1st Year Mathematics — MCQs & Practice Questions

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.

The powers of i repeat every four. 103 = 4 × 25 + 3, so i¹⁰³ = i³ = −i.

|z| = √((−5)² + 12²) = √(25 + 144) = √169 = 13.

1. Multiplies numerator and denominator by the conjugate 1 + 4i[1]
2. Denominator: (1 − 4i)(1 + 4i) = 1 + 16 = 17a² + b², because −16i² = +16[1]
3. Numerator: (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = −5 + 14i[1]
4. = −5/17 + (14/17)imust be split into a + bi form[1]

−5/17 + (14/17)i

1. z₁z₂ = (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 11 + 10i−3i² = +3, which combines with the 8[1]
2. |z₁z₂| = √(121 + 100) = √221[1]
3. |z₁| = √13 and |z₂| = √17[1]
4. |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.

1. Discriminant = (−6)² − 4(1)(25) = 36 − 100 = −64[1]
2. The discriminant is negative, so there are no real rootsthe conclusion must be stated, not just the number[1]
3. x = [6 ± √(−64)] / 2 with √(−64) = 8i[1]
4. x = 3 ± 4i[1]
5. Sum = (3 + 4i) + (3 − 4i) = 6 ✓the imaginary parts cancel[1]
6. Product = (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.

r = √(1 + 1) = √2. The point is in the fourth quadrant with α = tan⁻¹(1) = 45°, so θ = −45°.

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.

1. r = √(4 + 4) = √8 = 2√2[1]
2. Acute angle α = tan⁻¹(2/2) = 45°use the absolute values to get the acute angle[1]
3. Both parts are negative, so the point is in the third quadrant and θ = 45° − 180° = −135°accept 225° if the range is not restricted[1]
4. z = 2√2 (cos(−135°) + i sin(−135°))[1]

z = 2√2 cis(−135°)

1. r = √2 and θ = 45°, so 1 + i = √2 cis 45°first quadrant, so no adjustment needed[1]
2. De Moivre: (√2)⁸ cis(8 × 45°)[1]
3. (√2)⁸ = 2⁴ = 16 and 8 × 45° = 360°[1]
4. 16(cos 360° + i sin 360°) = 16a real answer, since the argument came back to a full turn[1]

16

1. 8 = 8(cos 0° + i sin 0°)8 lies on the positive real axis, so its argument is 0°[1]
2. Roots have modulus 8^(1/3) = 2[1]
3. Arguments (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]
4. Roots: 2 cis 0°, 2 cis 120°, 2 cis 240°the first is the obvious real root 2[1]
5. The arguments differ by 120° = 360°/3, so the roots are equally spacedthey form an equilateral triangle[1]
6. All three have modulus 2, so they lie on a circle of radius 2 centred at the origin[1]

(b) 2 cis 0°, 2 cis 120°, 2 cis 240° (c) equally spaced 120° apart on a circle of radius 2

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

The discriminant is b² − 4ac. It gives the number of real roots: two if positive, one repeated if zero, none if negative.

b² − 4ac = 0 → k² − 36 = 0 → k = ±6

(3, 5), a minimum, because the coefficient of the squared term is positive.

1. Factorising: (2x − 1)(x − 3) = 0[1]
2. x = 1/2 or x = 3[1]
3. Formula: identifies a = 2, b = −7, c = 3[1]
4. x = [7 ± √(49 − 24)] / 4[1]
5. = (7 ± 5)/4[1]
6. x = 3 or x = 1/2, the same pair[1]

x = 1/2 and x = 3

1. Takes out the factor: h = −5(t² − 4t)[1]
2. Completes the square inside: t² − 4t = (t − 2)² − 4[1]
3. h = −5[(t − 2)² − 4] = 20 − 5(t − 2)²[1]
4. Greatest height = 20 mthe squared term is zero there[1]
5. At t = 2 s[1]
6. Sets h = 0: t(20 − 5t) = 0[1]
7. 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

1. Equates the two expressions: x + k = x² + 3x + 4[1]
2. Rearranges to x² + 2x + 4 − k = 0[1]
3. A tangent means one repeated root, so b² − 4ac = 0[1]
4. 4 − 4(4 − k) = 0[1]
5. 4 − 16 + 4k = 0 → k = 3[1]
6. x² + 2x + 1 = 0 → x = −1, so the point is (−1, 2)[1]

k = 3, touching at (−1, 2)

3(x² − 4) = 3(x − 2)(x + 2)

3V = πr²h → r² = 3V/(πh) → r = √(3V/(πh))

8x⁹y⁶ ÷ 4x⁴y = 2x⁵y⁵

1. Multiplies to match one coefficient, e.g. first ×3 and second ×2[1]
2. 9x + 6y = 48 and 10x − 6y = 18[1]
3. Adds to eliminate y: 19x = 66[1]
4. x = 66/19not every set of coefficients gives whole numbers[1]
5. Substitutes back into either original equation[1]
6. y = (16 − 3x)/2 = 53/19[1]

x = 66/19 ≈ 3.47, y = 53/19 ≈ 2.79

1. Length is w + 3[1]
2. w(w + 3) = 88, so w² + 3w − 88 = 0[1]
3. Factorises: (w + 11)(w − 8) = 0[1]
4. w = −11 or w = 8[1]
5. Width = 8 m[1]
6. Length = 11 m[1]
7. A width cannot be negative, so w = −11 is rejected[1]

8 m by 11 m

1. Factorises the denominator: x² − 16 = (x − 4)(x + 4)[1]
2. Cancels the common factor (x + 4)[1]
3. = 1/(x − 4)[1]
4. x = −4cancelling removed a genuine restriction[1]
5. 1/(x − 4) = 1/2 → x − 4 = 2[1]
6. x = 6[1]

(a) 1/(x − 4) (b) x = −4 (c) x = 6

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.

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.

1. Writes A/(x − 3) + B/(x + 1) and multiplies up to 7x − 1 = A(x + 1) + B(x − 3)correct shape[1]
2. Substituting x = 3: 20 = 4A, so A = 5x = 3 kills the B term[1]
3. Substituting x = −1: −8 = −4B, so B = 2[1]
4. Answer 5/(x − 3) + 2/(x + 1)accept a verification by recombining[1]

5/(x − 3) + 2/(x + 1)

1. Shape A/(x − 2) + B/(x − 2)², giving x + 4 = A(x − 2) + Ba repeated factor needs both powers[1]
2. Substituting x = 2: 6 = B[1]
3. Comparing coefficients of x: 1 = Ano substitution can isolate A here[1]
4. Answer 1/(x − 2) + 6/(x − 2)²[1]

1/(x − 2) + 6/(x − 2)²

1. Numerator and denominator are both of degree 2, so the fraction is improper and partial fractions do not apply to it directlythe reason must reference the degrees[1]
2. Dividing: 2x² + 5x + 3 = 2(x² + 2x + 1) + (x + 1), so f(x) = 2 + (x + 1)/(x + 1)²x² + 2x + 1 = (x + 1)²[1]
3. The remainder simplifies: (x + 1)/(x + 1)² = 1/(x + 1)cancelling one factor, valid for x ≠ −1[1]
4. So the proper part is already a single partial fraction, 1/(x + 1)accept working through A/(x+1) + B/(x+1)² and finding B = 0[1]
5. f(x) = 2 + 1/(x + 1)[1]
6. Valid for x ≠ −1the excluded value must be stated[1]

f(x) = 2 + 1/(x + 1), for x ≠ −1

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).

1. S₁₀ = 5[2a + 9d] = 155, so 2a + 9d = 31divide through by 5 immediately to keep the numbers small[1]
2. S₂₀ = 10[2a + 19d] = 610, so 2a + 19d = 61[1]
3. Subtracting: 10d = 30, so d = 3the 2a terms cancel[1]
4. 2a + 27 = 31, so a = 2check: S₁₀ = 5(4 + 27) = 155 ✓[1]

a = 2, d = 3

1. Writes it as 0.47 + 0.0047 + 0.000047 + …each block of two digits is one term[1]
2. Identifies a = 0.47 and r = 0.01each term is the previous one divided by 100[1]
3. |r| < 1, so S∞ = 0.47 / (1 − 0.01) = 0.47 / 0.99the condition should be stated[1]
4. = 47/99[1]

47/99

1. Heights form a GP with 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. = 2.53 m (3 s.f.)[1]
3. Total distance = the initial 8 m drop + twice each subsequent rise, since each bounce goes up and comes back downthe factor of 2 is the mark most often missed[1]
4. Rises form a GP with first term 8(0.75) = 6 and r = 0.75, so their sum is 6/(1 − 0.75) = 24[1]
5. Total = 8 + 2(24) = 56 m[1]
6. Because |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

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 = 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.

1. 10 letters in total[1]
2. Repeats: S three times, T three times, I twiceA and C appear once each[1]
3. 10! / (3! 3! 2!)one factorial for each repeated letter[1]
4. = 3 628 800 / 72 = 50 400[1]

50 400

1. Order does not matter, so combinations are useda committee is a selection[1]
2. Case 3 women: ⁴C₃ × ⁶C₂ = 4 × 15 = 60choose the women, then the men to fill the rest[1]
3. Case 4 women: ⁴C₄ × ⁶C₁ = 1 × 6 = 6there are only 4 women, so this is the last case[1]
4. Total = 60 + 6 = 66add the cases, never multiply them[1]

66

1. Six distinct people in a row: 6! = 720order matters, and nothing is repeated[1]
2. Treat Ali and Sara as one block, giving 5 items to arrange: 5! = 120the glue technique[1]
3. Ali and Sara can swap within the block: × 2! = 2the internal arrangement is easy to forget[1]
4. Together in 120 × 2 = 240 ways[1]
5. Not together = total − togethercounting the complement is far quicker than counting directly[1]
6. = 720 − 240 = 480[1]

(a) 720 (b) 240 (c) 480

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.

There are 12 + 1 = 13 terms. The third term uses r = 2, so its coefficient is ¹²C₂ = 66.

T_(r+1) = ⁿCᵣ (3x)ⁿ⁻ʳ (−2)ʳ.

1. Basis: for n = 1, 2¹ = 2 > 1, so P(1) is trueboth sides evaluated[1]
2. Assume 2ᵏ > k for some k ≥ 1the hypothesis stated as an assumption[1]
3. Then 2ᵏ⁺¹ = 2 × 2ᵏ > 2k, using the hypothesisthe hypothesis is used here and must be visible[1]
4. Since 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.

1. General term T_(r+1) = ⁷Cᵣ (2)⁷⁻ʳ xʳ[1]
2. The power of x is r, so r = 4[1]
3. ⁷C₄ = 35 and 2³ = 87 − 4 = 3, so the 2 is cubed[1]
4. Coefficient = 35 × 8 = 280[1]

280

1. T_(r+1) = ⁶Cᵣ (x²)⁶⁻ʳ (−2/x)ʳthe minus stays with the 2[1]
2. Powers of x: x¹²⁻²ʳ × x⁻ʳ = x¹²⁻³ʳ[1]
3. Setting 12 − 3r = 0 gives r = 4[1]
4. Term = ⁶C₄ (−2)⁴ = 15 × 16 = 240the power 4 is even, so the result is positive[1]
5. The expansion has 6 + 1 = 7 terms[1]
6. A constant term exists only if the equation for r has a solution that is a whole number between 0 and n; otherwise no term has power zero and there is no constant term[1]

(b) 240 (c) 7 terms; a constant term requires an integer solution for r

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.

f(2) = 8 + 8 − 10 + 1 = 7.

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.)

1. f(1) = 2 + a + b − 6 = −6, so a + b = −2remainder theorem[1]
2. f(−2) = −16 + 4a − 2b − 6 = 0, so 4a − 2b = 22 and 2a − b = 11factor theorem: the remainder is zero[1]
3. Adding the two equations eliminates b: 3a = 9, so a = 3the b terms cancel because their coefficients are +1 and −1[1]
4. b = −2 − 3 = −5check: f(1) = 2 + 3 − 5 − 6 = −6 ✓ and f(−2) = −16 + 12 + 10 − 6 = 0 ✓[1]

a = 3, b = −5

1. f(3) = 27 − 36 + 3 + 6 = 0, so (x − 3) is a factorconfirming the division will be exact[1]
2. Division gives the quotient x² − x − 2by long or synthetic division[1]
3. x² − x − 2 = (x − 2)(x + 1)[1]
4. f(x) = (x − 3)(x − 2)(x + 1)[1]

(x − 3)(x − 2)(x + 1)

1. p(2) = 8 + 4p + 2q + 12 = 0, so 4p + 2q = −20 and 2p + q = −10factor theorem[1]
2. p(−3) = −27 + 9p − 3q + 12 = 0, so 9p − 3q = 15 and 3p − q = 5[1]
3. Adding: 5p = −5, so p = −1the q terms cancel[1]
4. q = −10 − 2(−1) = −8so p(x) = x³ − x² − 8x + 12[1]
5. The product of all three roots is −12 (from −constant/leading coefficient), and two roots are 2 and −3or divide out both known factors[1]
6. Third root = −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/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° = 1/2, cos 60° = 1/2, tan 45° = 1

1 − cos²θ = sin²θ, so the expression is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ.

1. Uses the cosine rule a² = b² + c² − 2bc cos Atwo sides and the included angle[1]
2. BC² = 49 + 81 − 2(7)(9)cos 52°[1]
3. BC = 7.29 cm[1]
4. Uses ½ab sin C = ½(7)(9)sin 52°[1]
5. = 24.8 cm²[1]
6. Sine rule: sin B = 9 sin 52° / 7.29 → B = 76.6°[1]

(a) 7.29 cm (b) 24.8 cm² (c) 76.6°

1. Diagram with the mast vertical, both angles marked at ground level and PQ = 45 m[1]
2. Lets the height be h and the distance from Q be d[1]
3. From Q: h = d tan 58°[1]
4. From P: h = (d + 45) tan 34°[1]
5. Equates: d tan 58° = (d + 45) tan 34°[1]
6. Solves: d(1.600 − 0.6745) = 30.35 → d = 32.8 m[1]
7. h = 32.8 × tan 58° = 52.5 m[1]

height ≈ 52.5 m, Q is ≈ 32.8 m from the foot

1. sin θ = 1/2[1]
2. θ = 30°[1]
3. Sine is also positive in the second quadrant[1]
4. θ = 180° − 30° = 150°[1]
5. No further solutions in the range, so θ = 30° and 150°[1]

θ = 30° and 150°

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

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.

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

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

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

1. Divides every term by the highest power, x²[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

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

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

|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