NDA Applications of Derivatives previous year questions with solutions

Answer: (a) 4π⁠√3/⁠27⁠

Method. differentiate term by term; the algebraic terms cancel, leaving dy/⁠dx = sin^-1 x/⁠(1 − x²)^3/⁠2

Write y = x sin^-1 x · (1 − x²)^-1/⁠2 + 12 ln(1 − x²).
Product rule on the first term: (sin^-1 x + x⁠√(1⁠ −⁠ x²))(1 − x²)^-1/⁠2 + x sin^-1 x · x(1 − x²)^-3/⁠2
= sin^-1 x⁠√(1⁠ −⁠ x²) + x1 − x² + x² sin^-1 x(1 − x²)^3/⁠2.
The second term gives 12 · -2x1 − x² = -x1 − x², which cancels the middle term.
So dydx = sin^-1 x · (1 − x²) + x²(1 − x²)^3/⁠2 = sin^-1 x(1 − x²)^3/⁠2.
At x = 0.5: sin^-1(0.5) = π6 and (1 − 14)^3/⁠2 = (34)^3/⁠2 = 3⁠√38.⁠
Slope = π6 · 83⁠√3⁠ = 4π9⁠√3⁠ = 4π⁠√327:⁠ option (a).

Answer: (c) 1

Method. differentiate dy/⁠dx = sin^-1 x · (1 − x²)^-3/⁠2 once more

From the previous item, dydx = sin^-1 x · (1 − x²)^-3/⁠2.
d²ydx² = 1⁠√(1⁠ −⁠ x²) · (1 − x²)^-3/⁠2 + sin^-1 x · 3x(1 − x²)^-5/⁠2 = 1(1 − x²)² + 3x sin^-1 x(1 − x²)^5/⁠2.
At x = 0: 11 + 0 = 1: option (c).

Answer: (c) θ sec³ θ

Method. substitute x = sin θ in dy/⁠dx = sin^-1 x/⁠(1 − x²)^3/⁠2

From the first item of the set, dydx = sin^-1 x(1 − x²)^3/⁠2.
With x = sin θ (θ ∈ (-π/⁠2, π/⁠2), so cos θ > 0): sin^-1 x = θ and 1 − x² = cos² θ, so (1 − x²)^3/⁠2 = cos³ θ.
dydx = θcos³ θ = θ sec³ θ: option (c).

Answer: (b) (-1, 0)

Method. equation of a tangent: y − y₁ = m(x − x₁)

y = e^x ⇒ dydx = e^x; at (0, 1) the slope is e⁰ = 1.
Tangent: y − 1 = 1(x − 0), i.e. y = x + 1.
On the x-axis y = 0 ⇒ x = -1. The tangent meets the x-axis at (-1, 0).

Answer: (d) π3 < θ < π2

Method. slope of the tangent = tan θ

y = x² ⇒ dydx = 2x = 2 at (1, 1), so tan θ = 2 with 0 < θ < π2.
tan π3 = ⁠√3⁠ ≈ 1.732 < 2, and tan increases on (0, π2), so θ > π3 (θ = tan^-1 2 ≈ 63.4°).
Hence π3 < θ < π2.

Answer: (a) 4.4 cm/⁠sec

Method. related rates: C = 2πr

C = 2πr, so dCdt = 2πdrdt.
dCdt = 2 × 227 × 0.7 = 4.4 cm/⁠sec: option (a).

Answer: (c) 6.03

Method. exact change y(10.1) − y(10)

y(10) = 3(100) + 2 = 302; y(10.1) = 3(102.01) + 2 = 308.03.
Total change Δy = 308.03 − 302 = 6.03: option (c).
(The differential dy = 6x dx = 6 × 10 × 0.1 = 6 is only the approximation; the total change is 6.03.)

Answer: (a) 1 and 2 only

Method. meaning of the derivative and the integral in geometry and kinematics

(1) dy/⁠dx at a point is the slope of the tangent there: correct.
(2) Velocity v satisfies dvdt = a(t), so v = ∫a(t)dt + c: correct.
(3) dsdt is the velocity, not the acceleration (acceleration is d²sdt²): wrong.
1 and 2 only: option (a).

Answer: (a) (-∞, 0)

Method. f'(x) is the integrand xe^x; f decreases where it is negative

f(x) = ∫ xe^x dx, so f'(x) = xe^x.
e^x > 0 for all x, so the sign of f'(x) is the sign of x.
f'(x) < 0 for x < 0 and f'(x) > 0 for x > 0, so f decreases on (-∞, 0) and increases on (0, ∞).
The interval in which f decreases is (-∞, 0): option (a). ((-∞, -1) is only part of it, and (-1, ∞) contains points where f increases.)

Answer: (c) π2

Method. write f as cos 2x and read off the length of one decreasing stretch

f(x) = 2cos²x − 1 = cos 2x, so f'(x) = -2 sin 2x.
f is decreasing where sin 2x ≥ 0, i.e. 2kπ ≤ 2x ≤ 2kπ + π, i.e. kπ ≤ x ≤ kπ + π2 (k an integer).
On the adjacent stretches [kπ + π2, (k + 1)π] f is increasing, so no interval of decrease is longer than these.
Longest interval of decrease has length π2: option (c).

Shortcut. cos 2x has period π and falls for half of each period: π/⁠2.

Answer: (b) {x < 2} ∪ {x > 3}

Method. sign of f'(x) = (x − 2)(x − 3); f increases at a point where f' > 0

f'(x) = x² - 5x + 6 = (x − 2)(x − 3).
f'(x) > 0 for x < 2 and for x > 3; f'(x) < 0 for 2 < x < 3.
At x = 2, f' changes from + to -, so f has a local maximum there; at x = 3, f' changes from − to +, a local minimum.
At a turning point f is neither increasing nor decreasing (no open interval around 2 or 3 on which f is monotonic), so 2 and 3 belong to neither S nor T.
T = {x < 2} ∪ {x > 3}: option (b).

The sets are read pointwise: f increases at x₀ only if it increases on an open interval about x₀, so the turning points 2 and 3 are excluded. The closed-interval form (-∞, 2] ∪ [3, ∞) describes intervals on which f increases, not the set of points where it increases.

Answer: (c) (2, 3)

Method. sign of f'(x) = (x − 2)(x − 3); f decreases at a point where f' < 0

From the previous item, f'(x) = (x − 2)(x − 3) < 0 exactly for 2 < x < 3.
x = 2 (local maximum) and x = 3 (local minimum) are turning points, where f is neither increasing nor decreasing, so they are excluded.
S = (2, 3): option (c).

The sets are read pointwise: f increases at x₀ only if it increases on an open interval about x₀, so the turning points 2 and 3 are excluded. The closed-interval form (-∞, 2] ∪ [3, ∞) describes intervals on which f increases, not the set of points where it increases.

Answer: (c) Statement-I is correct but Statement-II is not correct

Method. sign of f'(x) for f(x) = 1/⁠(x − 5)

f'(x) = -1(x − 5)².
(x − 5)² > 0 for every x ≠ 5, so f'(x) < 0 for all x ≠ 5.
Hence f is decreasing on x < 5 and on x > 5 (each interval separately). Statement-I is correct.
Statement-II says f'(x) > 0, which is false. Option (c).

Answer: (d) 1, 2 and 3

Method. sign of f'(x) for f(x) = x/⁠ln x, then compare f(9) and f(7)

f'(x) = ln x · 1 − x · 1x(ln x)² = ln x − 1(ln x)².
The denominator is positive for x > 1, so the sign of f'(x) is the sign of ln x − 1.
For x > e, ln x > 1, so f'(x) > 0: f is increasing on (e, ∞). Statement 1 is correct.
For 1 < x < e, 0 < ln x < 1, so f'(x) < 0: f is decreasing on (1, e). Statement 2 is correct.
7 and 9 both lie in (e, ∞) and 9 > 7, so f(9) > f(7): 9ln 9 > 7ln 7.
Multiplying by the positive number ln 7 · ln 9 gives 9 ln 7 > 7 ln 9. Statement 3 is correct.
All three are correct: option (d).

Answer: (d) 1, 2 and 3

Method. second-derivative test for f(x) = x/⁠ln x at x = e

Write f'(x) = ln x − 1(ln x)² = 1ln x − 1(ln x)².
f''(x) = -1x(ln x)² + 2x(ln x)³.
At x = e, ln e = 1: f''(e) = -1e + 2e = 1e. Statement 1 is correct.
f'(e) = 1 − 11 = 0 and f''(e) = 1e > 0, so f has a local minimum at x = e. Statement 2 is correct.
The local minimum value is f(e) = eln e = e. Statement 3 is correct.
All three are correct: option (d).

Answer: (a) [-283, 0]

Method. expand the determinant, then f'(x) ≤ 0

Expanding (or noting the matrix is xI + vv^T with v = (1, 2, 3)): f(x) = x³ + 14x².
Check at x = 1:  2 2 3 2 5 6 3 6 10  = 2(50 − 36) − 2(20 − 18) + 3(12 − 15) = 28 − 4 − 9 = 15 = 1 + 14. (holds)
f'(x) = 3x² + 28x = x(3x + 28) ≤ 0 ⇔ -283 ≤ x ≤ 0.
So f is decreasing on [-283, 0].

Shortcut. The matrix is xI + vv^T with v^Tv = 1 + 4 + 9 = 14, so its eigenvalues are x, x, x + 14 and f(x) = x²(x + 14) at once.

Answer: (c) x = 0

Method. second-derivative test

From the previous item, f(x) = x³ + 14x², f'(x) = x(3x + 28) = 0 ⇒ x = 0 or x = -283.
f''(x) = 6x + 28: f''(0) = 28 > 0 (local minimum); f''(-283) = -28 < 0 (local maximum).
The local minimum is at x = 0.

Answer: (a) 1 only

Method. sign of the first derivative

1. f(x) = ln x ⇒ f'(x) = 1x > 0 on (0, ∞), so f is increasing. Correct.
2. g'(x) = e^x + e^1/⁠x·(-1x²) = e^x - e^1/⁠xx².
At x = 2: g'(2) = e² - e^1/⁠24 ≈ 7.39 − 0.41 > 0, so g increases near x = 2 (g' < 0 only for 0 < x < 1). g is not decreasing on (0, ∞). Incorrect.
Only statement 1 is correct.

Answer: (b) (1, 6)

Method. f decreases where f'(x) < 0

f'(x) = x² - 7x + 6 = (x − 1)(x − 6).
The product is negative exactly when 1 < x < 6, and positive for x < 1 or x > 6.
So f decreases on (1, 6) (and increases on (-∞, 1) and (6, ∞)).

No correct option. This question is defective as printed in UPSC's paper.

Method. increasing on an interval ⇔ f'(x) ≥ 0 there

f'(x) = 2x − k. f increases on (1, ∞) iff 2x − k ≥ 0 for all x > 1, i.e. k ≤ 2x for all x > 1, i.e. k ≤ 2.
Boundary k = 2: f'(x) = 2(x − 1) > 0 for every x > 1, so f is (strictly) increasing on (1, ∞).
So the given condition implies k ≤ 2, and k = 2 is possible; then none of k < 2, 2 < k < 3, 3 < k < 4, k > 4 holds.
No option is necessarily correct.

The function is increasing on (1, ∞) exactly when k ≤ 2, and k = 2 itself works (f'(x) = 2(x − 1) > 0 for x > 1), so the printed k < 2 is not implied; reading the answer as k ≤ 2 yields option (a).

Answer: (b) 2 only

Method. sign of f'(x) = cos x on each interval

f'(x) = cos x.
1. On (0, π), cos x > 0 only for 0 < x < π2; cos x < 0 on (π2, π), where sin x decreases. So sin x does not increase on all of (0, π). Incorrect.
2. For 5π2 < x < 3π, write x = 2π + t with π2 < t < π; then cos x = cos t < 0, so sin x decreases there. Correct.
Only statement 2 is correct.

Answer: (c) Both 1 and 2

Method. sign of the first derivative

1. f(x) = ln x ⇒ f'(x) = 1x > 0 for every x > 0, so ln x increases on (0, ∞). Correct.
2. f(x) = tan x ⇒ f'(x) = sec² x > 0 for every x in (-π2, π2), so tan x increases there. Correct.
Both statements are correct.

Answer: (a) (-2, 3)

Method. factorise f'(x) and read its sign

f'(x) = 12x³ - 60x² - 24x + 288 = 12(x³ - 5x² - 2x + 24).
x = -2 is a root (-8 − 20 + 4 + 24 = 0), so f'(x) = 12(x + 2)(x² - 7x + 12) = 12(x + 2)(x − 3)(x − 4).
Sign of f': negative on (-∞, -2), positive on (-2, 3), negative on (3, 4), positive on (4, ∞).
f increases on (-2, 3): option (a). ((-3, -2) and (-4, -3) lie where f' < 0; (3, 4) also.)

Answer: (b) (3, 4)

Method. sign of the same factorised derivative

From the previous item, f'(x) = 12(x + 2)(x − 3)(x − 4).
f' < 0 exactly on (-∞, -2) and (3, 4); f' > 0 on (-2, 3) and on (4, ∞), which contains (4, 6) and (6, 9).
f decreases on (3, 4): option (b).

Answer: (a) T = (-∞, 2) ∪ (3, ∞) and S = (2, 3)

Method. sign of f'(x) for the cubic

f'(x) = x² - 5x + 6 = (x − 2)(x − 3).
f' > 0 for x < 2 or x > 3; f' < 0 for 2 < x < 3.
T = (-∞, 2) ∪ (3, ∞), S = (2, 3): option (a).

Answer: (a) It is increasing in the interval (0, π2)

Method. sign of f'(x) on (0, π/⁠2)

f'(x) = sin x + x cos x − sin x − cos x sin x = cos x(x − sin x).
On (0, π2): cos x > 0, and x − sin x > 0 (since x > sin x for x > 0).
So f'(x) > 0 throughout (0, π2): f is increasing there, option (a).
The same holds on (π4, π2), so (d) is false.

Answer: (a) (-∞, 2]

Method. sign of f'(x) for a quadratic

f'(x) = 2x − 5 < 0 for x < 52, so f decreases on (-∞, 52] and increases on [52, ∞).
(-∞, 2] lies inside (-∞, 52]: f decreases there.
[3, ∞) is in the increasing part; (2, 3) and (-∞, ∞) contain 52, where f turns.
Option (a).

Answer: (a) Both Statement I and Statement II are true and Statement II is the correct explanation of Statement I

Method. monotonic function compared with its value at 0

f(x) = x − sin x, f'(x) = 1 − cos x ≥ 0, with equality only at isolated points x = 2nπ.
So f is increasing for x > 0: Statement II is true.
f(0) = 0 and f is increasing, so f(x) > f(0) = 0 for x > 0, i.e. x > sin x: Statement I is true.
Statement I follows directly from Statement II: option (a).

Answer: (a) π3

Method. triple-angle identity, then sign of the derivative

f(x) = 3 sin x − 4 sin³ x = sin 3x.
f'(x) = 3 cos 3x > 0 when -π2 + 2nπ < 3x < π2 + 2nπ.
So f increases on intervals -π6 + 2nπ3 < x < π6 + 2nπ3, and decreases between them.
Each such interval has length π6 − (-π6) = π3.
The longest interval of increase has length π3: option (a).

Shortcut. sin 3x has period 2π3; it rises for half a period, π3.

Answer: (c) f(x) decreases in the interval

Method. sign of f'(x) on the interval

f'(x) = 1 − 1x² = x² - 1x².
For 0 < x < 1, x² - 1 < 0, so f'(x) < 0 throughout (0, 1).
Hence f decreases in the interval: option (c).

Answer: (c) minimum occurs at x = -1/⁠2

Method. first- and second-derivative tests for y = xe^2x

dydx = e^2x + 2xe^2x = e^2x(1 + 2x) = 0 gives x = -12 (e^2x is never 0).
d²ydx² = 2e^2x(1 + 2x) + 2e^2x = e^2x(4x + 4).
At x = -12: d²ydx² = e^-1 · 2 = 2e > 0, so it is a minimum.
The minimum occurs at x = -12: option (c).

Shortcut. y' = e^2x(1 + 2x) goes from − to + through x = -1/⁠2, so a minimum there.

Answer: (c) Statement-I is correct but Statement-II is not correct

Method. first- and second-derivative tests for f(x) = x² + 128/⁠x

f(x) = x³ + 128x = x² + 128x, x ≠ 0.
f'(x) = 2x − 128x² = 2(x³ - 64)x² = 0 gives x = 4.
f''(x) = 2 + 256x³, so f''(4) = 6 > 0: a (local) minimum at x = 4, with value f(4) = 16 + 32 = 48. Statement-I is correct.
Sign of f'(x) near 4 is the sign of x³ - 64: negative for x slightly less than 4, positive for x slightly more.
So f'(x) changes from negative to positive (that is what makes x = 4 a minimum); Statement-II, which says positive to negative, is wrong.
Option (c).

Statement-I means the local minimum; f has no least value (f → -∞ as x → 0⁻). Since Statement-II is false in any case and no option rejects both, (c) is the only consistent answer.

No correct option. This question is defective as printed in UPSC's paper.

Method. no extremum in x > 0 ⇔ f'(x) does not change sign for x > 0

f'(x) = 16x² - 8bx + 1, a quadratic with roots of product 116 > 0 and sum b2 ≥ 0.
If b > 1, the discriminant 64b² - 64 > 0: two distinct positive roots, f' changes sign at each, so f has a maximum and a minimum in x > 0.
If 0 ≤ b < 1, the discriminant is negative: f'(x) > 0 for all x, no extremum.
If b = 1, f'(x) = 16x² - 8x + 1 = (4x − 1)² ≥ 0, zero only at x = 14 with no sign change, so again no extremum.
The required set is 0 ≤ b ≤ 1, which no option prints.

The full set of non-negative b giving neither maximum nor minimum is 0 ≤ b ≤ 1 (at b = 1, f'(x) = (4x − 1)² does not change sign), but option (d) prints 0 ≤ b < 1 and omits b = 1, so no option is the required set; the intended reading (discriminant < 0, boundary ignored) yields option (d).

Answer: (a) 1 only

Method. first-derivative condition at an extremum (Fermat), and why the second-derivative test is not necessary

f is differentiable and has a local maximum at x = 0, so the tangent there is horizontal: f'(0) = 0. Statement 1 is correct.
Statement 2 does not follow. The second-derivative test runs one way only: f'(0) = 0 with f''(0) < 0 is enough for a maximum, but a maximum does not force f''(0) < 0.
Counter-example: f(x) = -x⁴ has a local maximum at x = 0 (f(x) ≤ 0 = f(0) for all x), yet f''(x) = -12x² gives f''(0) = 0, not < 0.
Besides, f is only given to be differentiable, so f''(0) need not even exist.
Only statement 1 is necessarily correct: option (a).

The second-derivative test is a sufficient condition, not a necessary one: f(x) = -x⁴ has a local maximum at 0 with f''(0) = 0, so statement 2 need not hold and (c) is not the answer.

Answer: (b) for all b and a = 0

Method. a differentiable function has y'(0) = 0 at a local maximum; the constant b only shifts the graph

y = 2f(x) + ax − b, so y'(x) = 2f'(x) + a.
From the previous item, f'(0) = 0, so y'(0) = a.
y is differentiable, so a relative maximum of y at x = 0 needs y'(0) = 0, i.e. a = 0. Hence (a) and (d), which allow a ≠ 0, fail.
With a = 0, y = 2f(x) − b. Since f(x) ≤ f(0) near 0, 2f(x) − b ≤ 2f(0) − b = y(0) near 0: y has a relative maximum at 0 whatever b is.
So the condition is a = 0 with b arbitrary; (c) is wrong because b need not be positive and a must be 0. Option (b).

Answer: (b) 2 only

Method. read the graph of |x| + 1 on [-3, 3]

For 0 < |x| ≤ 3, f(x) = |x| + 1, and f(0) = 1 = |0| + 1, so f(x) = |x| + 1 on the whole domain [-3, 3].
1. The largest value is 4, attained at x = 3 and also at x = -3. So the maximum is not attained only at x = 3. Incorrect.
2. f(x) > 1 = f(0) for all x ≠ 0, so x = 0 is a local minimum; f decreases on [-3, 0] and increases on [0, 3], so there is no other local minimum. Correct.
Only statement 2 is correct.

Answer: (b) 2

Method. critical points with a sign change of f'

f(x) = sin 4x + 2x ⇒ f'(x) = 4 cos 4x + 2.
f'(x) = 0 ⇒ cos 4x = -12. For 0 < x < π2, 0 < 4x < 2π, so 4x = 2π3 or 4π3, i.e. x = π6 or π3.
f''(x) = -16 sin 4x: f''(π6) = -16 sin 2π3 < 0 (local maximum), f''(π3) = -16 sin 4π3 > 0 (local minimum).
So f has 2 extreme values in the interval.

Answer: (a) 1 only

Method. first- and second-derivative tests

f'(x) = 1 − 1x² = 0 ⇒ x = ±1.
f''(x) = 2x³: f''(1) = 2 > 0 (local minimum), f''(-1) = -2 < 0 (local maximum).
Local minimum value f(1) = 2; local maximum value f(-1) = -2.
1. -2 < 2, so the local maximum value is less than the local minimum value. Correct.
2. The local maximum occurs at x = -1, not at x = 1. Incorrect.
Only statement 1 is correct.

No correct option. This question is defective as printed in UPSC's paper.

Method. no extremum ⇔ f'(x) never changes sign ⇔ discriminant of f'(x) ≤ 0

f'(x) = 3x² + 2x + k.
If the discriminant 4 − 12k > 0, f' has two distinct real roots and changes sign at each, giving a local maximum and a local minimum.
If 4 − 12k ≤ 0, i.e. 3k ≥ 1, then f'(x) ≥ 0 for all x and never changes sign, so there is no local extremum.
Boundary 3k = 1: f'(x) = 3x² + 2x + 13 = 3(x + 13)² ≥ 0, zero only at x = -13 with no sign change, so no extremum there either.
The exact condition is 3k ≥ 1, which no option prints.

The exact condition for no local extremum is 3k ≥ 1 (at 3k = 1, f'(x) = 3(x + 13)² does not change sign), but option (b) prints 3k > 1 and omits 3k = 1, so no option states the condition; the intended reading (discriminant < 0 with the boundary ignored) yields option (b).

Answer: (c) x = 1

Method. maximise the slope function dy/⁠dx

Slope m(x) = dydx = -3x² + 6x + 2.
dmdx = -6x + 6 = 0 ⇒ x = 1.
d²mdx² = -6 < 0, so the slope is greatest at x = 1 (where m = 5).

Shortcut. m(x) = -3(x − 1)² + 5 is greatest at x = 1.

Answer: (a) The local maximum value is larger than local minimum value

Method. local extrema of a cubic from f'(x) = 0

f(x) = x³ - x, f'(x) = 3x² - 1 = 0 gives x = ±1⁠√3;⁠ f''(x) = 6x.
x = -1⁠√3:⁠ f'' < 0, local maximum f = -13⁠√3⁠ + 1⁠√3⁠ = 23⁠√3.⁠
x = 1⁠√3:⁠ f'' > 0, local minimum f = 13⁠√3⁠ − 1⁠√3⁠ = -23⁠√3.⁠
23⁠√3⁠ > -23⁠√3:⁠ the local maximum value is larger, option (a).

Answer: (b) 1e

Method. first-derivative test for ln x / x

f(x) = ln xx, x > 0. f'(x) = 1 − ln xx².
f'(x) = 0 at ln x = 1, i.e. x = e; f' > 0 for x < e and f' < 0 for x > e, so x = e gives the maximum.
f(e) = 1e: option (b).

Answer: (d) Neither I nor II

Method. an increasing function on an open interval attains neither a maximum nor a minimum

f(x) = x is strictly increasing, and the interval (-1, 1) excludes its endpoints.
For any x in (-1, 1), the point x + 12 also lies in (-1, 1) and is larger than x, so no value is the greatest.
Similarly x − 12 lies in (-1, 1) and is smaller than x, so no value is the least.
The values approach 1 and -1 but never reach them. Neither statement is correct: option (d).

Answer: (b) ⁠√32⁠ + π12

Method. closed-interval method: compare f at critical points and endpoints

f'(x) = -2 sin 2x + 1 = 0 gives sin 2x = 12.
For x ∈ [-π/⁠2, π/⁠2], 2x ∈ [-π, π], so 2x = π6 or 5π6, i.e. x = π12 or 5π12.
f(π12) = cos π6 + π12 = ⁠√32⁠ + π12 ≈ 1.128
f(5π12) = cos 5π6 + 5π12 = -⁠√32⁠ + 5π12 ≈ 0.443
f(π2) = cos π + π2 = -1 + π2 ≈ 0.571; f(-π2) = cos(-π) − π2 = -1 − π2 ≈ -2.571
The greatest of these is ⁠√32⁠ + π12: option (b).

Answer: (a) -(1 + π2)

Method. closed-interval method: compare f at critical points and endpoints

From the previous item, the candidates are x = π12, 5π12 (where f'(x) = 1 − 2 sin 2x = 0) and the endpoints ±π2.
Values: ⁠√32⁠ + π12 ≈ 1.128, -⁠√32⁠ + 5π12 ≈ 0.443, -1 + π2 ≈ 0.571, -1 − π2 ≈ -2.571.
The least is f(-π2) = cos(-π) − π2 = -1 − π2 = -(1 + π2): option (a).

Answer: (a) tan² x = qp

Method. write f in terms of t = tan² x and minimise

sec² x = 1 + tan² x and cosec² x = 1 + cot² x. With t = tan² x > 0 (taking p, q > 0):
f = p²(1 + t) + q²(1 + 1t) = p² + q² + p²t + q²t.
dfdt = p² - q²t² = 0 ⇒ t² = q²p² ⇒ t = qp; d²fdt² = 2q²t³ > 0, so this is a minimum.
The minimum occurs when tan² x = qp (minimum value (p + q)²).

Shortcut. AM ≥ GM: p²t + q²t ≥ 2pq, with equality when p²t = q²t, i.e. t = qp.

Answer: (b) x = 4

Method. sum of squared deviations is least at the mean

f'(x) = 7Σj=1 2(x − j) = 2(7x − (1 + 2 + … + 7)) = 2(7x − 28).
f'(x) = 0 ⇒ x = 4, and f''(x) = 14 > 0, so x = 4 gives the minimum.

Shortcut. Σ(x − j)² is least at the mean of 1, 2, …, 7, which is 4.

Answer: (b) 12

Method. simplify to a sine of a double angle

tan x + cot x = sin xcos x + cos xsin x = sin² x + cos² xsin x cos x = 1sin x cos x.
So f(x) = sin x cos x = 12 sin 2x.
For 0 < x < π2, sin 2x ≤ 1 with equality at x = π4, so the maximum value is 12.

Shortcut. AM ≥ GM: tan x + cot x ≥ 2 for 0 < x < π2, so f(x) ≤ 12.

Answer: (b) π3

Method. set the derivative to zero and solve a quadratic in cos θ

f(θ) = sin θ + sin θ cos θ = sin θ + 12 sin 2θ.
f'(θ) = cos θ + cos 2θ = 2cos² θ + cos θ − 1 = (2cos θ − 1)(cos θ + 1).
In 0 < θ < π2, cos θ + 1 > 0, so f'(θ) = 0 ⇒ cos θ = 12 ⇒ θ = π3.
f'(θ) > 0 for θ < π3 (cos θ > 12) and f'(θ) < 0 for θ > π3, so θ = π3 gives the maximum.

Answer: (d) Neither 1 nor 2

Method. an increasing function on an open interval attains neither end value

f'(x) = 2x > 0 on (1, 2), so f increases strictly there.
As x → 1⁺, f(x) → 2 and as x → 2^-, f(x) → 5, but x = 1 and x = 2 are not in the open interval (1, 2).
So f takes every value strictly between 2 and 5 and never 2 or 5: it has no maximum and no minimum on (1, 2).
Neither statement is correct.

Answer: (a) 100

Method. maximum of a product with fixed sum

P = x(20 − x) = 20x − x².
dPdx = 20 − 2x = 0 ⇒ x = 10, and d²Pdx² = -2 < 0, so this is a maximum.
y = 10, P_max = 10 × 10 = 100.

Shortcut. AM ≥ GM: xy ≤ (x + y2)² = 100.

Answer: (b) 2abc

Method. eliminate y and minimise, or AM–⁠GM

Take a, b, c > 0 and x, y > 0, as intended. y = c²x, so f(x) = a²x + b²c²x.
f'(x) = a² - b²c²x² = 0 at x = bca; f''(x) = 2b²c²x³ > 0, a minimum.
f(bca) = abc + abc = 2abc: option (b).

Shortcut. AM–⁠GM: a²x + b²y ≥ 2⁠√(a²b²xy)⁠ = 2abc.

Answer: (a) (34)^1/⁠3

Method. minimise the inside quadratic, since the cube root is increasing

x(x − 1) + 1 = x² - x + 1 = (x − 12)² + 34.
On [0, 1] its least value is 34, at x = 12.
t ↦ t^1/⁠3 is increasing, so the minimum of the expression is (34)^1/⁠3: option (a).

Answer: (c) 16/⁠3

Method. maximum of a quadratic in sin θ over [-1, 1]

Let t = sin θ, -1 ≤ t ≤ 1. g(t) = 16t − 12t².
g'(t) = 16 − 24t = 0 at t = 23, which lies in [-1, 1]; g'' = -24 < 0, a maximum.
g(23) = 323 − 12 × 49 = 323 − 163 = 163: option (c).

Answer: (c) 16 square units

Method. maximise ½ab subject to a² + b² = 64

Let the legs be AB = a and BC = b. Then a² + b² = AC² = 64, and the area is Δ = 12ab.
Since (a − b)² ≥ 0, 2ab ≤ a² + b² = 64, so ab ≤ 32 and Δ ≤ 16.
Equality holds when a = b = 4⁠√2,⁠ the isosceles right triangle.
Maximum area = 16 square units: option (c).

Answer: (a) I only

Method. for a fixed perimeter the area l(10 − l) is greatest for the square

Perimeter 2(l + w) = 20, so w = 10 − l and area A = l(10 − l) = 10l − l², 0 < l < 10.
A′(l) = 10 − 2l = 0 gives l = 5, and A′'(l) = -2 < 0, so the maximum is at l = w = 5: the square. Statement I is correct.
The largest possible area is 5 × 5 = 25 cm² < 27 cm², so an area of 27 cm² cannot be formed. Statement II is wrong.
Option (a).

Shortcut. A = 25 − (l − 5)² ≤ 25, with equality only for the square.

Answer: (c) π/⁠3

Method. express the area in one variable (the hypotenuse) and maximise its square

Let AC = h (the hypotenuse, since ∠B = 90°). Then AB = 3 − h and BC² = h² - (3 − h)² = 6h − 9, with 32 < h < 3.
Area Δ = 12 · AB · BC = 12(3 − h)⁠√(6h⁠ −⁠ 9).
Maximise Δ² = 34(3 − h)²(2h − 3). Put g(h) = (3 − h)²(2h − 3):
g'(h) = -2(3 − h)(2h − 3) + 2(3 − h)² = 2(3 − h)(6 − 3h) = 6(3 − h)(2 − h).
For 3/⁠2 < h < 3, g' > 0 when h < 2 and g' < 0 when h > 2, so the area is greatest at h = 2.
Then AC = 2, AB = 1, and cos A = ABAC = 12, so ∠A = π3: option (c).

Shortcut. With AC = h and AB = h cos A, h(1 + cos A) = 3 and area = 92·sin A cos A(1 + cos A)²; checking the options, π/⁠3 gives ⁠√3/⁠2,⁠ the largest.

Answer: (a) ⁠√3/⁠2⁠ square unit

Method. area at the optimum found in the previous item

From the previous item the area is greatest when AC = 2 and AB = 3 − 2 = 1.
BC = ⁠√(AC²⁠ - AB²) = ⁠√(4⁠ −⁠ 1) = ⁠√3.⁠
Maximum area = 12 · AB · BC = 12 · 1 · ⁠√3⁠ = ⁠√32⁠ square unit: option (a).

Answer: (c) 8 square units

Method. rectangle in a circle: diagonal = diameter

The diagonal of the inscribed rectangle is a diameter, 4 units. Let the sides be 4 cos θ and 4 sin θ.
Area A = 16 sin θ cos θ = 8 sin 2θ, greatest when sin 2θ = 1 (θ = π4, a square).
A_max = 8 square units.

Shortcut. The largest rectangle in a circle is a square with diagonal 2r: area = (2r)²2 = 162 = 8.

Answer: (a) 4⁠√3⁠ cm

Method. fixed perimeter and fixed angle: area = r·s with r = (s − a) tan(A/⁠2)

Let A = 60° with opposite side a and the other sides b, c; a + b + c = 24, so s = 12.
Area = rs and r = (s − a) tan 30°, so Area = 12(12 − a)·1⁠√3:⁠ the area is greatest when a is least.
Cosine rule: a² = b² + c² - bc = (b + c)² - 3bc ≥ (b + c)² - 34(b + c)² = (b + c)²4, with equality iff b = c.
So a ≥ b + c2 = 24 − a2 ⇒ a ≥ 8, least when b = c = a = 8.
The best triangle is equilateral with side 8 cm; its altitude = ⁠√32·8⁠ = 4⁠√3⁠ cm.

Shortcut. With the angle fixed, symmetry (b = c) gives the maximum, making the triangle equilateral of side 8.

Answer: (d) k²4

Method. maximum of a product with fixed sum

Area A = lb = l(k − l).
dAdl = k − 2l = 0 ⇒ l = k2, and d²Adl² = -2 < 0, so this is a maximum.
Then b = k − k2 = k2, so A_max = k2·k2 = k²4.

Shortcut. AM ≥ GM: lb ≤ (l + b2)² = k²4.

Answer: (a) π : (π + 2)

Method. minimise total surface area of a half cylinder of fixed volume

Radius r, height (length) h. Volume V = 12πr²h is fixed, so h = 2Vπr².
Total surface = curved half πrh + rectangular base 2rh + two semicircular ends πr²:
S = (π + 2)rh + πr² = 2(π + 2)Vπr + πr².
dSdr = -2(π + 2)Vπr² + 2πr = -(π + 2)h + 2πr = 0, so (π + 2)h = π(2r); S′' > 0, a minimum.
h2r = ππ + 2: height: diameter = π : (π + 2), option (a).

Answer: (b) 2⁠√2⁠

Method. minimise distance using the parametric point (2t², 4t)

A point on y² = 8x is (2t², 4t).
D(t) = (2t² - 4)² + (4t − 2)² (square of the distance).
D′(t) = 2(2t² - 4)(4t) + 2(4t − 2)(4) = 16t³ - 32t + 32t − 16 = 16(t³ - 1).
D′ = 0 only at t = 1 (D′ < 0 before, > 0 after), giving the point (2, 4).
Minimum distance = ⁠√((2⁠ −⁠ 4)² + (4 −⁠ 2)²) = ⁠√8⁠ = 2⁠√2:⁠ option (b).

Answer: (c) 10 m

Method. maximise sector area with fixed perimeter

Radius r, angle θ (radians). Wire = 2r + rθ = 40, so rθ = 40 − 2r.
Area A = 12r²θ = 12r(40 − 2r) = 20r − r².
dAdr = 20 − 2r = 0 at r = 10; d²Adr² = -2 < 0, a maximum.
(θ = 2 rad, a valid sector.) Radius 10 m: option (c).

Answer: (c) π3

Method. maximise the area of an isosceles trapezium as a function of the base angle

The trapezium has the shorter parallel side and the two legs equal to 6; α is the acute angle between a leg and the longer parallel side.
Height = 6 sin α; longer parallel side = 6 + 2(6 cos α) = 6 + 12 cos α.
Area A = 12(6 + 6 + 12 cos α)(6 sin α) = 36 sin α(1 + cos α).
dAdα = 36(cos α + cos² α − sin² α) = 36(2cos² α + cos α − 1) = 36(2cos α − 1)(cos α + 1).
On (0, π2), this is zero only at cos α = 12; A′ > 0 before and < 0 after, so α = π3 gives the maximum: option (c).

Answer: (d) 12 cm

Method. longer parallel side at the optimum angle

From the previous item, the area is maximum at α = π3.
Fourth side = 6 + 12 cos α = 6 + 12 × 12 = 12 cm: option (d).

Answer: (c) 27⁠√3⁠ cm²

Method. evaluate the area function at the optimum angle

A = 36 sin α(1 + cos α) with α = π3.
A = 36 × ⁠√32⁠ × 32 = 27⁠√3⁠ cm²: option (c).
Check: parallel sides 6 and 12, height 6 sin π3 = 3⁠√3,⁠ area = 12(18)(3⁠√3)⁠ = 27⁠√3.⁠

Answer: (b) 2

Method. maximise volume of an open cylinder with fixed surface area

Radius r, height h. Surface (base + curved side, no lid): S = πr² + 2πrh, fixed.
h = S − πr²2πr, so V = πr²h = 12(Sr − πr³).
dVdr = 12(S − 3πr²) = 0 gives S = 3πr²; d²Vdr² = -3πr < 0, a maximum.
Then πr² + 2πrh = 3πr², so h = r.
Diameter 2r = 2h, so k = 2: option (b).

Answer: (c) 3⁠√3a²4⁠

Method. the largest triangle inscribed in a circle is equilateral

Let the triangle subtend central angles 2A, 2B, 2C (A + B + C = π) at the centre.
Area = 2a² sin A sin B sin C(from 12bc sin A with b = 2a sin B, c = 2a sin C). ln sin is concave on (0, π), so by Jensen sin A sin B sin C ≤ sin³π3 for A + B + C = π, with equality only at A = B = C = π3: the triangle is equilateral with side 2a sin π3 = ⁠√3a,⁠ area ⁠√34(√3a)²⁠ = 3⁠√3a²4.⁠
Then the triangle is equilateral with side ⁠√3a.⁠
Area = ⁠√34(√3a)²⁠ = 3⁠√3a²4:⁠ option (c).