2nd Year Mathematics — MCQs & Practice Questions

58 multiple-choice questions and 48 exam-style questions with mark schemes, organised by chapter, with answers you can check as you go. Free, no sign-up.

New to a topic? Read the 2nd Year Mathematics notes first, then come back to practise.

01

Functions and Limits

Multiple choice · 6

Q1For f(x) = (x² − 4)/(x − 2), the limit as x → 2 is:

  1. A0
  2. B2
  3. C4
  4. DDoes not exist
Show answer

Correct answer: C — 4

Factorise the numerator as (x−2)(x+2) and cancel to get x + 2, whose value at x = 2 is 4. The function itself is undefined at x = 2, but a limit never asks about the value at the point — only about the approach to it.

Q2The limit of f(x) as x → a exists if and only if:

  1. Af(a) is defined
  2. BThe left-hand and right-hand limits both exist and are equal
  3. Cf is continuous at a
  4. Df is differentiable at a
Show answer

Correct answer: B — The left-hand and right-hand limits both exist and are equal

Only the two-sided agreement is required. f(a) need not exist at all — the previous question is exactly that case. Continuity and differentiability are stronger conditions built on top of the limit, not requirements for it.

Q3lim(θ→0) sin θ / θ equals:

  1. A0
  2. B1
  3. C
  4. DUndefined
Show answer

Correct answer: B — 1

This is the standard limit, and it holds only when θ is in radians — the reason radians are used throughout calculus. Direct substitution gives 0/0, an indeterminate form, which is the signal that a standard result or a factorisation is needed.

Q4For f(x) = |x| / x, the limit as x → 0:

  1. AIs 0
  2. BIs 1
  3. CIs −1
  4. DDoes not exist
Show answer

Correct answer: D — Does not exist

From the right the function is constantly +1; from the left it is constantly −1. The one-sided limits disagree, so the two-sided limit does not exist. Both 1 and −1 are correct one-sided answers, which is what makes them tempting.

Q5Which condition is NOT required for f to be continuous at x = a?

  1. Af(a) must exist
  2. Blim(x→a) f(x) must exist
  3. Clim(x→a) f(x) must equal f(a)
  4. Df must be differentiable at a
Show answer

Correct answer: D — f must be differentiable at a

Differentiability is stronger than continuity, not required for it. f(x) = |x| is continuous at 0 but has no derivative there because of the corner. Every differentiable function is continuous; the converse fails.

Q6The expression 0/0 arising from direct substitution means:

  1. AThe limit is 0
  2. BThe limit is 1
  3. CThe limit does not exist
  4. DThe form is indeterminate and more work is needed
Show answer

Correct answer: D — The form is indeterminate and more work is needed

An indeterminate form is not an answer — it is an instruction to factorise, rationalise or cancel and try again. Depending on the function, 0/0 can resolve to any value at all, or to no limit.

Exam-style questions · 6

Q1[2 marks]
Explain what the statement lim(x→2) f(x) = 5 means.
Answer

As x is taken closer and closer to 2 from either side, the value of f(x) gets closer and closer to 5. It says nothing about the value of f(2) itself.

Q2[2 marks]
Evaluate lim(x→3) (x² − 9)/(x − 3).
Answer

Factorise: (x − 3)(x + 3)/(x − 3) = x + 3, so the limit is 6.

Q3[3 marks]
State the three conditions for a function to be continuous at x = a.
Answer

f(a) must exist; lim(x→a) f(x) must exist; and the two must be equal.

Q4[6 marks]
A function is defined by f(x) = (x² − 4)/(x − 2) for x ≠ 2, and f(2) = 3.
  1. Find lim(x→2) f(x). [3]
  2. State whether f is continuous at x = 2, giving a reason. [2]
  3. State the value f(2) would need for f to be continuous there. [1]
Mark scheme
  1. Factorises: x² − 4 = (x − 2)(x + 2)[1]
  2. Cancels to give x + 2 for x ≠ 2[1]
  3. Limit = 4[1]
  4. It is not continuous at x = 2[1]
  5. Because the limit is 4 but f(2) = 3, so they are not equal[1]
  6. f(2) would have to be 4a removable discontinuity[1]

(a) 4 (b) not continuous (c) 4

Q5[6 marks]
Evaluate the following limits, showing your method.
  1. lim(x→∞) (3x² + 5x)/(2x² − 1) [3]
  2. lim(x→0) (√(x + 4) − 2)/x [3]
Mark scheme
  1. Divides every term by the highest power, [1]
  2. (3 + 5/x)/(2 − 1/x²), and the fractions tend to 0[1]
  3. Limit = 3/2[1]
  4. Multiplies top and bottom by the conjugate √(x + 4) + 2[1]
  5. Numerator becomes (x + 4) − 4 = x, giving 1/(√(x + 4) + 2)[1]
  6. Limit = 1/4[1]

(a) 3/2 (b) 1/4

Q6[5 marks]
The gradient of the chord joining (2, 4) and (2 + h, (2 + h)²) on y = x² is to be investigated.
  1. Show that the gradient of the chord is 4 + h. [3]
  2. Explain what happens as h → 0 and what the result represents. [2]
Mark scheme
  1. Gradient = [(2 + h)² − 4] / h[1]
  2. Expands: (4 + 4h + h² − 4)/h = (4h + h²)/h[1]
  3. Cancels h to give 4 + hvalid because h ≠ 0[1]
  4. As h → 0 the gradient tends to 4[1]
  5. This is the gradient of the tangent at x = 2, that is, the derivative[1]

gradient → 4, the derivative of x² at x = 2

02

Differentiation

Multiple choice · 6

Q1What does f′(3) = 5 tell you about the graph of f?

  1. AThe graph passes through the point (3, 5)
  2. BAt x = 3 the curve is rising with slope 5
  3. CThe area under the curve up to x = 3 is 5
  4. DThe function equals 5 whenever x = 3
Show answer

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.

Q2Differentiate f(x) = 4x³ − 7x + 2.

  1. A12x² − 7
  2. B12x² − 7x
  3. C4x² − 7
  4. D12x³ − 7
Show answer

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.

Q3In the limit definition, why must h approach 0 rather than simply equal 0?

  1. ABecause 0 is not a real number
  2. BBecause the formula would give 0/0, which is undefined
  3. CBecause the secant line would become vertical
  4. DBecause f(x) must stay positive
Show answer

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.

Q4What is the derivative of sin(5x)?

  1. Acos(5x)
  2. B5 cos(5x)
  3. C5 sin(5x)
  4. D−5 cos(5x)
Show answer

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.

Q5A Riemann sum with n = 10 left rectangles underestimates the area under an increasing curve. What happens as n increases?

  1. AThe underestimate gets worse
  2. BThe estimate converges toward the true area
  3. CThe estimate overshoots and stays above
  4. DNothing changes; n does not matter
Show answer

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.

Q6Evaluate ∫₁³ 2x dx.

  1. A4
  2. B8
  3. C9
  4. D6
Show answer

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.

Exam-style questions · 6

Q1[2 marks]
Differentiate y = 5x³ − 2/x with respect to x.
Answer

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

Q2[3 marks]
Find the gradient of the curve y = x² − 4x + 1 at the point where x = 3.
Answer

dy/dx = 2x − 4, so at x = 3 the gradient is 2(3) − 4 = 2.

Q3[2 marks]
Explain how the second derivative distinguishes a maximum from a minimum.
Answer

At a stationary point, if d²y/dx² < 0 it is a maximum; if d²y/dx² > 0 it is a minimum.

Q4[7 marks]
A curve has equation y = x³ − 3x² − 9x + 5.
  1. Find dy/dx. [1]
  2. Find the coordinates of both stationary points. [4]
  3. Determine the nature of each. [2]
Mark scheme
  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)

Q5[8 marks]
An open-topped box is made from a square sheet of card of side 24 cm by cutting a square of side x cm from each corner and folding up the sides.
  1. Show that the volume is V = x(24 − 2x)². [2]
  2. Find the value of x that maximises the volume. [4]
  3. Calculate that maximum volume. [2]
Mark scheme
  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

Q6[5 marks]
The displacement of a particle is s = 2t³ − 9t² + 12t metres after t seconds.
  1. Find expressions for the velocity and the acceleration. [2]
  2. Find the times at which the particle is instantaneously at rest. [2]
  3. Find the acceleration at the later of those times. [1]
Mark scheme
  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

03

Integration

Multiple choice · 6

Q1∫ x³ dx equals:

  1. A3x² + C
  2. Bx⁴ + C
  3. Cx⁴/4 + C
  4. D4x⁴ + C
Show answer

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. ✓

Q2Why does the power rule for integration exclude n = −1?

  1. ANegative powers cannot be integrated
  2. BIt would require dividing by zero
  3. CThe result would be imaginary
  4. DThe integral does not exist
Show answer

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.

Q3∫₀² 3x² dx equals:

  1. A8
  2. B12
  3. C4
  4. D6
Show answer

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.

Q4The "+C" appears in indefinite integrals because:

  1. AIt makes the answer look complete
  2. BDifferentiation destroys constants, so many functions share one derivative
  3. CIt represents the area under the curve
  4. DIt cancels the dx
Show answer

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).

Q5Which technique is the reverse of the chain rule?

  1. AIntegration by parts
  2. BIntegration by substitution
  3. CPartial fractions
  4. DTrigonometric identities
Show answer

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.

Q6A velocity–time graph is integrated with respect to time. The result represents:

  1. AAcceleration
  2. BDisplacement
  3. CForce
  4. DPower
Show answer

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.

Exam-style questions · 6

Q1[2 marks]
Find ∫ (6x² − 4x + 3) dx.
Answer

2x³ − 2x² + 3x + c

Q2[3 marks]
Evaluate ∫₁³ (2x + 1) dx.
Answer

[x² + x]₁³ = (9 + 3) − (1 + 1) = 10

Q3[2 marks]
Explain what a definite integral represents geometrically.
Answer

The area between the curve and the x axis, between the two limits. Area below the axis counts as negative.

Q4[6 marks]
The curve y = x² − 4x crosses the x axis at two points.
  1. Find the two points of intersection. [2]
  2. Find the area enclosed between the curve and the x axis. [4]
Mark scheme
  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

Q5[7 marks]
A curve passes through the point (2, 9) and has gradient dy/dx = 3x² − 2x.
  1. Find the equation of the curve. [4]
  2. Find the area under the curve between x = 0 and x = 2. [3]
Mark scheme
  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

Q6[6 marks]
The region bounded by y = x + 2 and y = x² is to be found.
  1. Find the x coordinates of the points of intersection. [2]
  2. Write down the integral that gives the enclosed area. [2]
  3. Evaluate it. [2]
Mark scheme
  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

04

Introduction to Analytic Geometry

Multiple choice · 8

Q1The gradient of 6x + 3y − 9 = 0 is:

  1. A2
  2. B−2
  3. C6
  4. D−1/2
Show answer

Correct answer: B — −2

m = −a/b = −6/3 = −2. Rearranging gives y = −2x + 3, which agrees.

Q2The distance from (0, 0) to the line 3x + 4y − 10 = 0 is:

  1. A10
  2. B2
  3. C5
  4. D0.4
Show answer

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.

Q3Two lines have gradients 4 and −1/4. They are:

  1. Aparallel
  2. Bperpendicular
  3. Ccoincident
  4. Dat 45°
Show answer

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°.

Q4The line with x-intercept 2 and y-intercept 5 is:

  1. Ax/2 + y/5 = 1
  2. Bx/5 + y/2 = 1
  3. C2x + 5y = 1
  4. D5x + 2y = 1
Show answer

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 ✓.

Q5The centroid of the triangle with vertices (0,0), (6,0) and (0,9) is:

  1. A(3, 4.5)
  2. B(2, 3)
  3. C(6, 9)
  4. D(3, 3)
Show answer

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.

Q6The point dividing the segment from (0,0) to (10,0) in the ratio 3 : 2 is at:

  1. A(4, 0)
  2. B(6, 0)
  3. C(3, 0)
  4. D(5, 0)
Show answer

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.

Q7The perpendicular distance formula gives a negative value when:

  1. Athe point is below the line
  2. Bnever — the modulus makes it positive
  3. Cthe gradient is negative
  4. Dc is negative
Show answer

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.

Q8A line parallel to 2x − y + 3 = 0 through (1, 1) is:

  1. A2x − y − 1 = 0
  2. Bx + 2y − 3 = 0
  3. C2x − y + 3 = 0
  4. Dx − 2y + 1 = 0
Show answer

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.

Exam-style questions · 6

Q1[2 marks]
Find the gradient of the line 5x − 2y + 7 = 0.
Answer

Using m = −a/b = −5/(−2) = 5/2. Rearranging confirms it: 2y = 5x + 7, so y = (5/2)x + 3.5.

Q2[2 marks]
Write the equation of the line with x-intercept 4 and y-intercept −3.
Answer

Intercept form: x/4 + y/(−3) = 1, which rearranges to 3x − 4y = 12.

Q3[2 marks]
State the condition for two lines to be perpendicular, and give an example.
Answer

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.

Q4[4 marks]
Find the perpendicular distance from (−1, 4) to the line 3x + 4y − 10 = 0.
Mark scheme
  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

Q5[4 marks]
Find the acute angle between the lines y = 3x + 1 and y = −2x + 5.
Mark scheme
  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°

Q6[6 marks]
The triangle ABC has vertices A(1, 2), B(7, 4) and C(3, 8).
  1. Find the equation of the median from A.
  2. Find the coordinates of the centroid.
  3. Find the perpendicular distance from C to the line AB.
Mark scheme
  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

05

Linear Inequalities and Linear Programming

Multiple choice · 8

Q1The inequality x + y ≤ 5 is drawn with:

  1. Aa dashed line
  2. Ba solid line
  3. Cno line
  4. Dtwo lines
Show answer

Correct answer: B — a solid line

The ≤ sign includes the boundary itself, so the line is solid. A dashed line would mean the points on it are excluded.

Q2To test which side of 3x + 2y = 12 to shade, the easiest point is:

  1. A(4, 0)
  2. B(0, 6)
  3. C(0, 0)
  4. D(2, 3)
Show answer

Correct answer: C — (0, 0)

The origin makes the arithmetic trivial and the line does not pass through it. Options A and B lie on the line itself and so decide nothing.

Q3The feasible region is the set of points that:

  1. Asatisfy at least one constraint
  2. Bsatisfy every constraint
  3. Clie on the objective line
  4. Dmaximise the objective
Show answer

Correct answer: B — satisfy every constraint

Feasibility requires all constraints to hold at once; it is the intersection of the half-planes, not their union.

Q4The optimum value of a linear objective function occurs:

  1. Aat the centre of the region
  2. Bat a corner point
  3. Canywhere in the region
  4. Doutside the region
Show answer

Correct answer: B — at a corner point

The objective line sweeps across with a constant gradient, so it leaves a convex polygon at a vertex.

Q5If a problem omits x ≥ 0 and y ≥ 0, the likely consequence is:

  1. Ano change
  2. Ban unbounded region and a meaningless answer
  3. Ca smaller region
  4. Da dashed boundary
Show answer

Correct answer: B — an unbounded region and a meaningless answer

Without the non-negativity constraints the region can extend into quadrants where the variables are negative, which cannot represent a number of items produced.

Q6Corners of a region are best found by:

  1. Areading them off the graph
  2. Bsolving pairs of boundary equations
  3. Caveraging the intercepts
  4. Dtesting the origin
Show answer

Correct answer: B — solving pairs of boundary equations

A hand-drawn graph is not accurate enough for exact coordinates, and the marks are for exact values. Solving simultaneously gives them.

Q7P = 2x + 3y is evaluated at (0,0), (5,0), (3,4) and (0,6). The maximum is:

  1. A10
  2. B18
  3. C18 at two points
  4. D15
Show answer

Correct answer: C — 18 at two points

The values are 0, 10, 18 and 18. Two corners tie, which means the objective line is parallel to the edge joining them and every point on that edge is optimal.

Q8A constraint that never affects the feasible region is called:

  1. Abinding
  2. Bredundant
  3. Cinfeasible
  4. Doptimal
Show answer

Correct answer: B — redundant

A redundant constraint is automatically satisfied whenever the others are, so removing it changes nothing. Noticing one saves you drawing a line.

Exam-style questions · 6

Q1[2 marks]
Explain how to decide which side of the line 2x + 3y = 12 satisfies 2x + 3y < 12.
Answer

Test a point not on the line — the origin is easiest here. 2(0) + 3(0) = 0 < 12 is true, so the side containing the origin is the required region. The boundary is drawn dashed because the inequality is strict.

Q2[2 marks]
What is a feasible region?
Answer

The set of all points satisfying every constraint of the problem simultaneously — the overlap of all the half-planes, together with any non-negativity conditions.

Q3[2 marks]
Why is the optimum of a linear objective function always found at a corner of the feasible region?
Answer

The objective function drawn as a line keeps a fixed gradient as its value changes, so it sweeps across the region. The last point of contact before it leaves must be a vertex, because the region is a convex polygon.

Q4[4 marks]
Find the corner points of the region defined by x ≥ 0, y ≥ 0, x + y ≤ 6 and 2x + y ≤ 8.
Mark scheme
  1. Origin (0, 0) is a cornerthe two axes meet there[1]
  2. Intercepts on the axes: (4, 0) from 2x + y = 8, and (0, 6) from x + y = 6take the stricter intercept on each axis[1]
  3. The two slanted lines cross where x + y = 6 and 2x + y = 8; subtracting gives x = 2solve simultaneously rather than reading off the graph[1]
  4. y = 4, so the fourth corner is (2, 4)corners: (0,0), (4,0), (2,4), (0,6)[1]

(0, 0), (4, 0), (2, 4) and (0, 6)

Q5[4 marks]
Maximise P = 5x + 3y over the region with corners (0,0), (4,0), (2,4) and (0,6).
Mark scheme
  1. At (0, 0): P = 0[1]
  2. At (4, 0): P = 20; at (0, 6): P = 18[1]
  3. At (2, 4): P = 10 + 12 = 22[1]
  4. Maximum P = 22, occurring at (2, 4)the point must be stated as well as the value[1]

Maximum P = 22 at (2, 4)

Q6[6 marks]
A baker makes two kinds of cake. A plain cake needs 200 g of flour and 25 g of fat; a rich cake needs 100 g of flour and 50 g of fat. There are 5 kg of flour and 1 kg of fat available. The profit is Rs 20 on a plain cake and Rs 30 on a rich one.
  1. Write the constraints as inequalities.
  2. Find the corner points of the feasible region.
  3. Determine how many of each cake maximise the profit.
Mark scheme
  1. Let x = plain cakes, y = rich cakes. Flour: 200x + 100y ≤ 5000, so 2x + y ≤ 50convert kg to g before writing the inequality[1]
  2. Fat: 25x + 50y ≤ 1000, so x + 2y ≤ 40; and x ≥ 0, y ≥ 0the non-negativity conditions are worth a mark[1]
  3. Axis corners (0, 0), (25, 0) and (0, 20)from the stricter constraint on each axis[1]
  4. The two lines cross where 2x + y = 50 and x + 2y = 40: solving gives (20, 10)check: 40 + 10 = 50 ✓ and 20 + 20 = 40 ✓[1]
  5. Profit P = 20x + 30y: at (25,0) P = 500; at (0,20) P = 600; at (20,10) P = 400 + 300 = 700[1]
  6. Maximum profit Rs 700, making 20 plain and 10 rich cakesboth quantities and the profit required[1]

(c) 20 plain and 10 rich cakes, for a profit of Rs 700

06

Conic Sections

Multiple choice · 8

Q1The eccentricity of a parabola is:

  1. A0
  2. Bless than 1
  3. Cexactly 1
  4. Dgreater than 1
Show answer

Correct answer: C — exactly 1

A parabola is defined by being equidistant from the focus and the directrix, so the ratio of those distances is exactly 1.

Q2The centre of x² + y² − 4x + 10y + 4 = 0 is:

  1. A(4, −10)
  2. B(2, −5)
  3. C(−2, 5)
  4. D(−4, 10)
Show answer

Correct answer: B — (2, −5)

2g = −4 gives g = −2, and 2f = 10 gives f = 5. The centre is (−g, −f) = (2, −5).

Q3For the parabola x² = 8y, the focus is at:

  1. A(2, 0)
  2. B(0, 2)
  3. C(8, 0)
  4. D(0, 8)
Show answer

Correct answer: B — (0, 2)

Comparing with x² = 4ay gives 4a = 8, so a = 2. Since x is the squared variable, the parabola opens upward and the focus is on the y-axis at (0, 2).

Q4For the ellipse x²/9 + y²/25 = 1, the major axis is:

  1. Aalong the x-axis
  2. Balong the y-axis
  3. Cat 45°
  4. Dundefined
Show answer

Correct answer: B — along the y-axis

The larger denominator, 25, sits under y², so the ellipse is taller than it is wide and the major axis is vertical. The foci are then on the y-axis.

Q5For a hyperbola, c² equals:

  1. Aa² − b²
  2. Ba² + b²
  3. Cb² − a²
  4. Da²b²
Show answer

Correct answer: B — a² + b²

The hyperbola adds; the ellipse subtracts. That is why a hyperbola always has e > 1 — c is bigger than a.

Q6x² + y² + 2x + 2y + 10 = 0 represents:

  1. Aa circle of radius 10
  2. Ba circle of radius √8
  3. Cno real circle
  4. Da point
Show answer

Correct answer: C — no real circle

r² = g² + f² − c = 1 + 1 − 10 = −8, which is negative. No real points satisfy the equation, and saying so is the correct answer.

Q7The asymptotes of x²/4 − y²/9 = 1 are:

  1. Ay = ±(2/3)x
  2. By = ±(3/2)x
  3. Cy = ±(9/4)x
  4. Dy = ±x
Show answer

Correct answer: B — y = ±(3/2)x

y = ±(b/a)x with a = 2 and b = 3 gives y = ±(3/2)x. Note a comes from the term with the plus sign.

Q8An ellipse whose two foci coincide is:

  1. Aa parabola
  2. Ba circle
  3. Ca hyperbola
  4. Da straight line
Show answer

Correct answer: B — a circle

When c = 0 the eccentricity is 0, both foci sit at the centre, and a = b — the equation becomes x² + y² = a², a circle.

Exam-style questions · 6

Q1[2 marks]
Define the eccentricity of a conic and state its value for each of the four conics.
Answer

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.

Q2[2 marks]
Find the centre and radius of x² + y² + 8x − 6y = 0.
Answer

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

Q3[2 marks]
State the focus and directrix of the parabola y² = 12x.
Answer

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

Q4[4 marks]
Find the foci and eccentricity of the hyperbola x²/16 − y²/9 = 1.
Mark scheme
  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

Q5[4 marks]
Find the equation of the circle with centre (2, −3) that passes through the point (5, 1).
Mark scheme
  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

Q6[6 marks]
Consider the ellipse 9x² + 25y² = 225.
  1. Write the equation in standard form and state a and b.
  2. Find the coordinates of the vertices and the foci.
  3. Find the eccentricity and the length of the latus rectum, given that it equals 2b²/a.
Mark scheme
  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

07

Vectors

Multiple choice · 16

Q1Which of these is a vector quantity?

  1. AMass
  2. BTemperature
  3. CVelocity
  4. DTime
Show answer

Correct answer: C — Velocity

Velocity is speed together with a direction. Speed on its own would be a scalar, which is precisely the distinction being tested.

Q2If u = ⟨2, 3⟩ and v = ⟨5, −1⟩, then u + v is:

  1. A⟨7, 2⟩
  2. B⟨3, 4⟩
  3. C⟨10, −3⟩
  4. D⟨7, 4⟩
Show answer

Correct answer: A — ⟨7, 2⟩

Add componentwise: 2 + 5 = 7 and 3 + (−1) = 2.

Q3The magnitude of ⟨6, 8⟩ is:

  1. A14
  2. B10
  3. C48
  4. D√14
Show answer

Correct answer: B — 10

√(36 + 64) = √100 = 10. This is the 3–4–5 triple scaled by 2, which examiners use frequently.

Q4If A = (2, 7) and B = (5, 3), then AB is:

  1. A⟨7, 10⟩
  2. B⟨3, −4⟩
  3. C⟨−3, 4⟩
  4. D⟨3, 4⟩
Show answer

Correct answer: B — ⟨3, −4⟩

AB = b − a = ⟨5 − 2, 3 − 7⟩ = ⟨3, −4⟩. Option C is BA, the vector in the opposite direction.

Q5A unit vector in the direction of ⟨0, 4⟩ is:

  1. A⟨0, 4⟩
  2. B⟨0, 1⟩
  3. C⟨1, 0⟩
  4. D⟨0, 1/4⟩
Show answer

Correct answer: B — ⟨0, 1⟩

The magnitude is 4, so dividing gives ⟨0, 1⟩, which points the same way and has length 1.

Q6⟨6, −9⟩ is parallel to:

  1. A⟨2, −3⟩
  2. B⟨9, 6⟩
  3. C⟨−3, −2⟩
  4. D⟨3, 9⟩
Show answer

Correct answer: A — ⟨2, −3⟩

⟨6, −9⟩ = 3⟨2, −3⟩, a scalar multiple, so they are parallel. Option B is perpendicular to it, not parallel.

Q7If AB = DC in quadrilateral ABCD, then ABCD is:

  1. Aa rectangle
  2. Ba parallelogram
  3. Ca rhombus
  4. Da trapezium only
Show answer

Correct answer: B — a parallelogram

Equal vectors mean the sides are the same length and parallel, which defines a parallelogram. It might additionally be a rectangle or rhombus, but nothing here establishes that.

Q83⟨2, −1⟩ − 2⟨1, 4⟩ equals:

  1. A⟨4, −11⟩
  2. B⟨8, 5⟩
  3. C⟨4, 5⟩
  4. D⟨8, −11⟩
Show answer

Correct answer: A — ⟨4, −11⟩

3⟨2, −1⟩ = ⟨6, −3⟩ and 2⟨1, 4⟩ = ⟨2, 8⟩. Subtracting gives ⟨6 − 2, −3 − 8⟩ = ⟨4, −11⟩.

Q9The magnitude of 2i − 3j + 6k is:

  1. A5
  2. B7
  3. C11
  4. D49
Show answer

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.

Q10The dot product of two perpendicular vectors is:

  1. A1
  2. B0
  3. Ctheir magnitudes multiplied
  4. Dundefined
Show answer

Correct answer: B — 0

a · b = |a||b| cos 90° = 0. This is the standard test for perpendicularity.

Q11a × b is:

  1. Aa scalar
  2. Ba vector perpendicular to both
  3. Ca vector parallel to a
  4. Dalways zero
Show answer

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.

Q12If a × b = 0 for non-zero vectors, then a and b are:

  1. Aperpendicular
  2. Bparallel
  3. Cequal
  4. Dunit vectors
Show answer

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.

Q13i · j equals:

  1. A0
  2. B1
  3. Ck
  4. Di
Show answer

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.

Q14The area of the triangle formed by vectors a and b is:

  1. A|a × b|
  2. B½|a × b|
  3. Ca · b
  4. D½(a · b)
Show answer

Correct answer: B — ½|a × b|

The cross product magnitude gives the parallelogram area, and a triangle is half of that parallelogram.

Q15If a · (b × c) = 0, the three vectors are:

  1. Amutually perpendicular
  2. Bcoplanar
  3. Call zero
  4. Dunit vectors
Show answer

Correct answer: B — coplanar

The triple product is the volume of the solid they span. Zero volume means they all lie in one plane.

Q16Which is true?

  1. Aa · b = b · a and a × b = b × a
  2. Ba · b = b · a but a × b = −(b × a)
  3. Cboth anticommute
  4. Dboth are vectors
Show answer

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.

Exam-style questions · 12

Q1[2 marks]
Distinguish between a scalar and a vector, giving one example of each.
Answer

A scalar has magnitude only — for example a mass of 5 kg. A vector has both magnitude and direction — for example a displacement of 5 m due east.

Q2[2 marks]
If u = ⟨3, −4⟩, find |u| and a unit vector in the direction of u.
Answer

|u| = √(9 + 16) = 5, and the unit vector is ⟨3/5, −4/5⟩.

Q3[2 marks]
Given AB = ⟨4, −2⟩, write down BA and explain the relationship.
Answer

BA = ⟨−4, 2⟩. It is the same length but in the exactly opposite direction, so BA = −AB.

Q4[4 marks]
P is (1, 3), Q is (5, 6) and R is (9, 9). Show that P, Q and R are collinear.
Mark scheme
  1. PQ = ⟨5 − 1, 6 − 3⟩ = ⟨4, 3⟩head minus tail[1]
  2. QR = ⟨9 − 5, 9 − 6⟩ = ⟨4, 3⟩[1]
  3. PQ = QR, so PQ is a scalar multiple of QR with k = 1, hence they are parallel[1]
  4. They share the point Q, so P, Q and R lie on a single straight line and are collinearthe shared point is essential — parallel alone is not enough[1]

PQ = QR = ⟨4, 3⟩ and they share Q, so the three points are collinear (with Q the midpoint of PR).

Q5[4 marks]
Given a = ⟨2, 5⟩ and b = ⟨−3, 1⟩, find 2a − 3b and its magnitude.
Mark scheme
  1. 2a = ⟨4, 10⟩both components multiplied[1]
  2. 3b = ⟨−9, 3⟩[1]
  3. 2a − 3b = ⟨4 − (−9), 10 − 3⟩ = ⟨13, 7⟩subtracting a negative is where marks are lost[1]
  4. |2a − 3b| = √(169 + 49) = √218 ≈ 14.8exact surd or 3 s.f. both accepted[1]

⟨13, 7⟩, magnitude √218 ≈ 14.8

Q6[6 marks]
A quadrilateral has vertices A(1, 1), B(5, 2), C(6, 6) and D(2, 5).
  1. Find the vectors AB and DC.
  2. Show that ABCD is a parallelogram.
  3. Find the position vector of the point where the diagonals meet.
Mark scheme
  1. AB = ⟨5 − 1, 2 − 1⟩ = ⟨4, 1⟩[1]
  2. DC = ⟨6 − 2, 6 − 5⟩ = ⟨4, 1⟩take the vertices in the order that makes DC correspond to AB[1]
  3. AB = DC, so AB and DC are equal in length and parallel[1]
  4. One pair of opposite sides equal and parallel is sufficient for a parallelogramthe reason must be stated, not just the equality[1]
  5. The diagonals of a parallelogram bisect each other, so they meet at the midpoint of ACor the midpoint of BD, which gives the same point[1]
  6. Midpoint = (a + c)/2 = (⟨1,1⟩ + ⟨6,6⟩)/2 = ⟨3.5, 3.5⟩checking with BD: (⟨5,2⟩ + ⟨2,5⟩)/2 = ⟨3.5, 3.5⟩ ✓[1]

(a) AB = DC = ⟨4, 1⟩ (b) equal and parallel, so a parallelogram (c) ⟨3.5, 3.5⟩

Q7[2 marks]
Find the magnitude of v = 3i − 4j + 12k.
Answer

|v| = √(9 + 16 + 144) = √169 = 13.

Q8[2 marks]
State two differences between the dot product and the cross product.
Answer

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.

Q9[2 marks]
If a · b = 0 and neither vector is zero, what can you conclude?
Answer

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

Q10[4 marks]
Given a = i + 2j − k and b = 3i − j + 2k, find a · b and the angle between them.
Mark scheme
  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°

Q11[4 marks]
Find a × b for a = 2i + j + k and b = i − j + 3k, and hence the area of the triangle they form.
Mark scheme
  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

Q12[6 marks]
Three vectors are given: a = i + j, b = j + k and c = i + k.
  1. Find b × c.
  2. Hence find the scalar triple product a · (b × c).
  3. State what your answer tells you about the three vectors, and give the volume of the parallelepiped they span.
Mark scheme
  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

These questions come from the 2nd Year Mathematics lessons — each topic has its own notes, worked examples and an interactive diagram.