Application of Derivatives Class 12 NCERT Solutions Chapter 6 Ex 6.3

Concept used. For any real number t, the square t² ≥ 0, with equality only when t = 0. Consequently a function of the form A t² + C with A>0 has its minimum value C at t=0 and no maximum (it grows without bound as |t|→∞). With A<0 the situation flips: the function has its maximum value C at t=0 and no minimum. A pure cubic x³ is strictly increasing on R, so x³ + 1 has neither a maximum nor a minimum on R.

1. (i) Write f(x) = (2x-1)² + 3. The squared term (2x-1)² ≥ 0, so f(x) ≥ 0 + 3 = 3, with equality when 2x - 1 = 0, that is x = 12. As |x|→∞, (2x-1)² → ∞, so f(x)→∞.
   Minimum value = 3 at x = 12;   no maximum.
2. (ii) Complete the square in f(x) = 9x² + 12x + 2: f(x) = 9(x² + 43x) + 2 = 9(x + 23)² - 9·49 + 2 = 9(x + 23)² - 2. The squared term is ≥ 0, so f(x) ≥ -2, with equality at x = -23.
   Minimum value = -2 at x = -23;   no maximum.
3. (iii) Here f(x) = -(x-1)² + 10. Since (x-1)² ≥ 0, -(x-1)² ≤ 0 ⇒ f(x) ≤ 10, with equality at x = 1. As |x|→∞, -(x-1)² → -∞.
   Maximum value = 10 at x = 1;   no minimum.
4. (iv) g(x) = x³ + 1. The derivative g'(x) = 3x² ≥ 0 with equality only at x = 0, so g is strictly increasing on R. As x→∞, g→∞; as x→-∞, g→-∞. Hence neither a maximum nor a minimum value exists.

(i) min =3 at x=12, no max.   (ii) min =-2 at x=-23, no max.   (iii) max =10 at x=1, no min.   (iv) neither maximum nor minimum.

Concept used. The absolute value satisfies |u|≥ 0 for every real u with equality only at u = 0. The sine function is bounded by -1 ≤ sinθ ≤ 1 for every real θ. A function defined on an open interval may attain neither a maximum nor a minimum because the endpoints are not in the domain.

1. (i) f(x) = |x+2| - 1. Since |x+2| ≥ 0, f(x) ≥ 0 - 1 = -1, equality at x = -2. |x+2|→∞ as |x|→∞, so no maximum.
   Minimum = -1 at x = -2;   no maximum.
2. (ii) g(x) = -|x+1| + 3. As |x+1|≥ 0, -|x+1| ≤ 0, so g(x) ≤ 3, equality at x = -1. g→ -∞ as |x|→∞.
   Maximum = 3 at x = -1;   no minimum.
3. (iii) h(x) = sin(2x) + 5. Using -1(2x)≤ 1, 4 ≤ h(x) ≤ 6. sin(2x) = 1 at 2x = π2 + 2kπ, i.e. x = π4 + kπ; sin(2x) = -1 at x = -π4 + kπ.
   Maximum = 6, Minimum = 4.
4. (iv) f(x) = |sin 4x + 3|. Since -1≤ sin 4x ≤ 1, we get 2 ≤ sin 4x + 3 ≤ 4. The inner expression is always positive, so |sin 4x + 3| = sin 4x + 3, giving 2 ≤ f(x) ≤ 4. Maximum = 4 (when sin 4x = 1); Minimum = 2 (when sin 4x = -1).
5. (v) h(x) = x + 1 on the open interval (-1,1). As x → -1⁺, h → 0 but never equals 0 (since x = -1 is excluded); as x → 1^-, h → 2 but never equals 2. So h attains no minimum and no maximum on this open interval.

(i) min -1, no max.   (ii) max 3, no min.   (iii) max 6, min 4.   (iv) max 4, min 2.   (v) neither max nor min.

Concept used. Second Derivative Test. If f is twice differentiable and c is a critical point (f'(c) = 0), then:

• f''(c) < 0 ⇒ c is a point of local maximum with local maximum value f(c).
• f''(c) > 0 ⇒ c is a point of local minimum with local minimum value f(c).
• f''(c) = 0 ⇒ test is inconclusive; fall back on the First Derivative Test (sign change of f' across c).

1. (i) f(x) = x². Then f'(x) = 2x = 0 ⇒ x = 0. f''(x) = 2 > 0, so x = 0 is a local minimum. Local minimum value f(0) = 0.
2. (ii) g(x) = x³ - 3x. g'(x) = 3x² - 3 = 3(x-1)(x+1) = 0 ⇒ x = ± 1. g''(x) = 6x. At x = 1, g''(1) = 6 > 0: local min, value g(1) = 1 - 3 = -2. At x = -1, g''(-1) = -6 < 0: local max, value g(-1) = -1 + 3 = 2.
3. (iii) h(x) = sin x + cos x on (0,π2). h'(x) = cos x - sin x = 0 ⇒ tan x = 1 ⇒ x = π4. h''(x) = -sin x - cos x; at x = π4, h''(π4) = -22 - 22 = -2 < 0. Local max, value h(π4) = 22 + 22 = 2.
4. (iv) f(x) = sin x - cos x on (0, 2π). f'(x) = cos x + sin x = 0 ⇒ tan x = -1 ⇒ x = 3π4 or 7π4. f''(x) = -sin x + cos x. At x = 3π4: f''= -22 - 22 = -2 < 0 (local max). f(3π4) = 22 - (-22) = 2. At x = 7π4: f'' = -(-22) + 22 = 2 > 0 (local min). f(7π4) = -22 - 22 = -2.
5. (v) f(x) = x³ - 6x² + 9x + 15. f'(x) = 3x² - 12x + 9 = 3(x-1)(x-3) = 0 ⇒ x = 1, 3. f''(x) = 6x - 12. f''(1) = -6 < 0 (local max): f(1) = 1 - 6 + 9 + 15 = 19. f''(3) = 6 > 0 (local min): f(3) = 27 - 54 + 27 + 15 = 15.
6. (vi) g(x) = x2 + 2x, x>0. g'(x) = 12 - 2x² = 0 ⇒ x² = 4 ⇒ x = 2 (positive root only). g''(x) = 4x³; g''(2) = 48 = 12 > 0: local min, value g(2) = 1 + 1 = 2.
7. (vii) g(x) = 1x² + 2. g'(x) = -2x(x²+2)² = 0 ⇒ x = 0. g' changes sign from + (for x<0) to - (for x>0), so x = 0 is a local max (first-derivative test). Local max value g(0) = 12.
8. (viii) f(x) = x√1-x on (0,1). Compute f'(x) = √1-x + x · -12√1-x = 2(1-x) - x2√1-x = 2 - 3x2√1-x. Set f'(x) = 0 ⇒ 2 - 3x = 0 ⇒ x = 23. For 0 < x < 23, 2 - 3x > 0 ⇒ f' > 0. For 23 < x < 1, 2 - 3x < 0 ⇒ f' < 0. Sign changes +→-, so x = 23 is a local max. Value: f(23) = 23√1 - 23 = 23·13 = 233 = 239.

(i) min 0 at x=0.   (ii) max 2 at -1; min -2 at 1.   (iii) max 2 at π4.   (iv) max 2 at 3π4; min -2 at 7π4.   (v) max 19 at 1; min 15 at 3.   (vi) min 2 at x=2.   (vii) max 12 at x = 0.   (viii) max 239 at x = 23.

Concept used. A point c is a candidate for a local extremum only if f'(c) = 0 or f'(c) does not exist. A function whose derivative is strictly positive (or strictly negative) everywhere is strictly monotone, so it has no local extrema, and on an unbounded domain it has no absolute extrema either.

1. (i) f(x) = e^x. Differentiate: f'(x) = e^x. Since e^x > 0 for every real x, f'(x) ≠ 0 anywhere, and f' is defined everywhere. So f has no critical points and no extrema. (Strictly increasing, range (0,∞), unbounded above and bounded below by 0 but never reaches it.)
2. (ii) g(x) = log x on its domain (0,∞). Differentiate: g'(x) = 1x. Since x > 0, g'(x) > 0 for every x in the domain. g is strictly increasing on (0,∞), so no critical points and no extrema.
3. (iii) h(x) = x³ + x² + x + 1. Differentiate: h'(x) = 3x² + 2x + 1. Discriminant: Δ = 2² - 4· 3· 1 = 4 - 12 = -8 < 0. Since Δ < 0 and the leading coefficient is 3 > 0, h'(x) > 0 for every real x. So h' is never zero, h is strictly increasing on R, and has no extrema.

None of e^x, log x, x³+x²+x+1 has a maximum or minimum value.

Concept used. Absolute extrema on a closed interval [a,b]. If f is continuous on [a,b], then f attains its absolute maximum and absolute minimum on [a,b] (Extreme Value Theorem). To find them:

1. Find all x∈(a,b) at which f'(x) = 0 or f' does not exist (critical points).
2. Evaluate f at each critical point AND at the endpoints a and b.
3. The largest of those values is the absolute max; the smallest, the absolute min.

1. (i) f(x) = x³ on [-2, 2]. f'(x) = 3x² = 0 ⇒ x = 0. Evaluate: f(-2) = -8, f(0) = 0, f(2) = 8. Absolute max = 8 at x = 2; absolute min = -8 at x = -2.
2. (ii) f(x) = sin x + cos x on [0, π]. f'(x) = cos x - sin x = 0 ⇒ tan x = 1 ⇒ x = π4 (the only one in [0,π]). Values: f(0) = 0 + 1 = 1; f(π4) = 22+22 = 2; f(π) = 0 - 1 = -1. Absolute max = 2 at x = π4; absolute min = -1 at x = π.
3. (iii) f(x) = 4x - 12x² on [-2,92]. f'(x) = 4 - x = 0 ⇒ x = 4 ∈ [-2,92]. f(-2) = -8 - 2 = -10; f(4) = 16 - 8 = 8; f(92) = 18 - 818 = 144 - 818 = 638 = 7.875. Absolute max = 8 at x = 4; absolute min = -10 at x = -2.
4. (iv) f(x) = (x-1)² + 3 on [-3, 1]. f'(x) = 2(x-1) = 0 ⇒ x = 1 (endpoint). f(-3) = 16 + 3 = 19; f(1) = 0 + 3 = 3. Absolute max = 19 at x = -3; absolute min = 3 at x = 1.

(i) max=8 at 2, min=-8 at -2.   (ii) max=2 at π4, min=-1 at π.   (iii) max=8 at 4, min=-10 at -2.   (iv) max=19 at -3, min=3 at 1.

Concept used. For a smooth function on R, a local extremum at c requires p'(c) = 0 (critical point). The Second Derivative Test then classifies the critical point: p''(c) < 0 means local maximum.

1. Differentiate the profit function: p'(x) = -72 - 36x. Set p'(x) = 0: -72 - 36x = 0 ⇒ 36x = -72 ⇒ x = -2.
2. Second derivative: p''(x) = -36. p''(-2) = -36 < 0, so x = -2 is a local maximum.
3. Compute p(-2): p(-2) = 41 - 72(-2) - 18(-2)² = 41 + 144 - 72 = 113. Since p is a downward parabola (leading coefficient -18 < 0), this is the absolute maximum over all real x.

Maximum profit = 113 at x = -2.

Concept used. On a closed bounded interval, the absolute extrema of a continuous function are attained at (a) critical points inside the interval or (b) endpoints. Evaluate f at all candidates and pick the largest and smallest.

1. Let f(x) = 3x⁴ - 8x³ + 12x² - 48x + 25. Differentiate: f'(x) = 12x³ - 24x² + 24x - 48 = 12(x³ - 2x² + 2x - 4).
2. Factorise the cubic in the bracket. Group as x³ - 2x² + 2x - 4 = x²(x - 2) + 2(x - 2) = (x - 2)(x² + 2). So f'(x) = 12(x - 2)(x² + 2).
3. Set f'(x) = 0. Since x² + 2 > 0 for all real x, the only real root is x = 2, which lies in [0,3].
4. Evaluate at x = 0, 2, 3: aligned f(0) &= 25,
   f(2) &= 3(16) - 8(8) + 12(4) - 48(2) + 25
   &= 48 - 64 + 48 - 96 + 25 = -39,
   f(3) &= 3(81) - 8(27) + 12(9) - 48(3) + 25
   &= 243 - 216 + 108 - 144 + 25 = 16. aligned
5. Compare 25, -39, 16. Largest is 25 at x = 0; smallest is -39 at x = 2.

Maximum = 25 at x = 0; minimum = -39 at x = 2.

Concept used. The function sinθ attains its maximum value 1 when θ = π2 + 2kπ for integer k. For the composition sin 2x, set the inner argument 2x to one of those values and solve for x within the required domain.

1. Let f(x) = sin 2x. The maximum value of sin u is 1, attained when u = π2 + 2kπ. Take u = 2x: 2x = π2 + 2kπ ⇒ x = π4 + kπ.
2. Restrict to x ∈ [0, 2π]. The values of k that work are k = 0 (x = π4) and k = 1 (x = π4 + π = 5π4). (k = -1 gives x = -3π4 ∉ [0,2π]; k = 2 gives x = 9π4 ∉ [0,2π].)
3. Verify by direct calculation: sin(2·π4) = sinπ2 = 1, sin(2·5π4) = sin5π2 = 1.

Maximum value 1 is attained at x = π4 and x = 5π4.

Concept used. For any constants a, b, the expression asin x + bcos x can be rewritten as Rsin(x + ϕ) where R = √a² + b². The maximum value of Rsin(·) is R, since sin(·)≤ 1.

1. Identify a = 1 and b = 1. Compute the amplitude: R = √a² + b² = √1 + 1 = 2.
2. Write sin x + cos x = 2(12sin x + 12cos x) = 2 sin(x + π4), using sinπ4 = cosπ4 = 12.
3. Since sin(x + π4) ≤ 1, the maximum value of 2 sin(x + π4) is 2, attained when x + π4 = π2, i.e. x = π4 (and at every translate x = π4 + 2kπ).

Maximum value of sin x + cos x is 2.

Concept used. Absolute extrema on a closed interval = compare f at the critical points inside (a,b) and at the endpoints a, b. Work each interval separately.

1. Let f(x) = 2x³ - 24x + 107. Differentiate: f'(x) = 6x² - 24 = 6(x-2)(x+2). Critical points: x = 2 and x = -2.
2. Interval [1, 3]. Only x = 2 lies inside. Evaluate: aligned f(1) &= 2 - 24 + 107 = 85,
   f(2) &= 16 - 48 + 107 = 75,
   f(3) &= 54 - 72 + 107 = 89. aligned Largest is 89 at x = 3. Maximum on [1,3] is 89.
3. Interval [-3, -1]. Only x = -2 lies inside. Evaluate: aligned f(-3) &= -54 + 72 + 107 = 125,
   f(-2) &= -16 + 48 + 107 = 139,
   f(-1) &= -2 + 24 + 107 = 129. aligned Largest is 139 at x = -2. Maximum on [-3,-1] is 139.

Maximum on [1,3] is 89 at x = 3; maximum on [-3,-1] is 139 at x = -2.

Concept used. If f has a maximum (local or absolute) at an interior point c of a closed interval, and f is differentiable at c, then f'(c) = 0 (Fermat's theorem for interior extrema).

1. Let f(x) = x⁴ - 62x² + ax + 9. Differentiate: f'(x) = 4x³ - 124x + a.
2. Since x = 1 is an interior point of [0, 2] at which f attains a maximum, and f is a polynomial (hence differentiable), f'(1) = 0: f'(1) = 4(1)³ - 124(1) + a = 4 - 124 + a = a - 120. Set a - 120 = 0 ⇒ a = 120.
3. Verification. With a = 120, evaluate at the candidate points: f(0) = 9; f(1) = 1 - 62 + 120 + 9 = 68; f(2) = 16 - 248 + 240 + 9 = 17. Largest is 68 at x = 1. The condition holds.

a = 120.

Concept used. Absolute extrema on [a,b] = compare f at critical points (where f'(x) = 0) inside (a,b) and at the endpoints.

1. Let f(x) = x + sin 2x. Differentiate: f'(x) = 1 + 2cos 2x.
2. Solve f'(x) = 0: 1 + 2cos 2x = 0 ⇒ cos 2x = -12. The general solution is 2x = 2π3 + 2kπ or 2x = 4π3 + 2kπ, i.e. x = π3 + kπ or x = 2π3 + kπ.
3. Restrict to x ∈ (0, 2π): x = π3, 2π3, 4π3, 5π3.
4. Evaluate f at these critical points and at the endpoints 0, 2π: aligned f(0) &= 0 + 0 = 0,
   f(π3) &= π3 + sin2π3 = π3 + 32,
   f(2π3) &= 2π3 + sin4π3 = 2π3 - 32,
   f(4π3) &= 4π3 + sin8π3 = 4π3 + 32,
   f(5π3) &= 5π3 + sin10π3 = 5π3 - 32,
   f(2π) &= 2π + 0 = 2π. aligned (Used sin8π3 = sin(8π3 - 2π) = sin2π3 = 32 and similar reductions.)
5. Numerically with π ≈ 3.1416: the candidate values are 0, 1.91, 1.23, 5.06, 4.37, 6.28. Maximum = 2π at x = 2π; minimum = 0 at x = 0.

Maximum = 2π at x = 2π; minimum = 0 at x = 0.

Concept used. Optimisation in one variable. Use the constraint (x + y = 24) to write one variable in terms of the other, giving the objective (product) as a single-variable function. Maximise using the second-derivative test.

1. Let the two numbers be x and y. Constraint: x + y = 24, so y = 24 - x.
2. Product (objective): P(x) = x y = x(24 - x) = 24x - x². Since the numbers must be real (problem doesn't restrict positivity), x∈R.
3. Find critical points. P'(x) = 24 - 2x = 0 ⇒ x = 12.
4. Second derivative test: P''(x) = -2 < 0, so x = 12 is a local (and global, since P is a downward parabola) maximum.
5. Find y: y = 24 - 12 = 12. Maximum product: P(12) = 12· 12 = 144.

Both numbers are 12; maximum product = 144.

Concept used. Constrained optimisation in one variable: solve the constraint x + y = 60 for x, substitute into xy³, differentiate once for the critical point and once more to verify it is a maximum.

1. Let f(y) = xy³ with x = 60 - y and y ∈ (0, 60) (both positive). f(y) = (60 - y) y³ = 60 y³ - y⁴.
2. Differentiate: f'(y) = 180 y² - 4 y³ = 4 y²(45 - y).
3. Set f'(y) = 0. Roots: y = 0 (boundary, both numbers must be positive) and y = 45. So the interior critical point is y = 45.
4. Second derivative test: f''(y) = 360 y - 12 y² = 12y(30 - y). At y = 45: f''(45) = 12(45)(30 - 45) = 540·(-15) = -8100 < 0. Local maximum.
5. Compute the corresponding x: x = 60 - 45 = 15.
   Maximum value: f(45) = 15 · 45³ = 15 · 91125 = 1 366 875.

x = 15, y = 45; maximum xy³ = 1 366 875.

Concept used. Single-variable substitution (same idea as Q14). Set up the product as a function of one variable using the constraint x + y = 35, then maximise.

1. Substitute x = 35 - y (0 < y < 35): P(y) = (35 - y)² y⁵.
2. Differentiate using the product rule: P'(y) = 2(35 - y)(-1) y⁵ + (35 - y)² · 5y⁴. Factor out y⁴ (35 - y): P'(y) = y⁴ (35 - y)[-2y + 5(35 - y)] = y⁴ (35 - y)(175 - 7y). Simplify the bracket: -2y + 175 - 5y = 175 - 7y.
3. Set P'(y) = 0. Factors: y⁴ = 0 ⇒ y = 0 (excluded); 35 - y = 0 ⇒ y = 35 (excluded); 175 - 7y = 0 ⇒ y = 25. So y = 25.
4. Sign of P' around y = 25: For y slightly less than 25: y⁴ > 0, 35 - y > 0, 175 - 7y > 0, product positive (P' > 0). For y slightly greater than 25: 175 - 7y < 0, product negative (P' < 0). Sign +→-: y = 25 is a local maximum.
5. Then x = 35 - 25 = 10. Maximum: P(25) = (10)² (25)⁵ = 100 · 9 765 625 = 9.76× 10⁸ (no need to compute exactly).

x = 10, y = 25.

Concept used. Same substitution technique. Use x + y = 16 to eliminate y, then minimise the resulting one-variable function.

1. Let S(x) = x³ + y³ with y = 16 - x, 0 < x < 16. Then S(x) = x³ + (16 - x)³.
2. Differentiate: S'(x) = 3x² + 3(16 - x)² · (-1) = 3[x² - (16 - x)²].
3. Factor as a difference of squares: x² - (16 - x)² = [x - (16 - x)][x + (16 - x)] = (2x - 16)· 16 = 16(2x - 16). So S'(x) = 3 · 16(2x - 16) = 48(2x - 16) = 96(x - 8).
4. Set S'(x) = 0 ⇒ x = 8. Then y = 16 - 8 = 8.
5. Second derivative: S''(x) = 96 > 0 everywhere. So x = 8 is a local minimum (and the only interior critical point, hence the absolute minimum on (0,16)).
6. Minimum sum of cubes: S(8) = 8³ + 8³ = 512 + 512 = 1024.

Both numbers are 8; minimum sum of cubes = 1024.

Concept used. Geometric optimisation. Express the quantity to be optimised (here, volume) in terms of one variable (the side x of the cut-out square), identify the natural domain 0 < x < 9 (so that the base remains positive), and apply the second derivative test.

1. Let x cm be the side of each cut-out square (0 < x < 9). After folding, the resulting open box has:
   • base: a square of side (18 - 2x) cm;
   • height: x cm.
2. Volume: V(x) = (18 - 2x)² · x. Expand: V(x) = (324 - 72x + 4x²) x = 4x³ - 72x² + 324x.
3. Differentiate: V'(x) = 12x² - 144x + 324 = 12(x² - 12x + 27).
4. Solve V'(x) = 0: x² - 12x + 27 = 0 ⇒ x = 12±√144 - 1082 = 12± 62. So x = 9 or x = 3. x = 9 is excluded (collapses the base to zero). Take x = 3.
5. Second derivative test: V''(x) = 24x - 144, V''(3) = 72 - 144 = -72 < 0. So x = 3 is a local (and absolute, since the only interior critical point) maximum.
6. Maximum volume: V(3) = (18 - 6)² · 3 = 144 · 3 = 432 cm³.

Side of the cut-off square = 3 cm; maximum volume = 432 cm³.

Concept used. Same setup as Q17 with a rectangular sheet. The base becomes a (45 - 2x)×(24 - 2x) rectangle and the height is x, with 0 < x < 12 (so both base dimensions stay positive).

1. Let x cm be the side of each cut-out square (0 < x < 12). Volume: V(x) = (45 - 2x)(24 - 2x) x.
2. Expand: aligned (45 - 2x)(24 - 2x) &= 1080 - 90x - 48x + 4x² = 4x² - 138x + 1080. aligned So V(x) = x(4x² - 138x + 1080) = 4x³ - 138x² + 1080x.
3. Differentiate: V'(x) = 12x² - 276x + 1080 = 12(x² - 23x + 90).
4. Solve V'(x) = 0: x = 23 ± √529 - 3602 = 23 ± √1692 = 23 ± 132. So x = 18 or x = 5. x = 18 ∉ (0, 12). Take x = 5.
5. Second derivative test: V''(x) = 24x - 276, V''(5) = 120 - 276 = -156 < 0. So x = 5 is a local maximum.
6. Maximum volume: V(5) = (45 - 10)(24 - 10)· 5 = 35 · 14 · 5 = 2450 cm³.

Cut-off square side = 5 cm; maximum volume = 2450 cm³.

Concept used. A rectangle inscribed in a circle of radius r has its diagonal equal to the diameter 2r. Parametrise the rectangle by an angle and use one-variable optimisation; the optimum will give equal sides, i.e. a square.

1. Let the rectangle have sides x (length) and y (breadth). Since the rectangle is inscribed in a circle of radius r, its diagonal equals the diameter: x² + y² = (2r)² = 4r².
2. Area: A = xy. Eliminate y: y = √4r² - x². A(x) = x√4r² - x², 0 < x < 2r.
3. Differentiate A² for cleaner algebra. Let f(x) = A² = x²(4r² - x²). f'(x) = 2x(4r² - x²) + x²(-2x) = 8r² x - 2x³ - 2x³ = 8r² x - 4x³ = 4x(2r² - x²). Set f'(x) = 0: x = 0 (degenerate, excluded) or x² = 2r² ⇒ x = r2.
4. Second derivative: f''(x) = 8r² - 12x². At x = r2: f''(r2) = 8r² - 12(2r²) = 8r² - 24r² = -16r² < 0. So x = r2 is a maximum of f, hence of A (since A > 0).
5. Find y: y = √4r² - 2r² = √2r² = r2. So x = y = r2: the rectangle is a square.
6. Maximum area: A = (r2)(r2) = 2r².

The square (side r2, area 2r²) maximises the area of inscribed rectangles.

Concept used. A closed right circular cylinder with base radius r and height h has surface area S = 2π r² + 2π rh and volume V = π r² h. With S fixed, express h in terms of r, substitute into V, and maximise as a function of r.

1. Surface area constraint: S = 2π r² + 2π r h ⇒ h = S - 2π r²2π r. Here S is constant and r > 0 with r² < S2π (so h > 0).
2. Volume: V = π r² h = π r² · S - 2π r²2π r = r(S - 2π r²)2 = Sr2 - π r³.
3. Differentiate: dVdr = S2 - 3π r². Set dVdr = 0: 3π r² = S2 ⇒ r² = S6π.
4. Second derivative test: d² Vdr² = -6π r. At the critical r > 0, d² Vdr² < 0: local maximum.
5. Compute the optimal height. Substitute S = 6π r² into h = S - 2π r²2π r: h = 6π r² - 2π r²2π r = 4π r²2π r = 2r. That is, h = 2r, which means height = diameter of base.

Optimum cylinder has h = 2r (height equals diameter of base).

Concept used. Dual of Q20: now V = 100 cm³ is fixed; minimise the closed-can surface area S = 2π r² + 2π r h as a function of one variable.

1. Volume constraint: π r² h = 100 ⇒ h = 100π r².
2. Surface area: S(r) = 2π r² + 2π r · 100π r² = 2π r² + 200r.
3. Differentiate: S'(r) = 4π r - 200r². Set S'(r) = 0: 4π r = 200r² ⇒ 4π r³ = 200 ⇒ r³ = 50π. So r = (50π)^1/3 cm.
4. Second derivative test: S''(r) = 4π + 400r³. At the critical r > 0, both terms are positive, so S''(r) > 0: local minimum.
5. Find h. From r³ = 50π, r² = (50π)^2/3. Then h = 100π r² = 100π· (π50)^2/3 = 100π^1/3· 50^2/3 = 100(π· 50²)^1/3 = 2(50π)^1/3 = 2r. So again h = 2r: the height equals the diameter.

r = (50π)^1/3 cm, h = 2(50π)^1/3 cm = 2r.

Concept used. Let one piece of length form the circle's circumference and the other piece (28 - ) form the square's perimeter. Express the combined area as a function of and minimise.

1. Let the circular piece have length (with 0 < < 28).
   • Circle: circumference = 2π r = ⇒ r = 2π. Area = π r² = π · ²4π² = ²4π.
   • Square: perimeter = 4s = 28 - ⇒ s = 28 - 4. Area = s² = (28-)²16.
2. Combined area: A() = ²4π + (28-)²16.
3. Differentiate: A'() = 24π + 2(28-)(-1)16 = 2π - 28 - 8. Set A'() = 0: 2π = 28 - 8 ⇒ 8= 2π(28 - ). Expand: 8= 56π - 2π⇒ 8+ 2π= 56π ⇒ (8 + 2π) = 56π. = 56π8 + 2π = 56π2(4 + π) = 28π4 + π.
4. Second derivative test: A''() = 12π + 18 > 0, so the critical point is a minimum.
5. Lengths of the two pieces:
   • Circle: = 28ππ + 4 m.
   • Square: 28 - = 28 - 28ππ + 4 = 28(π + 4) - 28ππ + 4 = 112π + 4 m.

Circle's piece =28ππ + 4 m; square's piece =112π + 4 m.

Concept used. Inscribe a right circular cone in a sphere of radius R. Place the sphere's centre O on the cone's axis. Let the cone have height h measured from its apex along the axis and base radius a. The cone's apex sits at the ``north pole'' of the sphere, and the base circle (at depth h below the apex) is a horizontal slice of the sphere.

1. Set up coordinates. Place the apex at the top of the sphere. Let the base of the cone be at vertical distance h below the apex (so 0 < h < 2R). The centre of the sphere is at distance R below the apex (on the axis). So the base of the cone is at distance h - R below the centre.
2. The base circle has radius a. Since the base circle lies on the sphere, the right triangle (centre → base circle's centre → base circle's edge) gives a² + (h - R)² = R² ⇒ a² = R² - (h - R)² = 2Rh - h².
3. Volume of the cone: V = 13π a² h = 13π (2Rh - h²)h = π3(2Rh² - h³).
4. Differentiate with respect to h: V'(h) = π3(4Rh - 3h²) = π h3(4R - 3h). Set V'(h) = 0: h = 0 (degenerate) or h = 4R3.
5. Second derivative test: V''(h) = π3(4R - 6h); V''(4R3) = π3(4R - 8R) = -4π R3 < 0. Local (and absolute, since unique interior critical point) maximum.
6. Compute the maximum volume. With h = 4R3: a² = 2R·4R3 - (4R3)² = 8R²3 - 16R²9 = 24R² - 16R²9 = 8R²9. V_max = 13π· 8R²9· 4R3 = 32π R³81.
7. Compare with the sphere's volume V_sphere = 43π R³: V_maxV_sphere = 32π R³/814π R³/3 = 3281· 34 = 96324 = 827.

V_max = 827· V_sphere = 32π R³81.

Concept used. For a right circular cone of base radius r and altitude h, the slant height is = √r² + h², the curved (lateral) surface area is S = π r , and the volume is V = 13π r² h. With V fixed, eliminate h in favour of r inside S², differentiate.

1. Volume constraint: 13π r² h = V ⇒ h = 3Vπ r².
2. Curved surface: S = π r √r² + h² = π r√r² + 9V²π² r⁴. Minimise S² instead (which is monotone in S): S² = π² r² (r² + 9V²π² r⁴) = π² r⁴ + 9V²r².
3. Differentiate f(r) = π² r⁴ + 9V²r²: f'(r) = 4π² r³ - 18 V²r³. Set f'(r) = 0: 4π² r³ = 18 V²r³ ⇒ 4π² r⁶ = 18 V² ⇒ r⁶ = 9 V²2π².
4. Second derivative test: f''(r) = 12π² r² + 54 V²r⁴ > 0 for all r > 0. Hence the critical r is a minimum of S², hence of S.
5. Show h = r2. From the critical condition 4π² r⁶ = 18 V², take square roots (positive): 2π r³ = 3V2 ⇒ V = 2π r³32 = π r³23. And h = 3Vπ r² = 3π r²· π r³23 = r2.

For minimum curved surface at fixed volume, h = 2 r.

Concept used. For a right circular cone with slant height (constant), let α be the semi-vertical angle. Then base radius r = α and altitude h = α. The volume V = 13π r² h becomes a one-variable function of α.

1. Express dimensions in terms of α: r = α, h = α, where 0 < α < π2.
2. Volume: V(α) = 13π (α)² (α) = π³3sin²α.
3. Let g(α) = sin²α. Differentiate: g'(α) = 2sinα· cosα + sin²α·(-sinα) = 2sin²α - sin³α. Factor: g'(α) = sinα (2cos²α - sin²α).
4. Set g'(α) = 0. Since sinα > 0 on the open interval, 2cos²α - sin²α = 0 ⇒ sin²α = 2cos²α ⇒ tan²α = 2. So tanα = 2, i.e. α = tan^-12.
5. Confirm it is a maximum. Use sign analysis. Just left of the critical α, sin²α < 2cos²α, so g'(α) > 0. Just right of it, the inequality flips, so g'(α) < 0. Sign change +→-: local maximum.

Semi-vertical angle = tan^-12.

Concept used. Cone with base radius r, slant height , altitude h = √² - r², total surface area (curved + base): S = π r² + π r (constant), volume V = 13π r² h. Use the constraint to eliminate one variable, then optimise.

1. Surface constraint: S = π r² + π r= constant. Solve for : = S - π r²π r = Sπ r - r.
2. Square the volume to avoid the root in h = √² - r²: V² = π² r⁴ h²9 = π² r⁴ (² - r²)9. Compute ² - r²: ² = (Sπ r - r)² = S²π² r² - 2Sπ + r², so ² - r² = S²π² r² - 2Sπ.
3. Substitute: V² = π² r⁴9(S²π² r² - 2Sπ) = r² S²9 - 2π S r⁴9 = S r² (S - 2π r²)9.
4. Let f(r) = r²(S - 2π r²) = Sr² - 2π r⁴. Then f'(r) = 2Sr - 8π r³ = 2r(S - 4π r²). Set f'(r) = 0: r = 0 (excluded) or S = 4π r² ⇒ r² = S4π.
5. Second derivative: f''(r) = 2S - 24π r²; at r² = S4π, f''(r) = 2S - 6S = -4S < 0. Local max.
6. At this r, compute : = Sπ r - r. With S = 4π r², Sπ r = 4r, so = 4r - r = 3r.
7. Semi-vertical angle α: sinα = r = r3r = 13. So α = sin^-1(13).

Semi-vertical angle = sin^-1(13).

Concept used. Minimise the squared distance D(x, y) = x² + (y - 5)² subject to the curve x² = 2y. Substitute the constraint to obtain a one-variable function and minimise.

1. Parametrise the curve: any point on x² = 2y can be written as (x, x²2).
2. Squared distance from (0, 5): D(x) = x² + (x²2 - 5)².
3. Differentiate. Let u = x²2 - 5; then dudx = x and D'(x) = 2x + 2u· x = 2x(1 + u) = 2x(1 + x²2 - 5) = 2x(x²2 - 4).
4. Set D'(x) = 0: x = 0 or x²2 - 4 = 0 ⇒ x² = 8 ⇒ x = ± 22.
5. Evaluate D at candidates: D(0) = 0 + 25 = 25; D(± 22) = 8 + (4 - 5)² = 8 + 1 = 9. (Since (22)²2 = 4.)
6. The smaller value is 9 at x = ± 22, giving y = 82 = 4. So the nearest points are (± 22, 4). Choice (A) lists (22, 4).

(A) (22, 4).

Concept used. For a rational function p(x)q(x) with q(x) > 0 for all real x, the minimum can be found by differentiating using the quotient rule and solving f'(x) = 0.

1. Set f(x) = 1 - x + x²1 + x + x². Note 1 + x + x² > 0 for all real x (discriminant 1 - 4 = -3 < 0, leading coefficient > 0).
2. Differentiate using the quotient rule. Let N = 1 - x + x² and D = 1 + x + x². N' = -1 + 2x, D' = 1 + 2x. f'(x) = N'D - N D'D². Compute the numerator: aligned N'D - N D' &= (-1 + 2x)(1 + x + x²) - (1 - x + x²)(1 + 2x)
   &= [-1 - x - x² + 2x + 2x² + 2x³] - [1 + 2x - x - 2x² + x² + 2x³]
   &= [2x³ + x² + x - 1] - [2x³ - x² + x + 1]
   &= 2x² - 2. aligned
3. So f'(x) = 2x² - 2(1 + x + x²)². Set numerator = 0: x² = 1 ⇒ x = ± 1.
4. Evaluate: f(1) = 1 - 1 + 11 + 1 + 1 = 13; f(-1) = 1 + 1 + 11 - 1 + 1 = 31 = 3.
5. Behaviour at infinity: as |x|→∞, f(x)→ 1. So f has range bounded by the two critical values: minimum = 13 at x = 1 and maximum = 3 at x = -1.

(D) 13.

Concept used. The cube root u^1/3 is a strictly increasing function of u, so the maximum of [g(x)]^1/3 is the cube root of the maximum of g(x). Therefore maximise g(x) = x(x - 1) + 1 = x² - x + 1 on [0, 1] and cube root the result.

1. Let g(x) = x² - x + 1. Differentiate: g'(x) = 2x - 1. Set g'(x) = 0 ⇒ x = 12.
2. Second derivative test: g''(x) = 2 > 0, so x = 12 is a local minimum (not maximum). Local min value: g(12) = 14 - 12 + 1 = 34.
3. Since the only interior critical point is a minimum, the maximum on [0,1] is at an endpoint. Evaluate: g(0) = 0 - 0 + 1 = 1; g(1) = 1 - 1 + 1 = 1. Maximum of g on [0, 1] is 1.
4. Take the cube root: max[x(x - 1) + 1]^1/3 = 1^1/3 = 1.

(C) 1.