2nd Year Mathematics — MCQs & Practice Questions

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.

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.

Since x² ≥ 0 for every real x, the smallest value of f is 3, reached at x = 0. The range is f(x) ≥ 3.

1. fg(x) = f(x² − 3) = 2(x² − 3) + 1 = 2x² − 5substitute the whole of g into f[1]
2. gf(x) = g(2x + 1) = (2x + 1)² − 3 = 4x² + 4x − 2expand the bracket fully[1]
3. Set them equal: 2x² − 5 = 4x² + 4x − 2, so 2x² + 4x + 3 = 0[1]
4. Discriminant = 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

1. Let y = (2x − 3)/(x + 1) and multiply up: y(x + 1) = 2x − 3clear the fraction before rearranging[1]
2. xy + y = 2x − 3, so xy − 2x = −3 − ygather every term containing x on one side[1]
3. x(y − 2) = −(3 + y), so x = −(y + 3)/(y − 2)factorising out x is the key step[1]
4. 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

1. Need x − 4 ≥ 0, so the domain is x ≥ 4a square root requires a non-negative argument[1]
2. A square root is never negative, so the range is f(x) ≥ 0[1]
3. y = √(x − 4) gives y² = x − 4, so x = y² + 4squaring is safe here because y ≥ 0[1]
4. f⁻¹(x) = x² + 4[1]
5. Domain of f⁻¹ is x ≥ 0not all of ℝ — it must match the range of f[1]
6. The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f — the inverse swaps the two sets[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

Write it as 5x³ − 2x⁻¹. Then dy/dx = 15x² + 2x⁻², that is 15x² + 2/x².

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.

1. dy/dx = 3x² − 6x − 9[1]
2. Sets it to zero: 3(x² − 2x − 3) = 0[1]
3. Factorises: (x − 3)(x + 1) = 0, so x = 3 or x = −1[1]
4. At x = 3: y = 27 − 27 − 27 + 5 = −22[1]
5. At x = −1: y = −1 − 3 + 9 + 5 = 10[1]
6. d²y/dx² = 6x − 6; at x = 3 it is +12, a minimum[1]
7. At x = −1 it is −12, a maximum[1]

Minimum at (3, −22), maximum at (−1, 10)

1. The base is a square of side 24 − 2xa square is removed from both ends[1]
2. The height is x, so V = x(24 − 2x)²[1]
3. Expands: V = 576x − 96x² + 4x³[1]
4. dV/dx = 576 − 192x + 12x²[1]
5. Sets to zero: 12(x² − 16x + 48) = 0 → (x − 4)(x − 12) = 0[1]
6. x = 4; x = 12 is rejected because it leaves no base[1]
7. V = 4 × 16² = 4 × 256[1]
8. V = 1024 cm³[1]

x = 4 cm giving V = 1024 cm³

1. v = ds/dt = 6t² − 18t + 12[1]
2. a = dv/dt = 12t − 18[1]
3. Sets v = 0: 6(t² − 3t + 2) = 0 → (t − 1)(t − 2) = 0[1]
4. t = 1 s and t = 2 s[1]
5. At 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

2x³ − 2x² + 3x + c

[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.

1. Sets y = 0: x(x − 4) = 0[1]
2. x = 0 and x = 4[1]
3. Integrates: ∫(x² − 4x) dx = x³/3 − 2x²[1]
4. Evaluates: (64/3 − 32) − 0 = −32/3[1]
5. Recognises the curve is below the axis on this interval[1]
6. Area = 32/3 ≈ 10.7 square unitsthe magnitude is taken[1]

(a) x = 0 and x = 4 (b) 32/3 square units

1. Integrates: y = x³ − x² + c[1]
2. Substitutes the point: 9 = 8 − 4 + c[1]
3. c = 5[1]
4. y = x³ − x² + 5[1]
5. Integrates again: ∫₀² (x³ − x² + 5) dx = [x⁴/4 − x³/3 + 5x]₀²[1]
6. = 4 − 8/3 + 10[1]
7. = 34/3 ≈ 11.3 square units[1]

(a) y = x³ − x² + 5 (b) 34/3 square units

1. Sets x + 2 = x² → x² − x − 2 = 0[1]
2. (x − 2)(x + 1) = 0, so x = −1 and x = 2[1]
3. Recognises the line is above the curve between them[1]
4. Area = ∫₋₁² [(x + 2) − x²] dx[1]
5. = [x²/2 + 2x − x³/3]₋₁²[1]
6. = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 9/2[1]

x = −1 and 2; area = 9/2 square 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.

1. Separate: (1/y) dy = 3x² dxEach variable with its own differential.[1]
2. Integrate: ln|y| = x³ + cA single arbitrary constant is enough.[1]
3. x = 0, y = 2 gives ln 2 = cApplying the condition before rearranging keeps the algebra simple.[1]
4. ln(y/2) = x³, so y = 2e^(x³)Exponentiating both sides; y stays positive so the modulus can be dropped.[1]

y = 2e^(x³)

1. (a) Separate and integrate: (1/V) dV = −0.02 dt gives ln V = −0.02t + cV is positive throughout, so the modulus is unnecessary.[1]
2. At t = 0, V = 500, so c = ln 500, giving V = 500e^(−0.02t)The standard exponential-decay shape.[1]
3. (b) V = 500e^(−0.6) = 500 × 0.5488 = 274 litresTo 3 significant figures.[1]
4. (c) 250 = 500e^(−0.02t), so e^(−0.02t) = 0.5Halving means setting V to 250, not subtracting 250.[1]
5. −0.02t = ln 0.5, so t = 34.7 minutesln 0.5 is negative, and dividing by −0.02 makes t positive — a useful sign check.[1]
6. (d) As t → ∞, e^(−0.02t) → 0, so V → 0 but never reaches it.An exponential decay approaches zero asymptotically.[1]
7. So according to this model the tank never completely empties, which shows the model is unrealistic for very small volumes.The final mark is for the interpretation, including the comment on the model's limitation.[1]

(a) V = 500e^(−0.02t); (b) 274 litres; (c) 34.7 min; (d) V → 0 but never reaches it

1. Separate: 1/(1 + y²) dy = (1/x) dxRecognising that the right side factorises into a function of x times a function of y.[1]
2. Integrate: tan⁻¹ y = ln x + cThe integral of 1/(1 + y²) is arctan y — a standard result worth knowing on sight.[1]
3. x = 1, y = 1: tan⁻¹1 = ln 1 + c, so c = π/4ln 1 = 0, and tan⁻¹1 = π/4 in radians.[1]
4. y = tan(ln x + π/4)Rearranged with y as the subject, as required.[1]

y = tan(ln x + π/4)

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. Identifies a = 3, b = 4, c = −10, with the line already in general form[1]
2. Substitutes the point: 3(−1) + 4(4) − 10 = −3 + 16 − 10 = 3[1]
3. √(9 + 16) = 5[1]
4. d = |3|/5 = 0.6 unitsthe modulus makes the distance positive[1]

0.6 units

1. m₁ = 3 and m₂ = −2[1]
2. tan θ = |(−2 − 3)/(1 + (3)(−2))| = |−5/−5|both numerator and denominator are negative[1]
3. = 1[1]
4. θ = 45°[1]

45°

1. Midpoint of BC = ((7+3)/2, (4+8)/2) = (5, 6)a median joins a vertex to the midpoint of the opposite side[1]
2. Gradient of the median = (6 − 2)/(5 − 1) = 1, so y − 2 = 1(x − 1), giving y = x + 1[1]
3. Centroid = ((1+7+3)/3, (2+4+8)/3) = (11/3, 14/3)average of the three vertices[1]
4. Gradient of AB = (4 − 2)/(7 − 1) = 1/3, so AB is x − 3y + 5 = 0from y − 2 = (1/3)(x − 1), cleared of fractions[1]
5. Substituting C(3, 8): |3 − 24 + 5| = 16[1]
6. 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

Eccentricity is the constant ratio of a point's distance from the focus to its distance from the directrix. e = 0 circle, 0 < e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.

Here 2g = 8 and 2f = −6, so g = 4, f = −3, c = 0. Centre (−g, −f) = (−4, 3) and radius √(16 + 9 − 0) = 5.

Comparing with y² = 4ax gives 4a = 12, so a = 3. The focus is at (3, 0) and the directrix is the line x = −3.

1. a² = 16 and b² = 9, so a = 4, b = 3[1]
2. For a hyperbola c² = a² + b² = 16 + 9 = 25add, unlike the ellipse[1]
3. c = 5, so the foci are at (±5, 0)[1]
4. e = c/a = 5/4 = 1.25greater than 1, as required for a hyperbola[1]

Foci (±5, 0); e = 1.25

1. The radius is the distance from the centre to the given point[1]
2. r = √[(5 − 2)² + (1 + 3)²] = √(9 + 16) = 5[1]
3. Standard form (x − 2)² + (y + 3)² = 25the sign of k flips inside the bracket[1]
4. Expanding: x² + y² − 4x + 6y − 12 = 0accept either form unless one is specified[1]

(x − 2)² + (y + 3)² = 25, i.e. x² + y² − 4x + 6y − 12 = 0

1. Divide throughout by 225: x²/25 + y²/9 = 1the right-hand side must be 1 before anything can be read off[1]
2. a = 5, b = 3, and since a > b the major axis is along the x-axis[1]
3. Vertices (±5, 0)the ends of the major axis[1]
4. c² = 25 − 9 = 16, so foci at (±4, 0)subtract for an ellipse[1]
5. e = 4/5 = 0.8less than 1 ✓[1]
6. Latus rectum = 2(9)/5 = 3.6[1]

(a) x²/25 + y²/9 = 1, a = 5, b = 3 (b) vertices (±5,0), foci (±4,0) (c) e = 0.8, latus rectum 3.6

Write it as 5x³ − 2x⁻¹. Then dy/dx = 15x² + 2x⁻², that is 15x² + 2/x².

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.

1. dy/dx = 3x² − 6x − 9[1]
2. Sets it to zero: 3(x² − 2x − 3) = 0[1]
3. Factorises: (x − 3)(x + 1) = 0, so x = 3 or x = −1[1]
4. At x = 3: y = 27 − 27 − 27 + 5 = −22[1]
5. At x = −1: y = −1 − 3 + 9 + 5 = 10[1]
6. d²y/dx² = 6x − 6; at x = 3 it is +12, a minimum[1]
7. At x = −1 it is −12, a maximum[1]

Minimum at (3, −22), maximum at (−1, 10)

1. The base is a square of side 24 − 2xa square is removed from both ends[1]
2. The height is x, so V = x(24 − 2x)²[1]
3. Expands: V = 576x − 96x² + 4x³[1]
4. dV/dx = 576 − 192x + 12x²[1]
5. Sets to zero: 12(x² − 16x + 48) = 0 → (x − 4)(x − 12) = 0[1]
6. x = 4; x = 12 is rejected because it leaves no base[1]
7. V = 4 × 16² = 4 × 256[1]
8. V = 1024 cm³[1]

x = 4 cm giving V = 1024 cm³

1. v = ds/dt = 6t² − 18t + 12[1]
2. a = dv/dt = 12t − 18[1]
3. Sets v = 0: 6(t² − 3t + 2) = 0 → (t − 1)(t − 2) = 0[1]
4. t = 1 s and t = 2 s[1]
5. At 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

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.

1. f(1) = 1 + 2 − 7 = −4Substituting the lower end and evaluating.[1]
2. f(2) = 8 + 4 − 7 = 5Substituting the upper end. Both values must be shown, not just described.[1]
3. f is continuous and changes sign from negative to positive, so a root lies between 1 and 2.The concluding statement is a mark of its own, and must mention continuity and the sign change.[1]

f(1) = −4, f(2) = 5; f continuous with a sign change, so a root lies in (1, 2)

1. x₁ = ∛(7 − 3) = ∛4 = 1.5874Correct substitution of the starting value.[1]
2. x₂ = ∛(7 − 3.1748) = ∛3.8252 = 1.5637Feeding the previous value back into the same formula.[1]
3. x₃ = 1.5708, x₄ = 1.5687Continuing until successive values agree to the required accuracy.[1]
4. x₃ and x₄ both round to 1.57, so the root is 1.57 to 2 d.p.Stating the stopping justification and rounding only at the end.[1]

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).

1. (a) Let f(x) = x − cos x. f(0.7) = 0.7 − 0.7648 = −0.0648Rewriting as f(x) = 0 first is what makes a sign change meaningful. Calculator in radians.[1]
2. f(0.8) = 0.8 − 0.6967 = +0.1033, so f is continuous with a sign change and α lies between 0.7 and 0.8.Both values plus the concluding statement.[1]
3. (b) x₁ = cos 0.7 = 0.7648, x₂ = cos 0.7648 = 0.7215The sequence alternates either side of the root — a spiral rather than a staircase, which is still convergence.[1]
4. x₃ = 0.7508, x₄ = 0.7311, x₅ = 0.7444, … x₁₆ ≈ 0.7391This one converges slowly, so several more steps are needed than usual.[1]
5. α = 0.739 to 3 d.p.Rounding only once successive values agree at that accuracy.[1]
6. (c) The iteration has diverged: this rearrangement is unsuitable even though it is algebraically valid. The student should use the rearrangement that converges, x₍ₙ₊₁₎ = cos xₙ.The mark is for recognising divergence as a property of the rearrangement, not an arithmetic mistake.[1]

(a) sign change between 0.7 and 0.8; (b) α = 0.739; (c) the iteration diverges — use x = cos x instead

Period = 360°/3 = 120°. Amplitude 4, so the range is −4 ≤ y ≤ 4.

tan x = sin x / cos x, so wherever cos x = 0 — at 90°, 270° and every 180° thereafter — the function is undefined and the graph rises or falls without limit. Those values of x are excluded from the domain.

Odd, since sin(−x) = −sin x. Its graph has rotational symmetry of order 2 about the origin.

1. Amplitude = |2| = 2[1]
2. Period = 360°/3 = 120°[1]
3. Factorising, 3x + 90° = 3(x + 30°), so the phase shift is 30° to the leftshift = −C/B = −90/3 = −30, i.e. 30° left[1]
4. Centre line y = −1, so the range is −3 ≤ y ≤ 1D ± amplitude[1]

Amplitude 2, period 120°, shift 30° left, range −3 ≤ y ≤ 1

1. sin x = 0.5[1]
2. Principal value x = 30°from the calculator or from the exact value[1]
3. Second solution 180° − 30° = 150°sine is positive in the first and second quadrants[1]
4. Both lie in range, and the next would be 390°, outside it: x = 30°, 150°a check that no further solutions fit[1]

x = 30° and x = 150°

1. Amplitude 3; period 360/2 = 180°[1]
2. Maximum = 1 + 3 = 4; minimum = 1 − 3 = −2centre line plus and minus the amplitude[1]
3. Maximum occurs when cos 2x = 1, i.e. 2x = 0°, 360°, 720°the range for 2x is 0° to 720°, which is where the extra solutions come from[1]
4. x = 0°, 180°, 360°dividing the whole solution set by 2[1]
5. Sketch shows two complete cycles across the intervalbecause the period is half the interval[1]
6. Curve oscillates between −2 and 4 about the line y = 1, starting at its maximumcosine starts at its peak[1]

(a) amplitude 3, period 180°, max 4, min −2 (b) x = 0°, 180°, 360° (c) two cycles about y = 1

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.

1. Replace sin²θ using sin²θ = 1 − cos²θ: 2(1 − cos²θ) + 3cos θ = 3Reducing to a single trigonometric function is always the first move.[1]
2. Rearrange: 2 − 2cos²θ + 3cos θ − 3 = 0, so 2cos²θ − 3cos θ + 1 = 0A quadratic in cos θ. Multiplying by −1 makes the leading coefficient positive.[1]
3. Factorise: (2cos θ − 1)(cos θ − 1) = 0, so cos θ = ½ or cos θ = 1Both roots are valid since each lies between −1 and 1.[1]
4. cos θ = ½ gives θ = 60° and 300°; cos θ = 1 gives θ = 0° and 360°Cosine is positive in the first and fourth quadrants, so the second solution is 360 − 60. Both endpoints are included by the stated interval.[1]

θ = 0°, 60°, 300°, 360°

1. (a) R sin(θ − α) = R sin θ cos α − R cos θ sin α, so R cos α = 5 and R sin α = 12The minus in the target form matches the minus in the expression, so both comparison equations are positive.[1]
2. R = √(25 + 144) = 13, and tan α = 12/5 so α = 67.4°A 5-12-13 triangle. α is in the first quadrant as required.[1]
3. So the expression is 13 sin(θ − 67.4°)Worth stating explicitly before using it in part (b).[1]
4. (b) 13 sin(θ − 67.4°) = 6.5, so sin(θ − 67.4°) = 0.5Dividing by R reduces it to a standard equation.[1]
5. Interval for θ − 67.4° is −67.4° to 292.6°. Solutions: 30° and 150°Adjusting the interval before solving is what prevents solutions being missed.[1]
6. θ = 97.4° and θ = 217.4°Adding 67.4° to each, then checking both lie in the original range.[1]
7. (c) Maximum is 13, when sin(θ − 67.4°) = 1, so θ − 67.4° = 90° and θ = 157.4°The maximum of the combined wave is R, which is the main reason for using this form.[1]

(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.

|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.

Since a · b = |a||b| cos θ and neither magnitude is zero, cos θ = 0, so θ = 90° — the vectors are perpendicular.

1. a · b = (1)(3) + (2)(−1) + (−1)(2) = 3 − 2 − 2 = −1a negative dot product means an obtuse angle[1]
2. |a| = √(1 + 4 + 1) = √6[1]
3. |b| = √(9 + 1 + 4) = √14[1]
4. 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°

1. i component: (1)(3) − (1)(−1) = 3 + 1 = 4[1]
2. j component: −[(2)(3) − (1)(1)] = −5the minus sign on j is where marks are lost[1]
3. k component: (2)(−1) − (1)(1) = −3, so a × b = 4i − 5j − 3k[1]
4. Area = 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

1. b = ⟨0, 1, 1⟩ and c = ⟨1, 0, 1⟩; i component = (1)(1) − (1)(0) = 1[1]
2. j component = −[(0)(1) − (1)(1)] = 1; k component = (0)(0) − (1)(1) = −1so b × c = i + j − k[1]
3. a · (b × c) = (1)(1) + (1)(1) + (0)(−1)a = ⟨1, 1, 0⟩[1]
4. = 2[1]
5. The triple product is non-zero, so the three vectors are NOT coplanara zero value would have meant they all lay in one plane[1]
6. Volume = |2| = 2 cubic units[1]

(a) i + j − k (b) 2 (c) not coplanar; volume 2 cubic units