20 sin 60°, the vertical component — swapping sine and cosine is the most common slip here. Check by asking which component should be smaller: at a steep 60°, most of the force points upward.
Q3Which of these is a scalar?
Correct answer: B — Work
Work is a scalar, even though it is calculated from two vectors — the dot product of force and displacement returns a plain number. Momentum, torque and electric field all carry direction.
Q4A body is in equilibrium under three forces. This means:
Correct answer: B — The three forces form a closed triangle when drawn head to tail
Zero resultant means that placing the three vectors head to tail brings you back to the starting point — a closed triangle. They need not be equal in magnitude or perpendicular, and "no forces act" describes a different situation entirely.
Q5A 10 N weight hangs on a string. The tension in the string is 10 N. This is because:
Correct answer: C — The weight is in equilibrium, so the upward and downward forces must cancel
It follows from equilibrium: ΣF = 0 vertically, so tension must equal weight here. It is not a general rule — accelerate the weight upward and the tension exceeds 10 N. And the two forces act on the same object, so they are not an action–reaction pair.
Q6On an inclined plane at angle θ, the component of weight acting down the slope is:
Correct answer: B — mg sin θ
mg sin θ acts along the slope and mg cos θ presses perpendicular into it. Sanity check with θ = 0: a flat surface should give zero force along it, and sin 0° = 0. The cos version would wrongly give the full weight.
Q7A 10 kg block sits still on a rough floor with μ = 0.4 and no applied force. The friction acting on it is:
Correct answer: C — 0 N
Friction opposes motion or attempted motion. With nothing pushing the block there is nothing to oppose, so friction is zero. μR = 39.2 N is only the maximum available, not the actual value.
Q8For a block on a slope of angle θ, the normal reaction is:
Correct answer: C — mg cos θ
R balances the component of weight perpendicular to the surface, which is mg cos θ. At θ = 0 this correctly reduces to mg.
Q9A block just begins to slide when the slope reaches 30°. The coefficient of friction is:
Correct answer: B — 0.58
μ = tan 30° = 0.577. Using sin 30° = 0.5 gives the tempting wrong answer.
Q10Doubling the mass of a block on an adjustable slope changes the angle at which it slips by:
Correct answer: C — not at all
At the point of slipping mg sin θ = μmg cos θ, and mg cancels. The critical angle depends only on μ.
Q11Two particles are connected over a smooth pulley by a light inextensible string. Which is true?
Correct answer: C — both share one acceleration magnitude
An inextensible string forces both to move together, so the magnitudes of their accelerations are equal. A light string over a smooth pulley also gives equal tension throughout.
Q12Masses of 7 kg and 3 kg hang over a smooth pulley. The tension must be:
Correct answer: B — between 3g and 7g
If T were outside that range one of the particles could not accelerate in the observed direction. Checking this catches sign errors quickly.
Q13A block slides down a rough slope. The friction acting on it is:
Correct answer: B — exactly μR
Once sliding, friction takes its maximum value μR and acts up the slope. The inequality only applies while the block is still static.
Q14A particle on a rough table is pulled horizontally by a string. Its weight appears in the horizontal equation of motion:
Correct answer: C — not at all
The weight is vertical and is balanced by the normal reaction. It enters only indirectly, through R, which sets the size of the friction.
Q15An athlete runs exactly one lap of a 400 m circular track in 50 s. Their average velocity is:
Correct answer: B — 0 m s⁻¹
Average velocity is displacement ÷ time, and after a complete lap the displacement is zero — start and finish are the same point. The 8 m s⁻¹ answer is the average speed, which uses distance instead.
Q16A car moves at a constant 20 m s⁻¹ around a circular bend. Which statement is true?
Correct answer: C — It is accelerating because its direction is changing
Velocity is a vector, so changing direction changes the velocity even at constant speed — and a changing velocity is what acceleration means. The acceleration points toward the centre of the bend.
Q17An object has negative velocity and negative acceleration. It is:
Correct answer: B — Moving backwards and speeding up
Matching signs mean the acceleration acts in the same direction as the motion, so the object speeds up. The common error is reading "negative acceleration" as "deceleration", which is only true when the velocity is positive.
Q18On a velocity–time graph, the area between the line and the time axis represents:
Correct answer: B — Displacement
Velocity × time = displacement, and the area is that product accumulated. The gradient of the same graph gives acceleration — the two are the pair most often swapped.
Q19A horizontal line on a velocity–time graph means the object is:
Correct answer: B — Moving at constant velocity
Constant velocity, so zero acceleration. A stationary object would be a horizontal line sitting on the time axis itself, at v = 0 — a special case, not the general meaning.
Q20A stone is dropped from rest and falls for 3.0 s. Taking g = 10 m s⁻², how far does it fall?
Correct answer: C — 45 m
s = ut + ½at² with u = 0 gives s = ½ × 10 × 3.0² = 45 m. Answer B is the mistake of calculating v = at = 30 and calling it a distance; answer A comes from forgetting to square the time.
Q21Which equation of motion would you choose if the question gives u, a and s, and asks for v?
Correct answer: C — v² = u² + 2as
The unknown you neither have nor want is t, so use the equation that omits t. Options A and B both contain t and would need it to be found first — twice the work and twice the chance of an arithmetic slip.
Q22Two balls leave a table edge at the same moment — one dropped, one thrown horizontally at 5 m s⁻¹. Ignoring air resistance:
Correct answer: C — They land at the same time
Vertical and horizontal motion are independent. Both start with zero vertical velocity and fall the same height under the same g, so both take the same time. The thrown ball merely covers ground while it falls.
Q23At the highest point of a projectile's path, its velocity is:
Correct answer: B — Equal to the horizontal component of the launch velocity
Only the vertical component reaches zero at the top. Nothing acts horizontally, so that component is unchanged throughout the flight and is the whole of the velocity at the peak.
Q24A projectile is launched on level ground. Which launch angle gives the greatest range?
Correct answer: B — 45°
Range = u² sin 2θ / g, largest when sin 2θ = 1, so 2θ = 90° and θ = 45°. A 90° launch goes straight up and lands back at the launch point with zero range — the answer that catches anyone reasoning "higher must be further".
Q25The SUVAT equations may only be used when:
Correct answer: B — The acceleration is constant
Constant acceleration is the one condition. Free fall is a common case of it, not the requirement. Where acceleration varies, you need the gradient and area of a graph, or calculus.
Q26Air resistance acts on a projectile. Compared with the ideal path, the actual path has:
Correct answer: B — A shorter range and a steeper descent than ascent
Drag opposes the motion throughout, cutting both range and maximum height, and the descent becomes steeper than the ascent — so the path is no longer a symmetric parabola.
Q27The unit of momentum is:
Correct answer: B — kg m s⁻¹
p = mv gives kg × m s⁻¹. It is also equal to the newton second, since impulse and momentum share a unit.
Q28Momentum is conserved in:
Correct answer: C — All collisions with no external resultant force
Momentum is always conserved provided no external resultant force acts. Kinetic energy is the quantity that separates elastic from inelastic.
Q29Two objects stick together after colliding. The collision is:
Correct answer: B — Inelastic
Sticking together always means kinetic energy was lost to heat, sound and deformation. Momentum is still conserved.
Q30A 2 kg object moves at 3 m s⁻¹. Its momentum is:
Correct answer: B — 6 kg m s⁻¹
p = mv = 2 × 3 = 6 kg m s⁻¹. Dividing gives 0.67 and confuses momentum with something that has no physical meaning here.
Q31Airbags reduce injury mainly because they:
Correct answer: B — Increase the time of the collision, reducing the force
The change in momentum is fixed by the crash. Extending Δt reduces F, since F = Δp/Δt. Nothing "absorbs" momentum — it is transferred, not destroyed.
Q32A stationary object explodes into two fragments. Their total momentum afterwards is:
Correct answer: A — Zero
It was zero before, and momentum is conserved, so it must be zero after. The two fragments carry equal and opposite momenta.
Q33A ball of mass 0.2 kg hits a wall at 5 m s⁻¹ and rebounds at 5 m s⁻¹. The magnitude of its change in momentum is:
Correct answer: C — 2 kg m s⁻¹
The velocity changes from +5 to −5, a change of 10 m s⁻¹, so Δp = 0.2 × 10 = 2 kg m s⁻¹. Answer A treats the speeds as equal and therefore unchanged, which ignores direction.
Q34Force is best defined as the rate of change of:
Correct answer: B — Momentum
F = Δp/Δt is the general form of Newton's second law, and reduces to F = ma when the mass is constant.
Q35A bullet is fired horizontally at the same instant an identical bullet is dropped from the same height. Which lands first?
Correct answer: C — They land at the same time
Vertical and horizontal motion are independent. Both bullets start with zero vertical velocity and fall under the same gravity, so both take identical time to reach the ground. The fired bullet simply travels much further horizontally while doing it.
Q36A projectile is launched at 30°. Which other angle gives the same horizontal range at the same speed?
Correct answer: B — 60°
Range depends on sin(2θ), and sin(60°) = sin(120°), so 30° and 60° pair up. The 60° shot goes higher and stays in the air longer but moves more slowly across; the two effects cancel exactly. Maximum range is at 45°.
Q37You push a wall and it pushes back equally. Why does nothing accelerate?
Correct answer: B — The two forces act on different objects, and the wall is anchored to the Earth
Your push acts on the wall; the wall's push acts on you. They never appear on the same free-body diagram, so they cannot cancel. Nothing accelerates because the wall is bolted to the ground and friction on your feet balances the force on you.
Q38A 2 kg ball is dropped from 5 m. Ignoring air resistance, its speed just before landing is about:
Correct answer: B — 10 m/s
Energy conservation: mgh = ½mv², and the mass cancels from both sides. So v = √(2gh) = √(2 × 9.81 × 5) = √98.1 ≈ 9.9 m/s, which rounds to 10 m/s. Note that a 5 kg ball dropped from the same height arrives at exactly the same speed.
Q39A pendulum swings with friction. What happens to the total energy?
Correct answer: B — It converts to heat and sound, so the mechanical total falls
Energy is never destroyed. Friction converts the ordered kinetic energy into disordered thermal energy in the air and pivot. The mechanical total (KE + PE) falls, but the total including heat is unchanged.
Q40Two cars collide and lock together. Which quantity is definitely conserved?
Correct answer: B — Momentum only
Momentum is conserved in every collision, since no external horizontal force acts during the impact. Kinetic energy is not: this is a perfectly inelastic collision, and a large fraction of it goes into crumpling metal, heat and noise.
Q41A waiter carries a tray horizontally at constant speed across a room. The work done by the waiter on the tray is:
Correct answer: B — Zero
The supporting force is vertical and the movement is horizontal, so θ = 90° and cos 90° = 0. Effort is not the same as work — your arm gets tired maintaining the force, but no energy is transferred to the tray.
Q42A car doubles its speed. Its kinetic energy:
Correct answer: C — Quadruples
KE = ½mv², so v → 2v gives (2v)² = 4v². This is why braking distance grows roughly fourfold when speed doubles: the brakes must dissipate four times the energy.
Q43A 2 kg ball is dropped from 5 m. Taking g = 10 m s⁻² and ignoring air resistance, its speed on landing is:
Correct answer: B — 10 m s⁻¹
mgh = ½mv², and the mass cancels: v = √(2gh) = √(2 × 10 × 5) = 10 m s⁻¹. The 100 answer is v² left un-square-rooted, and the mass being given at all is a deliberate distraction.
Q44A motor lifts a 50 kg load 4 m in 10 s. Taking g = 10 m s⁻², its useful power output is:
Correct answer: B — 200 W
Work = mgh = 50 × 10 × 4 = 2000 J, and power = 2000 ÷ 10 = 200 W. The 2000 W answer is the work in joules read off as watts — always check whether you have divided by the time.
Q45A pendulum swings back and forth with friction present. Over time:
Correct answer: B — KE and PE both fall, and the difference becomes heat
Energy is never destroyed — it leaves the pendulum as heat and sound, so the mechanical total falls. The period, notably, stays essentially the same as the swing dies away, which is the isochronism that made pendulum clocks work.
Q46A machine takes in 500 J and delivers 350 J of useful output. Its efficiency is:
Correct answer: A — 70%
350 ÷ 500 × 100 = 70%. The remaining 150 J has not vanished; it has been dissipated as heat, sound and vibration. Any efficiency above 100% means you have divided the wrong way round.
A scalar has magnitude only, for example mass. A vector has both magnitude and direction, for example velocity.
Distance is the total path length, a scalar that only ever grows. Displacement is the straight line from start to finish, so returning to the starting point makes it zero.
The single force that has the same effect on the body as the two forces acting together.
R = √(8.0² + 6.0²)[1]R = 10.0 N[1]tan θ = 6.0 / 8.0[1]θ = 36.9° north of easta direction with no reference line scores nothing[1]10.0 N at 36.9° north of east
W sin θ for the component along the slope[1]= 250 × sin 20°[1]= 85.5 N[1]W cos θ = 250 × cos 20°[1]= 235 N[1](b) 85.5 N (c) 235 N
t = 30 / 1.2perpendicular components are independent[1]t = 25 s[1]distance = 0.90 × 25[1]= 22.5 m downstream[1]√(1.2² + 0.90²)[1]= 1.5 m s⁻¹[1](a) 25 s (b) 22.5 m (c) 1.5 m s⁻¹
F = μR gives the maximum friction available, which occurs only when the block is on the point of slipping or already sliding. A stationary block that is not on the point of moving experiences only as much friction as is needed to balance the other forces, so F < μR.
Resolving perpendicular to the slope: R = mg cos θ. On the point of sliding, friction is at its maximum and acts up the slope, so resolving along the slope gives mg sin θ = μR = μ mg cos θ. Dividing both sides by mg cos θ gives tan θ = μ.
It moves, with a = 0.737 m s⁻²
(a) 5.10 m s⁻²; (b) 28.2 N; (c) light and inextensible — equal tension, equal acceleration
Neither — they slip at the same angle. At the point of slipping tan θ = μ, and the mass cancels out of that derivation entirely. Since both blocks are of the same material on the same surface, μ is identical, so the critical angle is identical.
v² = u² + 2asthe equation without t, since t is not given in part (a)[1]10.0² = 4.0² + 2 × a × 42 → 100 = 16 + 84a → a = 1.0 m s⁻²unit required for the mark[1]v = u + at (or s = ½(u+v)t)[1]10.0 = 4.0 + 1.0t → t = 6.0 s[1](a) 1.0 m s⁻² (b) 6.0 s
g = 9.81 m s⁻² and ignore air resistance.s = ut + ½at² with u = 0"dropped from rest" is what tells you u = 0[1]s = ½ × 9.81 × 3.2² = 50.2 ≈ 50 m[1]v = u + ator v² = u² + 2as[1]v = 9.81 × 3.2 = 31.4 ≈ 31 m s⁻¹[1](a) 50 m (b) 31 m s⁻¹ (c) the speed would be less
(20 − 0) / 40gradient of a velocity–time graph is acceleration[1]= 0.50 m s⁻²[1]½ × 40 × 20 = 400; rectangle 60 × 20 = 1200; triangle ½ × 30 × 20 = 300all three areas needed[1]= 1900 m[1]= 1900 / 130 = 14.6 ≈ 15 m s⁻¹total distance ÷ total time, not the mean of the velocities[1](a) 0.50 m s⁻² (b) 1900 m (c) 15 m s⁻¹
g = 10 m s⁻² and ignore air resistance.u_y = 0: 45 = ½ × 10 × t²"thrown horizontally" means the initial vertical velocity is zero[1]t² = 9.0, t = 3.0 s[1]s_x = u_x t with constant horizontal velocity[1]s_x = 15 × 3.0 = 45 m[1](a) 3.0 s (b) 45 m (c) the two components are independent
Distance is the total length of the path travelled, a scalar. Displacement is the straight line from start to finish together with its direction, a vector.
Yes. A ball thrown vertically upward is momentarily at rest at the top of its flight, but gravity still acts, so its acceleration is g downward.
a = (v − u)/t = (0 − 25)/5.0[1]a = −5.0 m s⁻², a deceleration of 5.0 m s⁻²the negative sign or the word deceleration, not both required[1]s = ½(u + v)t or v² = u² + 2as[1]s = ½(25 + 0) × 5.0 = 62.5 m[1]deceleration 5.0 m s⁻², distance 62.5 m
p = mv[1]p = 1200 × 15 = 18 000 kg m s⁻¹the stationary car contributes nothing[1]v = 18 000 / 2000 = 9.0 m s⁻¹[1](a) 1.8 × 10⁴ kg m s⁻¹ (b) 9.0 m s⁻¹ (c) not conserved; inelastic
0 = 0.020 × 400 + 4.0 × v[1]v = −8.0 / 4.0 = −2.0 m s⁻¹, i.e. 2.0 m s⁻¹ backwardsdirection required[1]p = mv is fixed, a larger m gives a smaller v[1]2.0 m s⁻¹ backwards
−8.0 m s⁻¹the sign is the whole question[1]Δp = m(v − u) = 0.15 × (−8.0 − 12) = −3.0 kg m s⁻¹a change of 20 m s⁻¹, not 4[1]F = Δp/Δt[1]F = −3.0 / 0.050 = −60 N, i.e. 60 N away from the wall[1]60 N, directed away from the wall
The total momentum of a system before an interaction equals the total momentum after it, provided no resultant external force acts on the system.
It increases the time over which the ball is brought to rest. Since F = Δp/Δt and the change in momentum is fixed, a longer time means a smaller force on the hands.
Δp = m(v − u) with u = 0the ball starts at rest on the tee[1]Δp = 0.045 × 60 = 2.7 kg m s⁻¹[1]0.50 ms = 5.0 × 10⁻⁴ s[1]F = Δp/Δt[1]F = 2.7 / 5.0 × 10⁻⁴ = 5400 N[1]0.30 × 70 = 21 kg m s⁻¹the ball contributes nothing[1]21 = 0.30 × v + 2.7[1]v = 18.3 / 0.30 = 61 m s⁻¹[1](a) 2.7 kg m s⁻¹ (b) 5400 N (c) 61 m s⁻¹
For every action there is an equal and opposite reaction. The two forces are of the same type and act on different bodies.
By Newton's first law the passenger continues moving at the same velocity because no resultant force acts on them; the bus decelerates beneath them, so they move forward relative to it.
The product of mass and velocity, p = mv. A vector quantity. SI unit: kg m s⁻¹.
p = 2.0 × 3.0 + 4.0 × 0 = 6.0 kg m s⁻¹[1]= 6.0 kg, so 6.0 = 6.0 × vthey stick together, so they share one velocity[1]v = 1.0 m s⁻¹ in the original directiondirection expected for full marks[1]1.0 m s⁻¹ in the direction of the original motion
= 15 − 6.0 = 9.0 Nfriction opposes, so it subtracts[1]F = ma[1]a = F/m = 9.0 / 3.0[1]a = 3.0 m s⁻²unit required[1]3.0 m s⁻²
Work = force × distance moved in the direction of the force, W = Fs cos θ. It is zero when the force is perpendicular to the motion, since cos 90° = 0.
Energy cannot be created or destroyed, only transferred from one store to another. The total energy of a closed system remains constant.
No. Energy is transferred to heat and sound in the ball and the floor at each bounce, so less remains as gravitational potential energy. The total is unchanged.
E = mghthe useful energy is the gravitational potential energy gained[1]E = 300 × 10 × 12 = 36 000 J[1]P = E/t[1]P = 36 000 / 40 = 900 Wunit required[1]900 W
KE = ½mv²[1]KE = ½ × 1200 × 20² = 240 000 Jsquare the velocity before multiplying[1]P = 240 000 / 8.0 = 30 000 W = 30 kW[1]2.4 × 10⁵ J and 30 kW
Q1The median of 3, 7, 9, 15, 21 is:
Correct answer: A — 9
The data is already in order and there are five values, so the middle one is the third: 9. Option B is the mean, which is a different statistic.
Q2The class boundaries of the class 30–39 are:
Correct answer: B — 29.5 and 39.5
Boundaries lie half a unit outside the limits so that consecutive classes meet without a gap. The class 30–39 really covers everything from 29.5 up to 39.5.
Q3Which average is most affected by an extreme value?
Correct answer: C — Mean
The mean is calculated from every value, so a single very large or very small one shifts it substantially. The median depends only on position, and the mode only on frequency.
Q4For grouped data, the mean is calculated using:
Correct answer: B — class midpoints
The original values are lost after grouping, so every value in a class is assumed to sit at its midpoint. This is why the answer is an estimate rather than the exact mean.
Q5A shopkeeper wants to know which size of shirt to order most of. He should use the:
Correct answer: C — mode
The mode is the most frequently occurring size, which is exactly what "order most of" asks for. A mean size of 39.6 would not correspond to any shirt he can buy.
Q6The range of 14, 22, 9, 30, 18 is:
Correct answer: A — 21
Largest 30 minus smallest 9 gives 21. The range is a single number, not an interval, so writing "9 to 30" would not be accepted.
Q7Two data sets have the same mean but different standard deviations. This tells you:
Correct answer: B — one is more spread out than the other
The standard deviation measures spread. A larger one means values sit typically further from the mean, even though the centres of the two sets coincide.
Q8Why are deviations squared when calculating variance?
Correct answer: B — Because the deviations would otherwise total zero
By the definition of the mean, the positive and negative deviations cancel exactly, so their sum is always zero. Squaring makes them all positive so the spread survives the addition.
Q9On a histogram with unequal class widths, the vertical axis shows:
Correct answer: B — frequency density
Area must represent frequency, so height must be frequency ÷ width. Plotting raw frequency would exaggerate the wider classes.
Q10Cumulative frequency is plotted against:
Correct answer: C — the upper class boundary
The running total for a class is only complete at the top of that class, so the upper boundary is the correct horizontal position.
Q11For 60 observations, the median is read from the ogive at a cumulative frequency of:
Correct answer: A — 30
N/2 = 30. Q₁ would be read at 15 and Q₃ at 45.
Q12If Q₁ = 12 and Q₃ = 28, the interquartile range is:
Correct answer: B — 16
IQR = Q₃ − Q₁ = 28 − 12 = 16. Option C is the semi-interquartile range, which is half of it.
Q13A class 20–30 has frequency 24 and a class 30–50 has frequency 24. On a histogram:
Correct answer: B — the second bar is half as tall
Densities are 24/10 = 2.4 and 24/20 = 1.2, so the second bar is half as tall. Its double width keeps the areas equal, which is exactly the point of frequency density.
Q14The mean of grouped data is described as an estimate because:
Correct answer: B — individual values are replaced by class midpoints
Grouping discards the actual values, so every observation in a class is assumed to sit at its midpoint. That assumption is what makes the answer an estimate.
Q15If the mean is noticeably greater than the median, the distribution is:
Correct answer: B — skewed with a tail to the right
A few unusually large values pull the mean up while leaving the median where it is. That long right-hand tail is positive skew.
Q16Which measure of spread is least affected by an extreme value?
Correct answer: B — Interquartile range
The IQR is built only from Q₁ and Q₃, so the top and bottom quarters of the data — where any extreme value lives — have no effect on it.
Q17Points falling from top left to bottom right show:
Correct answer: B — negative correlation
As one variable increases the other decreases, which is negative correlation. Its strength depends on how closely the points follow the trend.
Q18A line of best fit should pass through:
Correct answer: C — the middle of the points and the two means
It summarises the trend, so it belongs among the points with roughly equal numbers above and below. There is no reason for it to pass through the origin.
Q19Predicting beyond the range of the data is called:
Correct answer: B — extrapolation
It assumes the relationship continues where nothing was measured, and is the less reliable of the two.
Q20A prediction from a line of best fit is most reliable when:
Correct answer: C — the correlation is strong and the value is inside the range
Strong correlation means the points lie close to the line, and staying inside the range means the pattern has actually been observed there.
Q21Ice cream sales correlate with drowning deaths because:
Correct answer: B — hot weather causes both
Temperature is a third factor driving both independently — the standard illustration that correlation is not causation.
Q22Towns with more doctors have more illness. The likely explanation is:
Correct answer: B — population size causes both
Bigger towns have more of everything. Comparing doctors and illness per head of population would test whether any real relationship remains.
Q23On a scatter diagram showing no correlation, you should:
Correct answer: B — draw no line of best fit
There is no trend to summarise, so a line would suggest a relationship the data does not support.
Q24Correlation between two variables means:
Correct answer: B — they tend to change together
It describes a tendency to move together and nothing more. Establishing a cause requires a controlled investigation.
Q25⁶P₂ equals:
Correct answer: B — 30
6 × 5 = 30. Option A is ⁶C₂, which does not distinguish the order of the two chosen items.
Q26How many ways can 4 people be seated around a round table?
Correct answer: C — 6
(4 − 1)! = 3! = 6. Rotating everyone one seat clockwise is not a new arrangement, so one person is fixed.
Q27A committee of 4 from 9 people can be chosen in:
Correct answer: B — 126 ways
⁹C₄ = 126, because a committee has no internal order. 3024 is ⁹P₄, which would treat the four roles as distinct.
Q28The number of arrangements of the letters in LEVEL is:
Correct answer: C — 30
5 letters with L twice and E twice, so 5!/(2!2!) = 120/4 = 30.
Q29¹²C₁₀ is quickest to evaluate as:
Correct answer: A — ¹²C₂ = 66
ⁿCᵣ = ⁿCₙ₋ᵣ, so ¹²C₁₀ = ¹²C₂ = (12 × 11)/2 = 66 — two lines instead of ten.
Q30How many 3-digit numbers can be formed from the digits 1–5 with no repeats?
Correct answer: C — 60
Order matters in a number, so ⁵P₃ = 5 × 4 × 3 = 60. Option B would allow repeated digits.
Q31Five books are arranged on a shelf with two particular ones always together. The number of arrangements is:
Correct answer: B — 48
Glue the pair into one block: 4 items arrange in 4! = 24 ways, and the pair can swap internally in 2 ways, giving 48.
Q32Choosing 3 flavours from 8 for a bowl of ice cream is a:
Correct answer: B — combination, ⁸C₃
The scoops end up in one bowl, so the order they were chosen in makes no difference — a selection, counted by ⁸C₃ = 56.
Q33A fair die is rolled. P(an even number) is:
Correct answer: C — 1/2
Three of the six faces (2, 4, 6) are even, so P = 3/6 = 1/2.
Q34If P(E) = 0.28, then P(E′) is:
Correct answer: A — 0.72
The complement rule gives 1 − 0.28 = 0.72. An event and its complement always account for the whole sample space, so they must total 1.
Q35Two coins are tossed. P(exactly one head) is:
Correct answer: B — 1/2
The sample space is {HH, HT, TH, TT}. Two of the four outcomes have exactly one head, so P = 2/4 = 1/2. Treating HT and TH as the same outcome gives the wrong answer 1/3.
Q36A card is drawn from 52. P(a red king) is:
Correct answer: A — 2/52
There are exactly two red kings, hearts and diamonds, so P = 2/52 = 1/26. Option B counts all four kings, including the black ones.
Q37Which value could NOT be a probability?
Correct answer: C — 5/4
5/4 = 1.25 exceeds 1, which is impossible: an event cannot be more than certain. Zero and one are both perfectly valid, representing impossible and certain events.
Q38A bag has 4 red and 6 blue balls. Two are drawn with replacement. P(both red) is:
Correct answer: A — 16/100
With replacement the bag is restored, so both draws have P(red) = 4/10 and the events are independent: (4/10)(4/10) = 16/100 = 4/25. Option B would be the answer without replacement.
Q39Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.45. P(A or B) is:
Correct answer: B — 0.75
Mutually exclusive means the overlap is zero, so the general rule reduces to simple addition: 0.3 + 0.45 = 0.75. Option A multiplies instead of adding.
Q40P(at least one head in three tosses of a fair coin) is:
Correct answer: C — 7/8
Use the complement: P(no heads) = (1/2)³ = 1/8, so P(at least one) = 1 − 1/8 = 7/8. Listing all the favourable cases would take seven lines to reach the same answer.
Q41Two events are mutually exclusive when:
Correct answer: B — they cannot occur together
Mutual exclusivity means the intersection is empty, so P(A ∩ B) = 0. It says nothing about the events being equally likely.
Q42A fair die is rolled twice. P(two sixes) is:
Correct answer: C — 1/36
The rolls are independent, so multiply: (1/6)(1/6) = 1/36. Adding would be the rule for OR, which is a different question.
Q43For independent events with P(A) = 0.5 and P(B) = 0.4, P(A ∪ B) is:
Correct answer: C — 0.7
P(A ∩ B) = 0.5 × 0.4 = 0.2, so P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7. Simply adding gives 0.9, which double-counts the overlap.
Q44From a bag of 3 red and 2 blue, two are drawn without replacement. P(both red) is:
Correct answer: B — 3/10
(3/5) × (2/4) = 6/20 = 3/10. Option A is the answer with replacement, and option C is the same value unsimplified — the paper expects lowest terms.
Q45P(at least one head in 4 tosses of a fair coin) is:
Correct answer: C — 15/16
P(no heads) = (1/2)⁴ = 1/16, so P(at least one) = 1 − 1/16 = 15/16.
Q46On a tree diagram, the probabilities on the branches from one node must:
Correct answer: A — sum to 1
Those branches cover every possibility at that stage, so their probabilities must total exactly 1. It is the fastest check for an error before you go further.
Q47P(B|A) = P(B) tells you that:
Correct answer: B — A and B are independent
The occurrence of A has left the probability of B unchanged, which is exactly the definition of independence.
Q48A coin has landed tails eight times. The probability of tails on the ninth toss is:
Correct answer: C — exactly 1/2
Tosses are independent, so the coin carries no memory of what has already happened. Expecting a correction is the gambler's fallacy.
Q49For any discrete random variable, ΣP(X = x) equals:
Correct answer: B — 1
The list of values covers every possibility and they are mutually exclusive, so the probabilities must total 1.
Q50E(X) is best described as:
Correct answer: B — the long-run average
It is the mean over many repetitions — the balance point. It need not be likely, or even possible.
Q51Var(X) is given by:
Correct answer: A — E(X²) − [E(X)]²
The mean of the squares minus the square of the mean. Reversing the order gives a negative number, which cannot be a variance.
Q52X takes values 1, 2, 3 with probabilities 0.5, 0.3, 0.2. E(X) is:
Correct answer: B — 1.7
1(0.5) + 2(0.3) + 3(0.2) = 0.5 + 0.6 + 0.6 = 1.7. The value 2.0 would be the unweighted average, which ignores the probabilities.
Q53Which situation is NOT binomial?
Correct answer: B — drawing 5 cards without replacement, counting aces
Drawing without replacement changes the probability with each draw and makes the trials dependent, breaking two of the four conditions.
Q54For X ~ B(30, 0.2), the mean and variance are:
Correct answer: B — 6 and 4.8
E(X) = np = 6 and Var(X) = np(1 − p) = 30 × 0.2 × 0.8 = 4.8. Forgetting the (1 − p) factor gives 6.
Q55P(X ⩾ 3) is equal to:
Correct answer: B — 1 − P(X ⩽ 2)
"At least 3" includes 3, so the complement is everything up to and including 2. Using P(X ⩽ 3) would wrongly remove the value 3 as well.
Q56A student computes Var(X) = −0.6. This means:
Correct answer: B — there is an arithmetic error
Variance is an average of squares and can never be negative. The usual causes are squaring the probabilities or forgetting to square the mean.
Q57For X ~ N(100, 36), the standard deviation is:
Correct answer: B — 6
The second parameter is the variance, so σ = √36 = 6. Using 36 directly is the most common error in the topic.
Q58For a continuous distribution, P(X < 7) compared with P(X ⩽ 7) is:
Correct answer: C — equal
P(X = 7) = 0 because a single point encloses no area, so including or excluding the endpoint changes nothing.
Q59If z = −1.5, then P(Z < −1.5) equals:
Correct answer: B — 1 − Φ(1.5)
By symmetry the area left of −1.5 equals the area right of +1.5, which is 1 − Φ(1.5). A probability can never be negative, ruling out two options immediately.
Q60X ~ N(50, 16) and x = 58. The z-score is:
Correct answer: B — 2
σ = √16 = 4, so z = (58 − 50)/4 = 2. Dividing by 16 instead of 4 gives 0.5.
Q61P(a < Z < b) is given by:
Correct answer: B — Φ(b) − Φ(a)
Both table values are areas to the left, so subtracting leaves the strip between them.
Q62About what proportion lies within two standard deviations of the mean?
Correct answer: C — 95%
The 68-95-99.7 rule gives about 95% within two standard deviations. 68% is one, and 99.7% is three.
Q63To find the mark exceeded by the top 5%, you need the z value:
Correct answer: A — 1.645
The top 5% lies above the 95th percentile, whose z-score is 1.645. The value 1.960 corresponds to the top 2.5%.
Q64Standardising a normal distribution changes:
Correct answer: B — only the labelling of the axis
The curve is identical; only the scale beneath it is renamed in standard deviations. That is precisely why a single table can serve every normal distribution.
Class limits are the values written in the table, such as 20–29. Class boundaries are the true dividing values, 19.5 and 29.5, obtained by going half a unit beyond each limit so that consecutive classes meet with no gap.
The mean uses every value in its calculation, so one very large value pulls it far above the bulk of the data and it no longer represents a typical member of the set. The median should be used instead, because it depends only on position and is unaffected by extremes.
The mode, which is size 8. The manager needs to know which size sells most often; a mean of 8.14 is not a size that can be ordered.
= 120/8 = 15[1]= (15 + 15)/2 = 15the data must be ordered first — this is where marks are lost[1]Mean = 15, median = 15, mode = 15
x̄ = 635 / 30 = 21.2 cm (1 d.p.)accept 21.17[1]Estimated mean ≈ 21.2 cm; modal class 20–24
Σx = 300, mean = 300/5 = 60[1]Σx = 300, mean = 300/5 = 60the means are identical, which is the point of the question[1]= 62 − 58 = 4[1]= 80 − 40 = 40[1](a) both 60 (b) Ali 4, Bilal 40 (c) Ali — same mean but far smaller spread
A bar chart displays discrete categories with gaps between the bars, and only the heights are meaningful. A histogram displays continuous data with no gaps, and it is the area of each bar that represents the frequency.
Because the cumulative total for a class counts every observation up to and including the top of that class. The count "36 or fewer" is only complete at the upper boundary, so plotting at the midpoint would claim the total was reached earlier than it was.
IQR = Q₃ − Q₁, the spread of the middle 50% of the data. Unlike the range it is unaffected by extreme values, since it depends only on the quartiles and ignores the top and bottom quarters entirely.
Σfx = 570[1]x̄ = 570 / 50 = 11.4 kgaccept 11.4 kg[1]Estimated mean 11.4 kg; modal class 10–15 kg
IQR = 68 − 42 = 26 marks[1]Semi-IQR = 26 / 2 = 13 marks[1]3/4 × 80 = 60 students scored 68 or less[1]80 − 60 = 20 students scored above 68a quarter of the data lies above Q₃, by definition[1]IQR 26 marks; semi-IQR 13 marks; 20 students above 68
N/2 = 50 to the curve and read down to the horizontal axisthe median is about 490 hours[1]Σfx = 1000 + 7500 + 17500 + 14000 + 9000 = 49 000[1]x̄ = 49 000 / 100 = 490 hours[1](a) 10, 35, 70, 90, 100 (b) read across at 50 and down to the axis (c) mean ≈ 490 h, almost equal to the median, so the data is nearly symmetrical
Negative correlation — as age increases, value decreases. It would likely be fairly strong, since age is one of the main determinants of a used car's price.
The points show no pattern — as one variable changes the other does not tend to change in any particular direction. The two appear unrelated, and no line of best fit should be drawn.
That is extrapolation: it assumes the relationship continues into a region where nothing was measured. The pattern may change or stop, and the prediction may be impossible — such as a predicted mark above 100%.
(a) reliable — inside the range and strong correlation (b) unreliable — extrapolation beyond the evidence
Unjustified — population size causes both. Compare per head of population instead.
(a) marks tend to fall as absence rises, with noticeable scatter (b) line through the middle, read up then across (c) a third factor such as illness, or reversed causation
A permutation is an arrangement in which the order matters, counted by ⁿPᵣ. A combination is a selection in which order is irrelevant, counted by ⁿCᵣ. Since each selection of r objects can be arranged in r! ways, ⁿPᵣ = r! ⁿCᵣ.
⁸C₃ and ⁸P₃.⁸C₃ = 8!/(3!5!) = 56 and ⁸P₃ = 8!/5! = 8 × 7 × 6 = 336. Note that 336 = 3! × 56.
(7 − 1)! = 6! = 720. One person is fixed to remove the rotations, and the remaining six are arranged around them.
10! / (3! 3! 2!)one factorial for each repeated letter[1]= 3 628 800 / 72 = 50 400[1]50 400
⁴C₃ × ⁶C₂ = 4 × 15 = 60choose the women, then the men to fill the rest[1]⁴C₄ × ⁶C₁ = 1 × 6 = 6there are only 4 women, so this is the last case[1]= 60 + 6 = 66add the cases, never multiply them[1]66
6! = 720order matters, and nothing is repeated[1]5! = 120the glue technique[1]× 2! = 2the internal arrangement is easy to forget[1]120 × 2 = 240 ways[1]= 720 − 240 = 480[1](a) 720 (b) 240 (c) 480
The sample space is the set of all possible outcomes of the experiment. For two coins, S = {HH, HT, TH, TT}, so n(S) = 4.
Two events are mutually exclusive if they cannot occur at the same time, so P(A and B) = 0. Example: on a single roll of a die, "getting a 2" and "getting a 5".
P(no rain) = 1 − 0.35 = 0.65, using the complement rule P(E′) = 1 − P(E).
n(S) = 5 + 4 + 3 = 12the total is the denominator throughout[1]P(green) = 4/12 = 1/3lowest terms expected[1]P(not blue) = 1 − 3/12 = 9/12 = 3/4accept counting 5 + 4 = 9 favourable outcomes directly[1](i) 1/3 (ii) 3/4
n(S) = 6 × 6 = 36ordered pairs, so (2,5) and (5,2) both count[1]P = 6/36 = 1/67 is the most likely total on two dice[1]P = 6/36 = 1/6[1](i) 1/6 (ii) 1/6
7/10 white and 3/10 black[1]P(both white) = (7/10) × (6/9)multiply along the branch[1]= 42/90 = 7/15[1](7/10)(3/9) + (3/10)(7/9)both orders are needed[1]= 21/90 + 21/90 = 42/90 = 7/15[1](b) 7/15 (c) 7/15
Mutually exclusive events cannot occur together, so P(A ∩ B) = 0. Independent events can occur together, but one occurring does not affect the probability of the other, so P(A ∩ B) = P(A) × P(B).
1/2. Successive tosses are independent, so the coin has no memory of previous results and the probability is unchanged. Expecting a tail to "even things out" is the gambler's fallacy.
No. Removing the first ball changes both the number of balls left and the composition of the bag, so the probability for the second draw depends on what the first was. The draws are dependent, and the second-stage probabilities are conditional.
P(heart) = 13/52 and P(face) = 12/52twelve face cards: three in each of four suits[1]P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 13/52 + 12/52 − 3/52[1]= 22/52 = 11/26[1]11/26
P(both) = 0.8 × 0.6[1]= 0.48[1]P(neither) = 0.2 × 0.4 = 0.08multiply the two failure probabilities[1]P(at least one) = 1 − 0.08 = 0.92accept the direct route 0.48 + 0.32 + 0.12 = 0.92[1](i) 0.48 (ii) 0.92
P(good) = 6/10, P(defective) = 4/10the two branches must sum to 1[1]5/9 good, 4/9 defective; after a defective: 6/9 good, 3/9 defectivedenominator 9 throughout, since one bulb has gone[1]P(both good) = (6/10) × (5/9)multiply along the branch[1]= 30/90 = 1/3[1]P(at least one defective) = 1 − P(both good)"at least one defective" is the exact opposite of "both good"[1]= 1 − 1/3 = 2/3[1](b) 1/3 (c) 2/3
Any two of: a fixed number of trials; each trial has only two outcomes; the trials are independent; the probability of success is constant for every trial.
c = 0.2, E(X) = 1.6, Var(X) = 0.84
P(X = 3) = 0.104, E(X) = 1.33
(a) 0.274; (b) 0.711; (c) mean 2.4, variance 2.112; (d) independence with constant p
E(X) is the long-run average of many rolls, not a value of a single roll. Averaging the six equally likely scores gives (1+2+3+4+5+6)/6 = 3.5. An average need not be one of the values being averaged.
Probability for a continuous variable is the area under the curve, and a single point has no width, so it encloses zero area. Consequently P(X < 5) and P(X ⩽ 5) are equal, unlike for a discrete variable.
0.945
723 hours
(a) two z equations; (b) μ = 30.9 cm, σ = 7.06 cm; (c) 0.270
The bracket in N(40, 9) gives the variance, so σ = √9 = 3, not 9. The correct standardisation is z = (46 − 40)/3 = 2.
Q1The Poisson distribution is appropriate when:
Correct answer: B — events occur at a constant rate over an interval
A fixed number of trials with two outcomes describes the binomial. Poisson applies when there is no n, only a rate.
Q2For Po(6), the variance is:
Correct answer: A — 6
For a Poisson the mean and variance are both λ. This equality is what distinguishes it from other discrete distributions.
Q3Defects occur at 1.5 per metre. For a 4-metre length, λ is:
Correct answer: C — 6
λ scales with the interval: 1.5 × 4 = 6. Failing to scale is the most common error in the topic.
Q4For Po(λ), P(X = 0) equals:
Correct answer: B — e^(−λ)
Putting r = 0 gives e^(−λ)λ⁰/0! = e^(−λ), since both λ⁰ and 0! equal 1.
Q5X ~ Po(3) and Y ~ Po(5) are independent. X + Y follows:
Correct answer: A — Po(8)
Independent Poissons add by adding their means, giving Po(8).
Q6B(600, 0.004) is best approximated by:
Correct answer: B — Po(2.4)
n is large and p small, so use Po(np) = Po(2.4). λ is too small here for a normal approximation.
Q7For Po(4), P(X ⩾ 2) is best found as:
Correct answer: B — 1 − P(0) − P(1)
The upper tail is infinite, so the complement must be used. "At least 2" includes 2, so only 0 and 1 are removed.
Q8A sample has mean 3.0 and variance 9.2. This suggests:
Correct answer: B — the events are clustered, so Poisson is inappropriate
Poisson requires the mean and variance to be roughly equal. A variance three times the mean indicates the events are not occurring singly and independently.
Q9If Var(X) = 4, then Var(5X) equals:
Correct answer: B — 100
Var(aX) = a²Var(X) = 25 × 4 = 100. Multiplying by 5 instead of 25 gives the tempting wrong answer of 20.
Q10If Var(X) = 7, then Var(X + 10) equals:
Correct answer: C — 7
Adding a constant slides the distribution without changing its spread, so the variance is unchanged.
Q11For independent X and Y with variances 9 and 16, Var(X − Y) is:
Correct answer: C — 25
Variances add even when the variables are subtracted: 9 + 16 = 25. A variance could never be negative, which immediately rules out the first option.
Q12X has variance σ². The variance of X₁ + X₂ + X₃ is:
Correct answer: A — 3σ²
Three independent observations give 3σ². Getting 9σ² would be the variance of 3X, a single observation tripled.
Q13The variance of the mean of 25 observations, each with variance 100, is:
Correct answer: B — 4
Var(X̄) = σ²/n = 100/25 = 4, so the standard deviation of the mean is 2 rather than 10.
Q14"The total weight of three boxes" corresponds to:
Correct answer: B — X₁ + X₂ + X₃
Three separate boxes are three independent observations. 3X would mean weighing one box and tripling the reading.
Q15X ~ N(20, 9) and Y ~ N(15, 16) independently. X + Y is distributed as:
Correct answer: A — N(35, 25)
Means add to 35 and variances add to 25. A linear combination of independent normals is itself normal.
Q16Every variance rule in this topic requires that the variables are:
Correct answer: B — independent
Independence is the essential condition. Without it a covariance term appears and the simple addition of variances is no longer valid.
Q17For a continuous random variable, probability corresponds to:
Correct answer: B — the area under f(x)
f is a density. Only the area between two values is a probability, which is why every formula uses an integral.
Q18Which is NOT required of a probability density function?
Correct answer: C — f(x) ⩽ 1 everywhere
The density may exceed 1; only the area is limited to 1. A density on a very short interval must be tall.
Q19E(X) for a continuous variable is:
Correct answer: B — ∫ x·f(x) dx
The discrete Σx·P(X = x) becomes ∫x·f(x)dx. The first option is the total probability, which is always 1.
Q20If F(x) = x³/27 on [0, 3], then f(x) is:
Correct answer: B — x²/9
Differentiating F gives 3x²/27 = x²/9. The third option is the same value unsimplified but is written before cancelling, so x²/9 is the expected form.
Q21The median m of a continuous variable satisfies:
Correct answer: B — F(m) = 0.5
The median splits the total area in half, so the cumulative function reaches 0.5 there.
Q22For a continuous variable, P(X < 4) compared with P(X ⩽ 4) is:
Correct answer: C — equal
P(X = 4) = 0 because a point has no width, so including the endpoint changes nothing.
Q23f(x) is defined only on [0, 5]. To find P(X > 3) you integrate from:
Correct answer: B — 3 to 5
Beyond 5 the density is zero, so the upper limit is the top of the range. Using ∞ risks an incorrect or divergent answer.
Q24Var(X) for a continuous variable is found by:
Correct answer: B — ∫x²f dx − [∫xf dx]²
It is E(X²) − [E(X)]², exactly as in the discrete case, with both expectations computed as integrals.
Q25The standard error of the sample mean is:
Correct answer: C — σ/√n
The variance of X̄ is σ²/n, so the standard deviation is σ/√n. Using σ itself ignores that averaging reduces variability.
Q26The central limit theorem allows you to assume X̄ is approximately normal when:
Correct answer: B — n is large
The theorem applies for large n whatever the population shape. If the population is already normal, X̄ is exactly normal and the theorem is not needed.
Q27The unbiased estimate of population variance divides by:
Correct answer: B — n − 1
Dividing by n underestimates, because deviations are taken from the sample mean. n − 1 corrects that bias exactly.
Q28A population has σ = 20. For a sample of 25, the standard error is:
Correct answer: B — 4
20/√25 = 20/5 = 4. Dividing by 25 rather than 5 gives 0.8.
Q29Increasing the confidence level from 95% to 99% makes the interval:
Correct answer: B — wider
A larger z is needed to capture more of the distribution, so the interval widens. More confidence costs precision.
Q30To halve the width of a confidence interval you should multiply the sample size by:
Correct answer: B — 4
The width is proportional to 1/√n, so n must be quadrupled for √n to double.
Q31A 95% confidence interval of (10, 14) means:
Correct answer: C — 95% of such intervals would contain μ
μ is a fixed number, so it is either inside this interval or not. The 95% refers to the long-run success rate of the procedure.
Q32When standardising a sample mean, the denominator should be:
Correct answer: B — σ/√n
A sample mean varies less than a single observation, so the standard error σ/√n is the correct denominator. Using σ makes z too small by a factor of √n.
Q33The null hypothesis is the statement that:
Correct answer: B — nothing unusual is happening
H₀ is the conservative position. The test looks for evidence against it, and H₁ is what the researcher suspects instead.
Q34A 5% significance level means the probability of:
Correct answer: B — rejecting H₀ when it is true is 0.05
It is the Type I error rate — the chance of a false alarm. It says nothing about the probability that H₀ itself is true.
Q35H₁: μ ≠ 50 at the 5% level requires a critical z of:
Correct answer: B — 1.960
The ≠ makes it two-tailed, so 5% splits into 2.5% per tail, giving 1.960.
Q36A p-value of 0.07 at the 5% level means:
Correct answer: B — do not reject H₀
0.07 > 0.05, so the data is not surprising enough. Note this is not the same as showing H₀ is true.
Q37A Type II error occurs when you:
Correct answer: B — fail to reject a false H₀
It is the missed detection — a real effect exists but the test did not find enough evidence for it.
Q38Lowering the significance level from 5% to 1%:
Correct answer: B — makes Type I rarer and Type II more likely
Demanding stronger evidence means fewer false alarms but more missed real effects. Only a larger sample reduces both together.
Q39When testing a sample mean, the denominator of the test statistic is:
Correct answer: B — σ/√n
The sample mean varies less than a single observation, so the standard error σ/√n is correct. Using σ makes z too small by a factor of √n.
Q40The final mark in a hypothesis test is usually for:
Correct answer: C — a conclusion stated in context
Examiners reserve a mark for interpreting the result in the situation described. "Reject H₀" alone does not earn it.
Any two of: events occur independently of one another; they occur at a constant average rate; and they occur singly rather than in clusters.
0.168
0.762
(a) B(400, 0.005); (b) Po(2); (c) 0.406; (d) 0.271
For a Poisson distribution the mean and variance are equal, both λ. A mean of 4.1 and variance of 4.0 are close enough to support that model. A variance of 12 would be far larger than the mean, so the events would not be occurring independently at a constant rate — they would be clustered, and a Poisson model would be inappropriate.
Mean: E(3X − 4) = 3(12) − 4 = 32. Variance: Var(3X − 4) = 3² × 5 = 45 — the −4 has no effect on the variance, and the 3 is squared.
Variances of independent variables add regardless of whether the variables are added or subtracted, so Var(X − Y) = 10 + 6 = 16. Subtracting makes the difference more variable, not less: an unusually large X combined with an unusually small Y produces an extreme difference, so both variabilities contribute.
0.100
(a) 0.309; (b) N(4.08, 0.0064), P = 0.191; (c) 16σ² versus 4σ²
X and Y must be independent. Without independence the variables move together to some degree, and a covariance term would have to be added, so the simple sum would no longer be correct.
f(x) is a density — a height, not a probability. Probability is the area under the curve, and a tall narrow curve can enclose an area of 1 while having heights above 1. Only the area is constrained to lie between 0 and 1.
k = 1/8, E(X) = 4/3
F(x) = x³/8; median = 1.59
(a) verified; (b) 0.64; (c) E(X) = 6.67, Var(X) = 5.56
Differentiate: f(x) = dF/dx = 2x/16 = x/8 for 0 ⩽ x ⩽ 4, and zero elsewhere.
A random sample is one in which every member of the population has an equal chance of being selected and the selections are independent. A non-random sample is biased — it systematically over- or under-represents part of the population — and since the direction and size of that bias are unknown, no calculation performed afterwards can remove it.
E(X̄) = μ and Var(X̄) = σ²/n. The standard deviation of X̄, called the standard error, is therefore σ/√n.
x̄ = 12, s² = 4
(a) (998.1, 1005.9) g; (b) 7.73 g; (c) n = 256; (d) repeated-sampling interpretation
The deviations are measured from the sample mean, which sits in the middle of the sample by construction. This makes the sum of squared deviations systematically smaller than it would be about the true population mean, so dividing by n would underestimate the variance. Dividing by n − 1 corrects the bias exactly.
A test only measures whether the data is surprising under H₀; it can never demonstrate that H₀ is true. Failing to find evidence may simply mean the sample was too small to detect a real effect. So the correct conclusion is that the evidence was insufficient, not that H₀ has been shown correct.
A Type I error occurs when H₀ is rejected even though it is actually true — a false alarm. Its probability is exactly the significance level of the test, α.
z = −2.26 < −1.645; reject H₀ — evidence the mean is below 40 hours
(a) H₀: p = 0.5, H₁: p ≠ 0.5; (c) z = 2.05 > 1.96, reject H₀; (d) missing a real bias
At 5%: 0.032 < 0.05, so H₀ is rejected — there is sufficient evidence against it. At 1%: 0.032 > 0.01, so H₀ would not be rejected. The same data supports different conclusions because the level sets how much evidence is demanded.
These questions come from the A Level Mathematics (9709) lessons — each topic has its own notes, worked examples and an interactive diagram.