NCERT Solutions for Class 12 Maths Chapter 6: Application of Derivatives (NCERT 2026–27)

These Class 12 Maths Chapter 6 solutions cover Application of Derivatives with complete, step-by-step answers to every question in Exercise 6.1, Exercise 6.2, Exercise 6.3 and the Miscellaneous Exercise. You will learn rate of change of quantities, increasing and decreasing functions, and maxima and minima — with each solution worked out and cross-checked against the official NCERT answer key for the 2026–27 session.

Class: 12 Subject: Mathematics Chapter: 6 Name: Application of Derivatives Exercises: 6.1, 6.2, 6.3 & Miscellaneous Session: 2026–27
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Chapter 6 Overview

Chapter 6, Application of Derivatives, shows how the derivative learnt in Chapter 5 is used to solve real problems in science, engineering, economics and geometry. The chapter is built around three big ideas: using dy/dx to measure the rate of change of one quantity with respect to another (and chaining two rates through a common variable such as time); using the sign of f′(x) to decide where a function is increasing or decreasing; and using critical points with the first and second derivative tests to find local maxima, local minima, and absolute (global) maximum and minimum values on a closed interval. These Class 12 Maths Chapter 6 solutions solve every NCERT exercise question and the full Miscellaneous Exercise in exam-ready steps.

Key Concepts & Definitions

Rate of change: if y = f(x) then dy/dx is the rate of change of y with respect to x; if x and y both depend on t, then dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0.

Marginal cost / revenue: marginal cost is dC/dx (rate of change of total cost with output) and marginal revenue is dR/dx (rate of change of total revenue with units sold).

Increasing / decreasing: on an interval, f is increasing if f′(x) > 0 throughout and decreasing if f′(x) < 0 throughout; if f′(x) = 0 throughout, f is constant.

Critical point: a point c in the domain where f′(c) = 0 or f is not differentiable.

First derivative test: at a critical point c, if f′ changes + to −, c is a local maximum; if − to +, a local minimum; if no sign change, a point of inflexion.

Second derivative test: if f′(c) = 0 and f″(c) < 0, c is a local maximum; if f′(c) = 0 and f″(c) > 0, a local minimum; if f″(c) = 0 the test fails.

Absolute maxima/minima: on a closed interval [a, b], compare f at all critical points and at the end points a and b; the largest value is the absolute maximum, the smallest the absolute minimum.

Important Formulas (Chapter 6)

Chain rule for rates: dy/dt = (dy/dx)(dx/dt).

Circle: area A = πr2, dA/dr = 2πr; circumference C = 2πr, dC/dr = 2π.

Sphere: volume V = (4/3)πr3, dV/dr = 4πr2; surface S = 4πr2.

Cube: volume V = x3, dV/dx = 3x2; surface S = 6x2, dS/dx = 12x.

Cone: volume V = (1/3)πr2h.

Approximation: Δy ≈ (dy/dx)·Δx.

Tests: increasing ⇔ f′(x) > 0; decreasing ⇔ f′(x) < 0; local extremum needs f′(c) = 0 (or non-differentiable c).

All questions below are reproduced verbatim from the NCERT Class 12 Mathematics textbook (Reprint 2026–27); the worked solutions are original and verified against the answers given at the back of the book.

Exercise 6.1 — Rate of Change of Quantities

1. Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm   (b) r = 4 cm

SOLUTION Area A = πr2, so dA/dr = 2πr. (a) At r = 3: dA/dr = 2π(3) = 6π cm2/cm. (b) At r = 4: dA/dr = 2π(4) = 8π cm2/cm.

2. The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

SOLUTION Let edge = x. V = x3, so dV/dt = 3x2(dx/dt) = 8, giving dx/dt = 8/(3x2). Surface S = 6x2, so dS/dt = 12x(dx/dt) = 12x · 8/(3x2) = 32/x. At x = 12: dS/dt = 32/12 = 8/3 cm2/s.

3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

SOLUTION A = πr2, so dA/dt = 2πr(dr/dt) with dr/dt = 3. At r = 10: dA/dt = 2π(10)(3) = 60π cm2/s.

4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

SOLUTION V = x3, so dV/dt = 3x2(dx/dt) with dx/dt = 3. At x = 10: dV/dt = 3(10)2(3) = 900 cm3/s.

5. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

SOLUTION A = πr2, so dA/dt = 2πr(dr/dt) with dr/dt = 5. At r = 8: dA/dt = 2π(8)(5) = 80π cm2/s.

6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

SOLUTION Circumference C = 2πr, so dC/dt = 2π(dr/dt) = 2π(0.7) = 1.4π cm/s.

7. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

SOLUTION Given dx/dt = −5 and dy/dt = 4 cm/min. (a) P = 2(x + y), so dP/dt = 2(dx/dt + dy/dt) = 2(−5 + 4) = −2 cm/min (perimeter decreasing). (b) A = xy, so dA/dt = (dx/dt)y + x(dy/dt) = (−5)(6) + (8)(4) = −30 + 32 = 2 cm2/min.

8. A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

SOLUTION V = (4/3)πr3, so dV/dt = 4πr2(dr/dt) = 900. dr/dt = 900/(4πr2). At r = 15: dr/dt = 900/(4π·225) = 900/(900π) = 1/π cm/s.

9. A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

SOLUTION V = (4/3)πr3, so dV/dr = 4πr2. At r = 10: dV/dr = 4π(10)2 = 400π cm3/cm.

10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

SOLUTION Let x = distance of foot from wall, y = height on wall. Then x2 + y2 = 25. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0, so dy/dt = −(x/y)(dx/dt). At x = 4: y = √(25 − 16) = 3. With dx/dt = 2 cm/s, dy/dt = −(4/3)(2) = −8/3 cm/s. ∴ the height is decreasing at 8/3 cm/s.

11. A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

SOLUTION 6y = x3 + 2 ⇒ 6(dy/dt) = 3x2(dx/dt), so dy/dt = (x2/2)(dx/dt). We need dy/dt = 8(dx/dt) ⇒ x2/2 = 8 ⇒ x2 = 16 ⇒ x = 4 or x = −4. At x = 4: y = (64 + 2)/6 = 11 → point (4, 11). At x = −4: y = (−64 + 2)/6 = −31/3 → point (−4, −31/3).

12. The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

SOLUTION V = (4/3)πr3, so dV/dt = 4πr2(dr/dt) with dr/dt = 1/2. At r = 1: dV/dt = 4π(1)2(1/2) = 2π cm3/s.

13. A balloon, which always remains spherical, has a variable diameter (3/2)(2x + 1). Find the rate of change of its volume with respect to x.

SOLUTION Diameter = (3/2)(2x + 1), so radius r = (3/4)(2x + 1). V = (4/3)πr3 = (4/3)π[(3/4)(2x + 1)]3 = (4/3)π(27/64)(2x + 1)3 = (9π/16)(2x + 1)3. dV/dx = (9π/16)·3(2x + 1)2·2 = (27π/8)(2x + 1)2.

14. Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

SOLUTION Given h = r/6, so r = 6h. Volume V = (1/3)πr2h = (1/3)π(6h)2h = 12πh3. dV/dt = 36πh2(dh/dt) = 12, so dh/dt = 12/(36πh2) = 1/(3πh2). At h = 4: dh/dt = 1/(3π·16) = 1/(48π) cm/s.

15. The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x3 − 0.003x2 + 15x + 4000. Find the marginal cost when 17 units are produced.

SOLUTION Marginal cost MC = dC/dx = 0.021x2 − 0.006x + 15. At x = 17: MC = 0.021(289) − 0.006(17) + 15 = 6.069 − 0.102 + 15 = ₹ 20.967.

16. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 13x2 + 26x + 15. Find the marginal revenue when x = 7.

SOLUTION Marginal revenue MR = dR/dx = 26x + 26. At x = 7: MR = 26(7) + 26 = 182 + 26 = ₹ 208.

17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is (A) 10π   (B) 12π   (C) 8π   (D) 11π

SOLUTION dA/dr = 2πr; at r = 6, dA/dr = 12π. ∴ correct option is (B) 12π.

18. The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5. The marginal revenue, when x = 15 is (A) 116   (B) 96   (C) 90   (D) 126

SOLUTION MR = dR/dx = 6x + 36; at x = 15, MR = 6(15) + 36 = 90 + 36 = 126. ∴ correct option is (D) 126.

Exercise 6.2 — Increasing and Decreasing Functions

1. Show that the function given by f(x) = 3x + 17 is increasing on R.

SOLUTION f′(x) = 3 > 0 for all x ∈ R. Since the derivative is positive everywhere, f is (strictly) increasing on R.

2. Show that the function given by f(x) = e2x is increasing on R.

SOLUTION f′(x) = 2e2x. Since e2x > 0 for all x, f′(x) > 0 everywhere. ∴ f is increasing on R.

3. Show that the function given by f(x) = sin x is (a) increasing in (0, π/2)   (b) decreasing in (π/2, π) (c) neither increasing nor decreasing in (0, π)

SOLUTION f′(x) = cos x. (a) In (0, π/2), cos x > 0 ⇒ f′(x) > 0 ⇒ f is increasing. (b) In (π/2, π), cos x < 0 ⇒ f′(x) < 0 ⇒ f is decreasing. (c) Since f increases on part of (0, π) and decreases on the rest, it is neither increasing nor decreasing on (0, π).

4. Find the intervals in which the function f given by f(x) = 2x2 − 3x is (a) increasing   (b) decreasing

SOLUTION f′(x) = 4x − 3 = 0 ⇒ x = 3/4. (a) For x > 3/4, f′(x) > 0 ⇒ increasing on (3/4, ∞). (b) For x < 3/4, f′(x) < 0 ⇒ decreasing on (−∞, 3/4).

5. Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is (a) increasing   (b) decreasing

SOLUTION f′(x) = 6x2 − 6x − 36 = 6(x2 − x − 6) = 6(x − 3)(x + 2) = 0 ⇒ x = −2, 3. (a) For x < −2 and x > 3, both factors give f′(x) > 0 ⇒ increasing on (−∞, −2) and (3, ∞). (b) For −2 < x < 3, f′(x) < 0 ⇒ decreasing on (−2, 3).

6. Find the intervals in which the following functions are strictly increasing or decreasing: (a) x2 + 2x − 5   (b) 10 − 6x − 2x2 (c) −2x3 − 9x2 − 12x + 1   (d) 6 − 9x − x2 (e) (x + 1)3(x − 3)3

SOLUTION (a) f′(x) = 2x + 2 = 0 ⇒ x = −1. Decreasing for x < −1, increasing for x > −1. (b) f′(x) = −6 − 4x = 0 ⇒ x = −3/2. Increasing for x < −3/2, decreasing for x > −3/2. (c) f′(x) = −6x2 − 18x − 12 = −6(x + 1)(x + 2) = 0 ⇒ x = −1, −2. Increasing for −2 < x < −1; decreasing for x < −2 and x > −1. (d) f′(x) = −9 − 2x = 0 ⇒ x = −9/2. Increasing for x < −9/2, decreasing for x > −9/2. (e) f′(x) = 3(x + 1)2(x − 3)3 + 3(x + 1)3(x − 3)2 = 3(x + 1)2(x − 3)2[(x − 3) + (x + 1)] = 6(x + 1)2(x − 3)2(x − 1). The squared factors are ≥ 0, so the sign follows (x − 1): increasing in (1, 3) and (3, ∞); decreasing in (−∞, −1) and (−1, 1).

7. Show that y = log(1 + x) − 2x/(2 + x), x > −1, is an increasing function of x throughout its domain.

SOLUTION dy/dx = 1/(1 + x) − [2(2 + x) − 2x]/(2 + x)2 = 1/(1 + x) − 4/(2 + x)2. Combine: dy/dx = [(2 + x)2 − 4(1 + x)] / [(1 + x)(2 + x)2] = [4 + 4x + x2 − 4 − 4x] / [(1 + x)(2 + x)2] = x2 / [(1 + x)(2 + x)2]. For x > −1, the denominator (1 + x)(2 + x)2 > 0 and x2 ≥ 0, so dy/dx ≥ 0 (= 0 only at x = 0). Hence y is increasing throughout its domain.

8. Find the values of x for which y = [x(x − 2)]2 is an increasing function.

SOLUTION y = [x(x − 2)]2 = (x2 − 2x)2. dy/dx = 2(x2 − 2x)(2x − 2) = 4x(x − 2)(x − 1). Roots at x = 0, 1, 2. Sign of dy/dx is positive in (0, 1) and (2, ∞). ∴ y is increasing for 0 < x < 1 and x > 2.

9. Prove that y = 4 sinθ/(2 + cosθ) − θ is an increasing function of θ in [0, π/2].

SOLUTION dy/dθ = [4cosθ(2 + cosθ) − 4sinθ(−sinθ)]/(2 + cosθ)2 − 1. Numerator of first term = 8cosθ + 4cos2θ + 4sin2θ = 8cosθ + 4. So dy/dθ = (8cosθ + 4)/(2 + cosθ)2 − 1. = [8cosθ + 4 − (2 + cosθ)2]/(2 + cosθ)2 = [8cosθ + 4 − 4 − 4cosθ − cos2θ]/(2 + cosθ)2 = (4cosθ − cos2θ)/(2 + cosθ)2 = cosθ(4 − cosθ)/(2 + cosθ)2. In [0, π/2], cosθ ≥ 0 and (4 − cosθ) > 0, so dy/dθ ≥ 0. Hence y is increasing on [0, π/2].

10. Prove that the logarithmic function is increasing on (0, ∞).

SOLUTION Let f(x) = log x. Then f′(x) = 1/x. For x > 0, 1/x > 0, so f′(x) > 0 throughout (0, ∞). Hence log x is increasing on (0, ∞).

11. Prove that the function f given by f(x) = x2 − x + 1 is neither strictly increasing nor decreasing on (−1, 1).

SOLUTION f′(x) = 2x − 1 = 0 ⇒ x = 1/2, which lies in (−1, 1). For −1 < x < 1/2, f′(x) < 0 (decreasing); for 1/2 < x < 1, f′(x) > 0 (increasing). Since it both decreases and increases on (−1, 1), f is neither strictly increasing nor decreasing there.

12. Which of the following functions are decreasing on (0, π/2)? (A) cos x   (B) cos 2x   (C) cos 3x   (D) tan x

SOLUTION (A) (cos x)′ = −sin x < 0 on (0, π/2) ⇒ decreasing throughout. ✓ (B) (cos 2x)′ = −2sin 2x; for 0 < x < π/2, 0 < 2x < π, sin 2x > 0, so derivative < 0 ⇒ decreasing throughout. ✓ (C) (cos 3x)′ = −3sin 3x; here 3x ranges over (0, 3π/2), where sin 3x changes sign ⇒ not decreasing throughout. (D) (tan x)′ = sec2x > 0 ⇒ increasing, not decreasing. ∴ correct options are (A) and (B).

13. On which of the following intervals is the function f given by f(x) = x100 + sin x − 1 decreasing? (A) (0, 1)   (B) (π/2, π)   (C) (0, π/2)   (D) None of these

SOLUTION f′(x) = 100x99 + cos x. On (0, 1): both 100x99 > 0 and cos x > 0 ⇒ f′ > 0 (increasing). On (π/2, π): 100x99 is large positive (x > 1) and dominates cos x (−1 to 0) ⇒ f′ > 0. On (0, π/2): 100x99 + cos x > 0 ⇒ increasing. So f is not decreasing on any of these. ∴ correct option is (D) None of these.

14. For what values of a the function f given by f(x) = x2 + ax + 1 is increasing on [1, 2]?

SOLUTION f′(x) = 2x + a. For f increasing on [1, 2] we need 2x + a ≥ 0 for all x in [1, 2]. The minimum of 2x on [1, 2] is at x = 1, giving 2 + a ≥ 0 ⇒ a ≥ −2. ∴ a > −2 (a ≥ −2 makes it increasing throughout).

15. Let I be any interval disjoint from [−1, 1]. Prove that the function f given by f(x) = x + 1/x is increasing on I.

SOLUTION f′(x) = 1 − 1/x2 = (x2 − 1)/x2. If I is disjoint from [−1, 1], then |x| > 1, so x2 > 1 and x2 − 1 > 0; also x2 > 0. Hence f′(x) > 0 on I, so f is increasing on I.

16. Prove that the function f given by f(x) = log sin x is increasing on (0, π/2) and decreasing on (π/2, π).

SOLUTION f′(x) = (1/sin x)·cos x = cot x. On (0, π/2): cot x > 0 ⇒ f′(x) > 0 ⇒ increasing. On (π/2, π): cot x < 0 ⇒ f′(x) < 0 ⇒ decreasing.

17. Prove that the function f given by f(x) = log |cos x| is decreasing on (0, π/2) and increasing on (3π/2, 2π).

SOLUTION f′(x) = (1/cos x)·(−sin x) = −tan x. On (0, π/2): tan x > 0 ⇒ f′(x) = −tan x < 0 ⇒ decreasing. On (3π/2, 2π): tan x < 0 ⇒ f′(x) = −tan x > 0 ⇒ increasing.

18. Prove that the function given by f(x) = x3 − 3x2 + 3x − 100 is increasing in R.

SOLUTION f′(x) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2. Since (x − 1)2 ≥ 0, f′(x) ≥ 0 for all x ∈ R (= 0 only at x = 1). Hence f is increasing on R.

19. The interval in which y = x2e−x is increasing is (A) (−∞, ∞)   (B) (−2, 0)   (C) (2, ∞)   (D) (0, 2)

SOLUTION dy/dx = 2xe−x + x2(−e−x) = e−x(2x − x2) = e−x·x(2 − x). e−x > 0 always; x(2 − x) > 0 for 0 < x < 2. ∴ y is increasing on (0, 2) — option (D).

Exercise 6.3 — Maxima and Minima

1. Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = (2x − 1)2 + 3   (ii) f(x) = 9x2 + 12x + 2 (iii) f(x) = −(x − 1)2 + 10   (iv) g(x) = x3 + 1

SOLUTION (i) (2x − 1)2 ≥ 0, least 0 at x = 1/2 ⇒ minimum value = 3; no maximum. (ii) f(x) = 9x2 + 12x + 2 = (3x + 2)2 − 2, least −2 at x = −2/3 ⇒ minimum value = −2; no maximum. (iii) −(x − 1)2 ≤ 0, greatest 0 at x = 1 ⇒ maximum value = 10; no minimum. (iv) g(x) = x3 + 1 is unbounded both ways ⇒ neither maximum nor minimum.

2. Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = |x + 2| − 1   (ii) g(x) = −|x + 1| + 3 (iii) h(x) = sin(2x) + 5   (iv) f(x) = |sin 4x + 3| (v) h(x) = x + 1, x ∈ (−1, 1)

SOLUTION (i) |x + 2| ≥ 0, least 0 at x = −2 ⇒ minimum value = −1; no maximum value. (ii) −|x + 1| ≤ 0, greatest 0 at x = −1 ⇒ maximum value = 3; no minimum value. (iii) sin 2x ∈ [−1, 1] ⇒ minimum value = 4, maximum value = 6. (iv) sin 4x ∈ [−1, 1] so sin 4x + 3 ∈ [2, 4] (always positive) ⇒ |sin 4x + 3| ∈ [2, 4] ⇒ minimum = 2, maximum = 4. (v) On the open interval (−1, 1) the increasing linear function attains neither maximum nor minimum value.

3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (i) f(x) = x2   (ii) g(x) = x3 − 3x (iii) h(x) = sin x + cos x, 0 < x < π/2 (iv) f(x) = sin x − cos x, 0 < x < 2π (v) f(x) = x3 − 6x2 + 9x + 15   (vi) g(x) = x/2 + 2/x, x > 0 (vii) g(x) = 1/(x2 + 2)   (viii) f(x) = x√(1 − x), 0 < x < 1

SOLUTION (i) f′(x) = 2x = 0 ⇒ x = 0; f″(0) = 2 > 0 ⇒ local minimum at x = 0, value 0. (ii) g′(x) = 3x2 − 3 = 0 ⇒ x = ±1; g″(x) = 6x. At x = 1, g″ > 0 ⇒ local minimum, value −2; at x = −1, g″ < 0 ⇒ local maximum, value 2. (iii) h′(x) = cos x − sin x = 0 ⇒ x = π/4; h″(x) = −sin x − cos x < 0 there ⇒ local maximum at x = π/4, value √2. (iv) f′(x) = cos x + sin x = 0 ⇒ tan x = −1 ⇒ x = 3π/4, 7π/4. f″(x) = −sin x + cos x. At x = 3π/4, f″ < 0 ⇒ local maximum, value √2; at x = 7π/4, f″ > 0 ⇒ local minimum, value −√2. (v) f′(x) = 3x2 − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1, 3. f″(x) = 6x − 12. At x = 1, f″ < 0 ⇒ local maximum, value 19; at x = 3, f″ > 0 ⇒ local minimum, value 15. (vi) g′(x) = 1/2 − 2/x2 = 0 ⇒ x2 = 4 ⇒ x = 2 (x > 0). g″(x) = 4/x3 > 0 ⇒ local minimum at x = 2, value 2. (vii) g(x) = 1/(x2 + 2); g′(x) = −2x/(x2 + 2)2 = 0 ⇒ x = 0. g′ changes + to − through 0 ⇒ local maximum at x = 0, value 1/2. (viii) f(x) = x√(1 − x); f′(x) = √(1 − x) − x/(2√(1 − x)) = (2 − 3x)/(2√(1 − x)) = 0 ⇒ x = 2/3. Sign of f′ changes + to − ⇒ local maximum at x = 2/3, value (2√3)/9.

4. Prove that the following functions do not have maxima or minima: (i) f(x) = ex   (ii) g(x) = log x (iii) h(x) = x3 + x2 + x + 1

SOLUTION (i) f′(x) = ex > 0 for all x; the derivative is never zero, so no critical point and no maximum/minimum. (ii) g′(x) = 1/x > 0 on the domain (x > 0); never zero, so no maximum/minimum. (iii) h′(x) = 3x2 + 2x + 1; its discriminant = 4 − 12 = −8 < 0, so h′(x) > 0 for all x and never zero. Hence no maximum/minimum.

5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f(x) = x3, x ∈ [−2, 2]   (ii) f(x) = sin x + cos x, x ∈ [0, π] (iii) f(x) = 4x − (1/2)x2, x ∈ [−2, 9/2]   (iv) f(x) = (x − 1)2 + 3, x ∈ [−3, 1]

SOLUTION (i) f′(x) = 3x2 = 0 ⇒ x = 0. Values: f(−2) = −8, f(0) = 0, f(2) = 8 ⇒ absolute max 8, absolute min −8. (ii) f′(x) = cos x − sin x = 0 ⇒ x = π/4. Values: f(0) = 1, f(π/4) = √2, f(π) = −1 ⇒ absolute max √2, absolute min −1. (iii) f′(x) = 4 − x = 0 ⇒ x = 4. Values: f(−2) = −8 − 2 = −10, f(4) = 16 − 8 = 8, f(9/2) = 18 − 81/8 = 7.875 ⇒ absolute max 8, absolute min −10. (iv) f′(x) = 2(x − 1) = 0 ⇒ x = 1. Values: f(−3) = 16 + 3 = 19, f(1) = 3 ⇒ absolute max 19, absolute min 3.

6. Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 − 72x − 18x2.

SOLUTION p′(x) = −72 − 36x = 0 ⇒ x = −2. p″(x) = −36 < 0 ⇒ maximum. p(−2) = 41 − 72(−2) − 18(4) = 41 + 144 − 72 = 113. Maximum profit = 113 units.

7. Find both the maximum value and the minimum value of 3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3].

SOLUTION f′(x) = 12x3 − 24x2 + 24x − 48 = 12(x3 − 2x2 + 2x − 4) = 12(x − 2)(x2 + 2) = 0 ⇒ x = 2 (x2 + 2 ≠ 0). Values: f(0) = 25; f(2) = 48 − 64 + 48 − 96 + 25 = −39; f(3) = 243 − 216 + 108 − 144 + 25 = 16. ∴ maximum value 25 at x = 0, minimum value −39 at x = 2.

8. At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

SOLUTION sin 2x is maximum (= 1) when 2x = π/2 or 2x = 5π/2, i.e. x = π/4 or x = 5π/4. ∴ maximum value 1 occurs at x = π/4 and x = 5π/4.

9. What is the maximum value of the function sin x + cos x?

SOLUTION sin x + cos x = √2 sin(x + π/4), whose maximum is √2. ∴ maximum value = √2 (at x = π/4).

10. Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].

SOLUTION f′(x) = 6x2 − 24 = 6(x2 − 4) = 0 ⇒ x = ±2. On [1, 3]: x = 2 lies inside. f(1) = 85, f(2) = 16 − 48 + 107 = 75, f(3) = 54 − 72 + 107 = 89 ⇒ maximum 89 at x = 3. On [−3, −1]: x = −2 lies inside. f(−3) = −54 + 72 + 107 = 125, f(−2) = −16 + 48 + 107 = 139, f(−1) = −2 + 24 + 107 = 129 ⇒ maximum 139 at x = −2.

11. It is given that at x = 1, the function x4 − 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

SOLUTION f′(x) = 4x3 − 124x + a. For a maximum (interior point) at x = 1, f′(1) = 0. 4(1) − 124(1) + a = 0 ⇒ a − 120 = 0 ⇒ a = 120.

12. Find the maximum and minimum values of x + sin 2x on [0, 2π].

SOLUTION f′(x) = 1 + 2cos 2x = 0 ⇒ cos 2x = −1/2 ⇒ 2x = 2π/3, 4π/3, 8π/3, 10π/3 ⇒ x = π/3, 2π/3, 4π/3, 5π/3. Evaluate f at these and the end points; the overall extremes occur at the end points: f(0) = 0 and f(2π) = 2π + sin 4π = 2π. ∴ maximum value 2π at x = 2π and minimum value 0 at x = 0.

13. Find two numbers whose sum is 24 and whose product is as large as possible.

SOLUTION Let numbers be x and 24 − x. Product P = x(24 − x) = 24x − x2. P′(x) = 24 − 2x = 0 ⇒ x = 12; P″ = −2 < 0 ⇒ maximum. ∴ the numbers are 12 and 12.

14. Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.

SOLUTION x = 60 − y. Let P = (60 − y)y3 = 60y3 − y4. P′(y) = 180y2 − 4y3 = 4y2(45 − y) = 0 ⇒ y = 45 (y > 0). P″ < 0 there ⇒ maximum. ∴ y = 45, x = 15 ⇒ x = 15, y = 45.

15. Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.

SOLUTION x + y = 35, so x = 35 − y. Let P = (35 − y)2y5. log P = 2log(35 − y) + 5log y; differentiating and setting to 0: −2/(35 − y) + 5/y = 0 ⇒ 5(35 − y) = 2y ⇒ 175 = 7y ⇒ y = 25. ∴ y = 25, x = 10 ⇒ x = 10, y = 25 (this gives the maximum).

16. Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

SOLUTION Let numbers be x and 16 − x. S = x3 + (16 − x)3. S′(x) = 3x2 − 3(16 − x)2 = 0 ⇒ x2 = (16 − x)2 ⇒ x = 16 − x ⇒ x = 8. S″ > 0 there ⇒ minimum. ∴ the numbers are 8 and 8.

17. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

SOLUTION Let side cut = x. Box has base (18 − 2x) and height x. V = x(18 − 2x)2. V′(x) = (18 − 2x)2 + x·2(18 − 2x)(−2) = (18 − 2x)(18 − 6x) = 0 ⇒ x = 9 or x = 3. (x = 9 gives zero base, reject.) V″(3) < 0 ⇒ maximum. ∴ side to be cut = 3 cm.

18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

SOLUTION Let side cut = x. V = x(45 − 2x)(24 − 2x) = 4x3 − 138x2 + 1080x. V′(x) = 12x2 − 276x + 1080 = 12(x2 − 23x + 90) = 12(x − 5)(x − 18) = 0 ⇒ x = 5 or 18. (x = 18 makes 24 − 2x negative, reject.) V″(5) = 24(5) − 276 = −156 < 0 ⇒ maximum. ∴ side to be cut = 5 cm.

19. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

SOLUTION Let the circle have diameter 2a (radius a). A rectangle with sides x and y inscribed in it has x2 + y2 = (2a)2 = 4a2, so y = √(4a2 − x2). Area A = x√(4a2 − x2). Maximise A2 = x2(4a2 − x2). d(A2)/dx = 8a2x − 4x3 = 0 ⇒ x2 = 2a2. Then y2 = 4a2 − 2a2 = 2a2 = x2, so x = y. The second derivative is negative, confirming a maximum. Hence the rectangle of maximum area is a square.

20. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

SOLUTION Let radius r, height h, fixed surface S = 2πr2 + 2πrh, so h = (S − 2πr2)/(2πr). Volume V = πr2h = (r/2)(S − 2πr2) = (Sr/2) − πr3. dV/dr = S/2 − 3πr2 = 0 ⇒ S = 6πr2. Then h = (6πr2 − 2πr2)/(2πr) = 2r = diameter. d2V/dr2 = −6πr < 0 ⇒ maximum. Hence height = diameter of base.

21. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

SOLUTION V = πr2h = 100, so h = 100/(πr2). Surface S = 2πr2 + 2πrh = 2πr2 + 200/r. dS/dr = 4πr − 200/r2 = 0 ⇒ r3 = 50/π ⇒ r = (50/π)1/3. d2S/dr2 = 4π + 400/r3 > 0 ⇒ minimum. Then h = 100/(πr2) = 2r = 2(50/π)1/3. ∴ radius = (50/π)1/3 cm and height = 2(50/π)1/3 cm.

22. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

SOLUTION Let one piece (circle) = x m, the other (square) = (28 − x) m. Circle: 2πr = x ⇒ r = x/(2π), area = x2/(4π). Square: side = (28 − x)/4, area = (28 − x)2/16. A = x2/(4π) + (28 − x)2/16. dA/dx = x/(2π) − (28 − x)/8 = 0 ⇒ 8x = 2π(28 − x) ⇒ x(8 + 2π) = 56π ⇒ x = 28π/(4 + π). ∴ circle piece = 28π/(π + 4) cm and square piece = 112/(π + 4) cm (d2A/dx2 > 0, so area is minimum).

23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

SOLUTION Let the cone have base radius r and height h, with apex and base on a sphere of radius R. If the centre is at distance (h − R) from the base, then r2 = R2 − (h − R)2 = 2Rh − h2. V = (1/3)πr2h = (1/3)π(2Rh2 − h3). dV/dh = (1/3)π(4Rh − 3h2) = 0 ⇒ h = 4R/3. Then r2 = 2R(4R/3) − (4R/3)2 = 8R2/3 − 16R2/9 = 8R2/9. V = (1/3)π(8R2/9)(4R/3) = 32πR3/81. Volume of sphere = (4/3)πR3; ratio = (32πR3/81) ÷ (4πR3/3) = 8/27. Hence the largest cone is 8/27 of the sphere’s volume.

24. Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.

SOLUTION Volume V = (1/3)πr2h is fixed, so h = 3V/(πr2). Curved surface L = πrl where l = √(r2 + h2). Minimise L2 = π2r2(r2 + h2) = π2r4 + π2r2(9V2/(π2r4)) = π2r4 + 9V2/r2. d/dr = 4π2r3 − 18V2/r3 = 0 ⇒ r6 = 9V2/(2π2). Substituting V = (1/3)πr2h gives r6 = (9/2π2)(π2r4h2/9) = r4h2/2 ⇒ h2 = 2r2 ⇒ h = √2 r.

25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan−1√2.

SOLUTION Let slant height l be fixed and semi-vertical angle θ. Then r = l sinθ, h = l cosθ. V = (1/3)πr2h = (1/3)πl3sin2θ cosθ. dV/dθ = (1/3)πl3(2sinθ cos2θ − sin3θ) = (1/3)πl3sinθ(2cos2θ − sin2θ). Set 2cos2θ = sin2θ ⇒ tan2θ = 2 ⇒ tanθ = √2. The second derivative is negative, so V is maximum. Hence θ = tan−1√2.

26. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin−1(1/3).

SOLUTION Total surface S = πr2 + πrl is fixed. Using V2 = (1/9)π2r4h2 with l2 = r2 + h2, and maximising under the constraint, one obtains the condition S = 4πr2, i.e. πrl = 3πr2, so l = 3r. The semi-vertical angle θ satisfies sinθ = r/l = r/(3r) = 1/3. ∴ θ = sin−1(1/3).

27. The point on the curve x2 = 2y which is nearest to the point (0, 5) is (A) (2√2, 4)   (B) (2√2, 0)   (C) (0, 0)   (D) (2, 2)

SOLUTION Point (x, y) with x2 = 2y. Distance squared D = x2 + (y − 5)2 = 2y + (y − 5)2. dD/dy = 2 + 2(y − 5) = 0 ⇒ y = 4, so x2 = 8 ⇒ x = 2√2. ∴ nearest point is (2√2, 4) — option (A).

28. For all real values of x, the minimum value of (1 − x + x2)/(1 + x + x2) is (A) 0   (B) 1   (C) 3   (D) 1/3

SOLUTION Let y = (1 − x + x2)/(1 + x + x2). dy/dx = [(−1 + 2x)(1 + x + x2) − (1 − x + x2)(1 + 2x)]/(1 + x + x2)2. Numerator simplifies to 2x2 − 2 = 2(x2 − 1) = 0 ⇒ x = ±1. At x = 1, y = 1/3; at x = −1, y = 3. ∴ minimum value is 1/3 — option (D).

29. The maximum value of [x(x − 1) + 1]1/3, 0 ≤ x ≤ 1 is (A) (1/3)1/3   (B) 1/2   (C) 1   (D) 0

SOLUTION Let g(x) = x(x − 1) + 1 = x2 − x + 1. g′(x) = 2x − 1 = 0 ⇒ x = 1/2, giving g = 3/4 (a minimum). At end points g(0) = 1 and g(1) = 1, which are the largest values on [0, 1]. So the cube root [g(x)]1/3 is largest = 11/3 = 1. ∴ maximum value is 1 — option (C).

Miscellaneous Exercise on Chapter 6

1. Show that the function given by f(x) = (log x)/x has maximum at x = e.

SOLUTION f′(x) = (1 − log x)/x2. Setting f′(x) = 0 gives log x = 1 ⇒ x = e. For x < e, log x < 1 ⇒ f′ > 0; for x > e, log x > 1 ⇒ f′ < 0. The sign changes + to −, so x = e is a maximum. ∴ f has a maximum at x = e (maximum value 1/e).

2. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

SOLUTION Let each equal side = a. Height = √(a2 − b2/4), so area A = (b/2)√(a2 − b2/4). dA/dt = (b/2)·(a/√(a2 − b2/4))·(da/dt), with da/dt = −3. When a = b: √(b2 − b2/4) = (b√3)/2. So dA/dt = (b/2)·(b/((b√3)/2))·(−3) = (b/2)·(2/√3)·(−3) = −√3 b. ∴ the area is decreasing at the rate of √3 b cm2/s.

3. Find the intervals in which the function f given by f(x) = (4 sin x − 2x − x cos x)/(2 + cos x) is (i) increasing (ii) decreasing.

SOLUTION On simplification f′(x) = [cos x(4 − cos x)]/(2 + cos x)2. The denominator and (4 − cos x) are always positive, so the sign of f′ follows cos x. (i) Increasing where cos x > 0, i.e. on 0 ≤ x < π/2 and 3π/2 < x < 2π. (ii) Decreasing where cos x < 0, i.e. on π/2 < x < 3π/2.

4. Find the intervals in which the function f given by f(x) = x3 + 1/x3, x ≠ 0 is (i) increasing (ii) decreasing.

SOLUTION f′(x) = 3x2 − 3/x4 = 3(x6 − 1)/x4. Since x4 > 0, the sign follows (x6 − 1). x6 − 1 > 0 ⇔ |x| > 1; x6 − 1 < 0 ⇔ 0 < |x| < 1. (i) Increasing for x < −1 and x > 1. (ii) Decreasing for −1 < x < 1 (x ≠ 0).

5. Find the maximum area of an isosceles triangle inscribed in the ellipse x2/a2 + y2/b2 = 1 with its vertex at one end of the major axis.

SOLUTION Take vertex at (a, 0) and base vertices at (a cosθ, ±b sinθ). Base = 2b sinθ, height = a − a cosθ. Area A = (1/2)(2b sinθ)(a − a cosθ) = ab sinθ(1 − cosθ). dA/dθ = ab(cosθ − cos2θ … ) leads to cosθ = −1/2 in the relevant range, giving θ = 2π/3. Substituting gives the maximum area A = (3√3/4) ab square units.

6. A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

SOLUTION Let base be x by y. Volume = 2xy = 8 ⇒ xy = 4, so y = 4/x. Base area = xy = 4 (cost 70×4 = 280). Sides area = 2×(2x) + 2×(2y) = 4(x + y). Cost C = 280 + 45·4(x + y) = 280 + 180(x + 4/x). dC/dx = 180(1 − 4/x2) = 0 ⇒ x = 2, y = 2. C = 280 + 180(2 + 2) = 280 + 720 = Rs 1000.

7. The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

SOLUTION Let circle radius r, square side s. Then 2πr + 4s = k, so s = (k − 2πr)/4. Total area A = πr2 + s2. dA/dr = 2πr + 2s(ds/dr) = 2πr + 2s(−π/2) = 2πr − πs = 0 ⇒ s = 2r. d2A/dr2 > 0, so the area is least when s = 2r, i.e. side of square = double the radius.

8. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

SOLUTION Let rectangle width = 2x (= diameter), height = y. Perimeter: 2x + 2y + πx = 10 ⇒ y = (10 − 2x − πx)/2. Area A = 2xy + (1/2)πx2 = x(10 − 2x − πx) + (π/2)x2 = 10x − 2x2 − (π/2)x2. dA/dx = 10 − 4x − πx = 0 ⇒ x = 10/(π + 4). So width 2x = 20/(π + 4) m, and height y = 10/(π + 4) m. ∴ length (width) = 20/(π + 4) m and breadth (height) = 10/(π + 4) m.

9. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is (a2/3 + b2/3)3/2.

SOLUTION Let the hypotenuse make angle θ with one side. The hypotenuse length L = a/sinθ + b/cosθ (sum of the two segments). dL/dθ = −a cosθ/sin2θ + b sinθ/cos2θ = 0 ⇒ tan3θ = a/b ⇒ tanθ = (a/b)1/3. Then sinθ = a1/3/√(a2/3 + b2/3), cosθ = b1/3/√(a2/3 + b2/3). Substituting gives L = (a2/3 + b2/3)3/2, which is the minimum.

10. Find the points at which the function f given by f(x) = (x − 2)4(x + 1)3 has (i) local maxima   (ii) local minima (iii) point of inflexion

SOLUTION f′(x) = 4(x − 2)3(x + 1)3 + 3(x − 2)4(x + 1)2 = (x − 2)3(x + 1)2[4(x + 1) + 3(x − 2)] = (x − 2)3(x + 1)2(7x − 2). Critical points: x = 2, x = −1, x = 2/7. (i) At x = 2/7, f′ changes + to − ⇒ local maxima at x = 2/7. (ii) At x = 2, f′ changes − to + ⇒ local minima at x = 2. (iii) At x = −1, the squared factor gives no sign change ⇒ point of inflexion at x = −1.

11. Find the absolute maximum and minimum values of the function f given by f(x) = cos2x + sin x, x ∈ [0, π].

SOLUTION f(x) = (1 − sin2x) + sin x = 1 + sin x − sin2x. f′(x) = cos x(1 − 2sin x) = 0 ⇒ cos x = 0 (x = π/2) or sin x = 1/2 (x = π/6, 5π/6). Values: f(0) = 1; f(π/6) = 1 + 1/2 − 1/4 = 5/4; f(π/2) = 1 + 1 − 1 = 1; f(5π/6) = 5/4; f(π) = 1. ∴ absolute maximum = 5/4 (at x = π/6 and 5π/6) and absolute minimum = 1.

12. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3.

SOLUTION Let cone height = h, base radius = R. With centre at distance (h − r) below the apex, R2 = r2 − (h − r)2 = 2rh − h2. V = (1/3)πR2h = (1/3)π(2rh2 − h3). dV/dh = (1/3)π(4rh − 3h2) = 0 ⇒ h = 4r/3. d2V/dh2 = (1/3)π(4r − 6h) < 0 at h = 4r/3 ⇒ maximum. Hence altitude = 4r/3.

13. Let f be a function defined on [a, b] such that f′(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).

SOLUTION Take any x1 < x2 in (a, b). By the Mean Value Theorem there is c ∈ (x1, x2) with f(x2) − f(x1) = f′(c)(x2 − x1). Since f′(c) > 0 and (x2 − x1) > 0, we get f(x2) − f(x1) > 0, i.e. f(x1) < f(x2). As this holds for all such pairs, f is increasing on (a, b).

14. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also find the maximum volume.

SOLUTION Let cylinder height = 2h (so half-height h), radius r. Then r2 + h2 = R2, r2 = R2 − h2. V = πr2(2h) = 2πh(R2 − h2). dV/dh = 2π(R2 − 3h2) = 0 ⇒ h = R/√3, so height = 2h = 2R/√3. Maximum volume = 2π(R/√3)(R2 − R2/3) = 2π(R/√3)(2R2/3) = 4πR3/(3√3) cubic units.

15. Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is (4/27)πh3tan2α.

SOLUTION Cone base radius = h tanα. If cylinder has height y and radius x, by similar triangles x = (h − y)tanα. V = πx2y = πtan2α(h − y)2y. dV/dy = πtan2α[(h − y)2 − 2y(h − y)] = πtan2α(h − y)(h − 3y) = 0 ⇒ y = h/3. So cylinder height = h/3 (one-third of cone). Maximum volume = πtan2α(2h/3)2(h/3) = πtan2α(4h2/9)(h/3) = (4/27)πh3tan2α.

16. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of (A) 1 m/h   (B) 0.1 m/h   (C) 1.1 m/h   (D) 0.5 m/h

SOLUTION V = πr2h with r = 10, so V = 100πh. dV/dt = 100π(dh/dt) = 314. dh/dt = 314/(100π) = 314/314 = 1 m/h (using π ≈ 3.14). ∴ correct option is (A) 1 m/h.

Common Mistakes to Avoid

Watch out for these

  • Forgetting the chain rule in related-rate problems — always link the two rates through the common variable (usually time).
  • Mixing up diameter and radius (Q.13, sphere/balloon problems) before substituting into the volume formula.
  • Treating a point where f′(c) = 0 as automatically a maximum or minimum — check the sign change or the second derivative; it may be a point of inflexion.
  • For absolute maxima/minima on a closed interval, forgetting to evaluate f at the end points as well as the critical points.
  • Rejecting valid roots, or keeping geometrically impossible ones (e.g. a negative length or a cut larger than the sheet) in optimisation word problems.
  • In increasing/decreasing problems, ignoring the domain restriction (e.g. x > −1, x ≠ 0) when stating the intervals.

Practice MCQs & Assertion–Reason

1. The rate of change of the area of a circle with respect to its radius at r = 5 cm is:

(a) 5π    (b) 10π    (c) 25π    (d) 2π

2. The marginal revenue when R(x) = 3x2 + 36x + 5 and x = 5 is:

(a) 36    (b) 66    (c) 96    (d) 126

3. The function f(x) = 2x3 − 3x2 − 36x + 7 is decreasing in:

(a) (−∞, −2)    (b) (−2, 3)    (c) (3, ∞)    (d) R

4. The function f(x) = e2x is:

(a) decreasing on R    (b) increasing on R    (c) constant    (d) neither

5. The local minimum value of g(x) = x3 − 3x is:

(a) 2    (b) −2    (c) 0    (d) 1

6. The absolute maximum value of f(x) = x3 on [−2, 2] is:

(a) 2    (b) 4    (c) 8    (d) −8

7. Two numbers whose sum is 24 and product is maximum are:

(a) 10, 14    (b) 12, 12    (c) 8, 16    (d) 6, 18

8. The maximum value of sin x + cos x is:

(a) 1    (b) 2    (c) √2    (d) 1/√2

9. The interval in which y = x2e−x is increasing is:

(a) (−∞, ∞)    (b) (−2, 0)    (c) (2, ∞)    (d) (0, 2)

10. For f(x) = x3, x = 0 is a:

(a) local maximum    (b) local minimum    (c) point of inflexion    (d) end point

Answer key: 1-(b), 2-(b), 3-(b), 4-(b), 5-(b), 6-(c), 7-(b), 8-(c), 9-(d), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: f(x) = 3x + 17 is increasing on R.

Reason: A function is increasing on an interval if its derivative is positive there.

A-R 2. Assertion: f(x) = x3 − 3x2 + 3x − 100 is increasing on R.

Reason: f′(x) = 3(x − 1)2 ≥ 0 for all real x.

A-R 3. Assertion: x = 0 is a point of local minimum of f(x) = x3.

Reason: f′(0) = 0 for f(x) = x3.

A-R 4. Assertion: On a closed interval [a, b], a continuous function attains both its absolute maximum and absolute minimum.

Reason: Absolute extrema can occur only at end points of the interval.

A-R 5. Assertion: The maximum value of sin x + cos x is √2.

Reason: sin x + cos x can be written as √2 sin(x + π/4).

Answer key: 1-(A), 2-(A), 3-(D), 4-(C), 5-(A).

Quick Revision Summary

  • dy/dx is the rate of change of y with respect to x; for parametric quantities use dy/dx = (dy/dt)/(dx/dt).
  • Marginal cost = dC/dx and marginal revenue = dR/dx.
  • f is increasing where f′(x) > 0 and decreasing where f′(x) < 0.
  • Critical points are where f′(c) = 0 or f is not differentiable.
  • First derivative test: sign change + → − means maximum; − → + means minimum; no change means inflexion.
  • Second derivative test: f′(c) = 0 with f″(c) < 0 ⇒ maximum; f″(c) > 0 ⇒ minimum; f″(c) = 0 ⇒ test fails.
  • For absolute extrema on [a, b], compare f at all critical points and at both end points.

How to score full marks in this chapter

Write the formula and the relationship between variables clearly before differentiating in word problems — examiners award marks for correct set-up. Always state which test you are using (first or second derivative) and justify the maximum/minimum with a sign check or the sign of f″. In absolute-extrema questions, list the value of f at every critical point and end point in a small table so no case is missed, and discard geometrically impossible roots with a one-line reason.

Frequently Asked Questions

What is Class 12 Maths Chapter 6 Application of Derivatives about?

Chapter 6 applies the derivative to three areas: finding the rate of change of quantities (including related rates and marginal cost/revenue), determining where a function is increasing or decreasing using the sign of f′(x), and finding local and absolute maxima and minima using the first and second derivative tests.

How many exercises are there in Class 12 Maths Chapter 6?

There are three numbered exercises — Exercise 6.1 (rate of change), Exercise 6.2 (increasing and decreasing functions) and Exercise 6.3 (maxima and minima) — plus a Miscellaneous Exercise on Chapter 6, all solved on this page.

What is the difference between local and absolute maximum?

A local (relative) maximum is the largest value of the function in a small neighbourhood around a point, while an absolute (global) maximum is the largest value over the whole interval considered. On a closed interval, absolute extrema are found by comparing the function’s values at all critical points and at the two end points.

Are these Class 12 Maths Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Class 12 Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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