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NCERT Solutions · Class 12 · Chapter 6: Application of Derivatives

NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise

The Miscellaneous Exercise on Application of Derivatives. Mixed practice: set up the function to optimise from the geometry, find critical points, confirm max or min with a derivative test, and read the question for what is asked (a dimension, a value, or an angle). In Q16 the numbers are built on \(\pi = 3.14\) (314 = 100 × 3.14), so take \(\pi = 3.14\) there, as the options do.

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Miscellaneous Exercise questions and solutions

Miscellaneous Exercise, Question 1

Show that \(f(x) = \dfrac{\log x}{x}\) has a maximum at \(x = e\).
Show solution
  1. \(f'(x) = \dfrac{1 - \log x}{x^2} = 0\) at \(x = e\): positive before, negative after.
Answer: Shown (maximum value \(\tfrac1e\))

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Miscellaneous Exercise, Question 2

An isosceles triangle has fixed base b; its equal sides decrease at 3 cm/s. How fast is the area decreasing when the equal sides equal the base?
Show solution
  1. Equal sides a: height \(\sqrt{a^2 - \tfrac{b^2}{4}}\), area \(A = \tfrac{b}{4}\sqrt{4a^2 - b^2}\).
  2. \[\begin{aligned}\dfrac{dA}{dt} &= \dfrac{b}{4} \cdot \dfrac{4a}{\sqrt{4a^2 - b^2}}\dfrac{da}{dt} \\ &= \dfrac{ab}{\sqrt{4a^2 - b^2}}(-3)\end{aligned}\]
  3. At \(a = b\): \(\dfrac{-3b^2}{\sqrt3 b} = -\sqrt3 b\).
Answer: Decreasing at \(\sqrt3\,b\) cm\(^2\)/s

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Miscellaneous Exercise, Question 3

Find where \(f(x) = \dfrac{4\sin x - 2x - x\cos x}{2 + \cos x}\) is (i) increasing (ii) decreasing (take \(0 \le x \le 2\pi\)).
Show solution
  1. \(f(x) = \dfrac{4\sin x}{2 + \cos x} - x\), so \[f'(x) = \dfrac{\cos x(4 - \cos x)}{(2 + \cos x)^2}\] (as in Ex 6.2 Q9).
  2. \(4 - \cos x > 0\), so the sign of \(f'\) is the sign of \(\cos x\).
Answer: (i) Increasing on \(\left(0, \tfrac{\pi}{2}\right)\) and \(\left(\tfrac{3\pi}{2}, 2\pi\right)\) (ii) decreasing on \[\left(\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right)\]

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Miscellaneous Exercise, Question 4

Find where \(f(x) = x^3 + \dfrac{1}{x^3}\), \(x \ne 0\), is (i) increasing (ii) decreasing.
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  1. \[\begin{aligned}f'(x) &= 3x^2 - \dfrac{3}{x^4} \\ &= \dfrac{3(x^6 - 1)}{x^4}\end{aligned}\]: positive when \(|x| > 1\), negative when \(0 < |x| < 1\).
Answer: (i) \((-\infty, -1)\) and \((1, \infty)\) (ii) \((-1, 0)\) and \((0, 1)\)

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Miscellaneous Exercise, Question 5

Find the maximum area of an isosceles triangle inscribed in \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with its vertex at one end of the major axis.
Show solution
  1. Vertex \((a, 0)\) (take \(a > b\)); the base is the chord through \((a\cos\theta, \pm b\sin\theta)\).
  2. \[\begin{aligned}A &= \tfrac12 \cdot 2b\sin\theta \cdot (a - a\cos\theta) \\ &= ab\sin\theta(1 - \cos\theta)\end{aligned}\]
  3. \[\begin{aligned}A'(\theta) &= ab(\cos\theta - \cos 2\theta) \\ &= ab(1 - \cos\theta)(1 + 2\cos\theta)\end{aligned}\], \(0 < \theta < \pi\). Since \(1 - \cos\theta > 0\), \(A' = 0\) only at \(\cos\theta = -\tfrac12\), i.e. \(\theta = \tfrac{2\pi}{3}\).
  4. \(A' > 0\) before \(\tfrac{2\pi}{3}\) and \(A' < 0\) after, so this is the maximum: \[A = ab \cdot \tfrac{\sqrt3}{2} \cdot \tfrac32\]
Answer: \(\dfrac{3\sqrt3}{4}ab\)

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Miscellaneous Exercise, Question 6

An open tank with rectangular base has depth 2 m and volume 8 m\(^3\). The base costs Rs 70/m\(^2\) and the sides Rs 45/m\(^2\). Find the least cost.
Show solution
  1. Base \(l \times w\) with \(2lw = 8\), so \(lw = 4\).
  2. Cost \[= 70(4) + 45 \cdot 2 \cdot 2(l + w) = 280 + 180\left(l + \dfrac4l\right)\]
  3. \[\begin{aligned}&\dfrac{d}{dl}\left(l + \dfrac4l\right) = 1 - \dfrac{4}{l^2} = 0 \\ \Rightarrow\ &l = 2 = w\end{aligned}\]: cost \(280 + 180 \cdot 4\).
Answer: Rs 1000

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Miscellaneous Exercise, Question 7

The perimeters of a circle and a square add to k. Prove the sum of their areas is least when the side of the square is twice the radius.
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  1. \[\begin{aligned}&2\pi r + 4a = k \\ \Rightarrow\ &a = \dfrac{k - 2\pi r}{4}\end{aligned}\]; \(A = \pi r^2 + \dfrac{(k - 2\pi r)^2}{16}\).
  2. \[\begin{aligned}&A'(r) = 2\pi r - \dfrac{\pi(k - 2\pi r)}{4} = 0 \\ \Rightarrow\ &k - 2\pi r = 8r \\ \Rightarrow\ &a = 2r\end{aligned}\]; \(A'' = 2\pi + \tfrac{\pi^2}{2} > 0\).
Answer: Proved

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Miscellaneous Exercise, Question 8

A window is a rectangle topped by a semicircle, total perimeter 10 m. Find the dimensions that let in the most light.
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  1. Width x (semicircle radius \(\tfrac{x}{2}\)), rectangle height y: \(x + 2y + \dfrac{\pi x}{2} = 10\).
  2. Area \[\begin{aligned}A &= xy + \dfrac{\pi x^2}{8} \\ &= 5x - \dfrac{x^2}{2} - \dfrac{\pi x^2}{8}\end{aligned}\]; \[\begin{aligned}&A' = 5 - x - \dfrac{\pi x}{4} = 0 \\ \Rightarrow\ &x = \dfrac{20}{\pi + 4}\end{aligned}\]
  3. Then \[\begin{aligned}y &= \dfrac{10 - x\left(1 + \frac{\pi}{2}\right)}{2} \\ &= \dfrac{10}{\pi + 4}\end{aligned}\]
Answer: Width \(\dfrac{20}{\pi + 4}\) m, height of the rectangular part \(\dfrac{10}{\pi + 4}\) m

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Miscellaneous Exercise, Question 9

A point on the hypotenuse of a right triangle is at distances a and b from the two legs. Show the least hypotenuse is \(\left(a^{2/3} + b^{2/3}\right)^{3/2}\).
Show solution
  1. Let the hypotenuse make angle \(\theta\) with the leg that is b away: the two parts of the hypotenuse are \(a\sec\theta\) and \(b\csc\theta\), so \(L = a\sec\theta + b\csc\theta\).
  2. \[\begin{aligned}&L' = a\sec\theta\tan\theta - b\csc\theta\cot\theta = 0 \\ \Rightarrow\ &\tan^3\theta = \dfrac{b}{a}\end{aligned}\]; L is least there (\(L \to \infty\) at both ends).
  3. With \[\tan\theta = \left(\tfrac{b}{a}\right)^{1/3}\]: \[\sec\theta = \dfrac{\sqrt{a^{2/3} + b^{2/3}}}{a^{1/3}}\], \[\csc\theta = \dfrac{\sqrt{a^{2/3} + b^{2/3}}}{b^{1/3}}\], so \[L = \sqrt{a^{2/3} + b^{2/3}}\left(a^{2/3} + b^{2/3}\right)\]
Answer: Shown

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Miscellaneous Exercise, Question 10

Find the points where \(f(x) = (x - 2)^4(x + 1)^3\) has (i) a local maximum (ii) a local minimum (iii) a point of inflexion.
Show solution
  1. \[\begin{aligned}f'(x) &= 4(x - 2)^3(x + 1)^3 + 3(x - 2)^4(x + 1)^2 \\ &= (x - 2)^3(x + 1)^2(7x - 2)\end{aligned}\]
  2. At \(\tfrac27\): \(f'\) goes + to − (max). At 2: − to + (min). At \(-1\): \((x + 1)^2\) keeps its sign, so no extremum; f' does not change sign, a point of inflexion.
Answer: (i) \(x = \tfrac27\) (ii) \(x = 2\) (iii) \(x = -1\)

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Miscellaneous Exercise, Question 11

Find the absolute maximum and minimum of \(f(x) = \cos^2 x + \sin x\), \(x \in [0, \pi]\).
Show solution
  1. \(f'(x) = \cos x(1 - 2\sin x) = 0\): \[x = \tfrac{\pi}{6}, \tfrac{\pi}{2}, \tfrac{5\pi}{6}\]
  2. \(f(0) = 1\), \[\begin{aligned}f\left(\tfrac{\pi}{6}\right) &= \tfrac34 + \tfrac12 \\ &= \tfrac54\end{aligned}\], \(f\left(\tfrac{\pi}{2}\right) = 1\), \(f\left(\tfrac{5\pi}{6}\right) = \tfrac54\), \(f(\pi) = 1\).
Answer: Maximum \(\tfrac54\) (at \(\tfrac{\pi}{6}\) and \(\tfrac{5\pi}{6}\)), minimum 1 (at 0, \(\tfrac{\pi}{2}\), \(\pi\))

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Miscellaneous Exercise, Question 12

Show the altitude of the right circular cone of maximum volume inscribed in a sphere of radius r is \(\tfrac{4r}{3}\).
Show solution
  1. As in Ex 6.3 Q23: height \(h = r + x\), base radius\(^2\) \(r^2 - x^2\), \(V = \tfrac{\pi}{3}(r^2 - x^2)(r + x)\).
  2. \[\begin{aligned}&V' = \tfrac{\pi}{3}(r + x)(r - 3x) = 0 \\ \Rightarrow\ &x = \tfrac{r}{3}\end{aligned}\] (maximum), so \(h = \tfrac{4r}{3}\).
Answer: Shown

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Miscellaneous Exercise, Question 13

f is defined on \([a, b]\) with \(f'(x) > 0\) for all \(x \in (a, b)\). Prove f is increasing on \((a, b)\).
Show solution
  1. Take \(a < x_1 < x_2 < b\). f is differentiable, hence continuous, on \([x_1, x_2]\), so by the mean value theorem there is c in \((x_1, x_2)\) with \(f(x_2) - f(x_1) = f'(c)(x_2 - x_1)\).
  2. \(f'(c) > 0\) and \(x_2 - x_1 > 0\), so \(f(x_2) > f(x_1)\).
Answer: Proved

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Miscellaneous Exercise, Question 14

Show the height of the cylinder of maximum volume inscribed in a sphere of radius R is \(\tfrac{2R}{\sqrt3}\), and find that volume.
Show solution
  1. Height h, radius \(\rho\) with \(\rho^2 = R^2 - \tfrac{h^2}{4}\): \[V = \pi\left(R^2 - \tfrac{h^2}{4}\right)h\]
  2. \[\begin{aligned}&V'(h) = \pi\left(R^2 - \tfrac{3h^2}{4}\right) = 0 \\ \Rightarrow\ &h = \dfrac{2R}{\sqrt3}\end{aligned}\]; \(V'' = -\tfrac{3\pi h}{2} < 0\).
  3. \[V = \pi \cdot \tfrac{2R^2}{3} \cdot \tfrac{2R}{\sqrt3}\]
Answer: \(h = \dfrac{2R}{\sqrt3}\); maximum volume \(\dfrac{4\pi R^3}{3\sqrt3}\)

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Miscellaneous Exercise, Question 15

Show the cylinder of greatest volume inscribed in a cone of height h and semi-vertical angle \(\alpha\) has height \(\tfrac{h}{3}\) and volume \(\tfrac{4}{27}\pi h^3\tan^2\alpha\).
Show solution
  1. Cylinder height H: its radius is \((h - H)\tan\alpha\), so \(V = \pi(h - H)^2\tan^2\alpha \cdot H\).
  2. \[\begin{aligned}&V'(H) = \pi\tan^2\alpha(h - H)(h - 3H) = 0 \\ \Rightarrow\ &H = \tfrac{h}{3}\end{aligned}\] (\(H = h\) gives zero volume); V rises then falls there.
  3. \[V = \pi \cdot \tfrac{4h^2}{9}\tan^2\alpha \cdot \tfrac{h}{3}\]
Answer: Shown

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Miscellaneous Exercise, Question 16

A cylindrical tank of radius 10 m is filled with wheat at 314 m\(^3\)/h. The depth increases at: (A) 1 m/h (B) 0.1 m/h (C) 1.1 m/h (D) 0.5 m/h
Show solution
  1. \(V = \pi(10)^2h\): \[\begin{aligned}\dfrac{dV}{dt} &= 100\pi\dfrac{dh}{dt} \\ &= 314\end{aligned}\]
  2. Taking \(\pi = 3.14\) (as the numbers intend): \[\begin{aligned}\dfrac{dh}{dt} &= \dfrac{314}{100 \times 3.14} \\ &= 1\end{aligned}\]
Answer: (A) 1 m/h

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Application of Derivatives has 40 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

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Application of Derivatives in our sample papers: Sample paper 1 (questions 11, 23, 37) · Sample paper 2 (questions 11, 12, 23, 33) · Sample paper 3 (questions 10, 23, 37) · Sample paper 4 (questions 11, 12, 23, 29, 33) · Sample paper 5 (questions 10, 33, 37).

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Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.