NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

Exercise 6.3, Question 2

Find the maximum and minimum values, if any.

(i) \(f(x) = |x + 2| - 1\)

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\(|x + 2| \ge 0\), zero at \(x = -2\); unbounded above.

Answer: Minimum \(-1\) (at \(x = -2\)); no maximum

(ii) \(g(x) = -|x + 1| + 3\)

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\(-|x + 1| \le 0\), zero at \(x = -1\).

Answer: Maximum 3 (at \(x = -1\)); no minimum

(iii) \(h(x) = \sin 2x + 5\)

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\(-1 \le \sin 2x \le 1\).

Answer: Maximum 6, minimum 4

(iv) \(f(x) = |\sin 4x + 3|\)

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\(2 \le \sin 4x + 3 \le 4\), positive, so the modulus changes nothing.

Answer: Maximum 4, minimum 2

(v) \(h(x) = x + 1\), \(x \in (-1, 1)\)

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h is increasing and the open interval has no end points: the values approach 0 and 2 but never reach them.

Answer: Neither maximum nor minimum

Exercise 6.3, Question 3

Find the local maxima and minima and the local maximum and minimum values.

(i) \(f(x) = x^2\)

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\(f'(x) = 2x = 0\) at 0; \(f''(0) = 2 > 0\).

Answer: Local minimum at \(x = 0\), value 0

(ii) \(g(x) = x^3 - 3x\)

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\(g'(x) = 3x^2 - 3 = 0\) at \(\pm1\); \(g''(x) = 6x\).
\(g''(-1) < 0\): max, \(g(-1) = 2\). \(g''(1) > 0\): min, \(g(1) = -2\).

Answer: Local maximum at \(x = -1\) (value 2); local minimum at \(x = 1\) (value \(-2\))

(iii) \(h(x) = \sin x + \cos x\), \(0 < x < \tfrac{\pi}{2}\)

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\(h'(x) = \cos x - \sin x = 0\) at \(\tfrac{\pi}{4}\); \[h''\left(\tfrac{\pi}{4}\right) = -\sqrt2 < 0\]

Answer: Local maximum at \(x = \tfrac{\pi}{4}\), value \(\sqrt2\)

(iv) \(f(x) = \sin x - \cos x\), \(0 < x < 2\pi\)

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\[\begin{aligned}&f'(x) = \cos x + \sin x = 0 \\ \Rightarrow\ &\tan x = -1\end{aligned}\]: \(x = \tfrac{3\pi}{4}, \tfrac{7\pi}{4}\).
\(f''(x) = -\sin x + \cos x\): \[f''\left(\tfrac{3\pi}{4}\right) = -\sqrt2\] (max), \(f''\left(\tfrac{7\pi}{4}\right) = \sqrt2\) (min).

Answer: Local maximum at \(\tfrac{3\pi}{4}\) (value \(\sqrt2\)); local minimum at \(\tfrac{7\pi}{4}\) (value \(-\sqrt2\))

(v) \(f(x) = x^3 - 6x^2 + 9x + 15\)

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\(f'(x) = 3(x - 1)(x - 3)\); \(f''(x) = 6x - 12\).
\(f''(1) < 0\): max, \(f(1) = 19\). \(f''(3) > 0\): min, \(f(3) = 15\).

Answer: Local maximum at \(x = 1\) (value 19); local minimum at \(x = 3\) (value 15)

(vi) \(g(x) = \dfrac{x}{2} + \dfrac{2}{x}\), \(x > 0\)

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\(g'(x) = \tfrac12 - \dfrac{2}{x^2} = 0\) at \(x = 2\); \(g''(2) = \dfrac{4}{8} > 0\).

Answer: Local minimum at \(x = 2\), value 2

(vii) \(g(x) = \dfrac{1}{x^2 + 2}\)

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\(g'(x) = -\dfrac{2x}{(x^2 + 2)^2}\): positive for \(x < 0\), negative for \(x > 0\).

Answer: Local maximum at \(x = 0\), value \(\tfrac12\)

(viii) \(f(x) = x\sqrt{1 - x}\), \(0 < x < 1\)

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\[\begin{aligned}f'(x) &= \sqrt{1 - x} - \dfrac{x}{2\sqrt{1 - x}} \\ &= \dfrac{2 - 3x}{2\sqrt{1 - x}}\end{aligned}\], zero at \(x = \tfrac23\), + to −.
\[\begin{aligned}f\left(\tfrac23\right) &= \tfrac23\sqrt{\tfrac13} \\ &= \dfrac{2\sqrt3}{9}\end{aligned}\]

Answer: Local maximum at \(x = \tfrac23\), value \(\dfrac{2\sqrt3}{9}\)

Exercise 6.3, Question 4

Prove that these functions have no maxima or minima.

(i) \(f(x) = e^x\)

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\(f'(x) = e^x\) is never 0 (and always positive): no critical points, f is increasing.

Answer: Proved

(ii) \(g(x) = \log x\)

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\(g'(x) = \tfrac1x \ne 0\) for \(x > 0\).

Answer: Proved

(iii) \(h(x) = x^3 + x^2 + x + 1\)

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\(h'(x) = 3x^2 + 2x + 1\) has discriminant \(4 - 12 < 0\), so \(h'(x) > 0\) always.

Answer: Proved

Exercise 6.3, Question 5

Find the absolute maximum and minimum values on the given interval.

(i) \(f(x) = x^3\), \(x \in [-2, 2]\)

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\(f' = 3x^2 \ge 0\): f increases, so the extremes are at the ends.

Answer: Maximum 8 (at 2), minimum \(-8\) (at \(-2\))

(ii) \(f(x) = \sin x + \cos x\), \(x \in [0, \pi]\)

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Critical point \(x = \tfrac{\pi}{4}\): \(f = \sqrt2\). Ends: \(f(0) = 1\), \(f(\pi) = -1\).

Answer: Maximum \(\sqrt2\) (at \(\tfrac{\pi}{4}\)), minimum \(-1\) (at \(\pi\))

(iii) \(f(x) = 4x - \tfrac12x^2\), \(x \in \left[-2, \tfrac92\right]\)

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\(f'(x) = 4 - x = 0\) at 4: \(f(4) = 8\).
Ends: \(f(-2) = -10\), \[\begin{aligned}f\left(\tfrac92\right) &= 18 - \tfrac{81}{8} \\ &= \tfrac{63}{8}\end{aligned}\]

Answer: Maximum 8 (at 4), minimum \(-10\) (at \(-2\))

(iv) \(f(x) = (x - 1)^2 + 3\), \(x \in [-3, 1]\)

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Critical point \(x = 1\) (an end): \(f(1) = 3\). Other end: \(f(-3) = 19\).

Answer: Maximum 19 (at \(-3\)), minimum 3 (at 1)

Exercise 6.3, Question 6

Find the maximum profit if \(p(x) = 41 - 72x - 18x^2\).

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\(p'(x) = -72 - 36x = 0\) at \(x = -2\); \(p'' = -36 < 0\).
\(p(-2) = 41 + 144 - 72\).

Answer: Maximum profit 113 (at \(x = -2\), as the formula stands)

Exercise 6.3, Question 7

Find the maximum and minimum of \(3x^4 - 8x^3 + 12x^2 - 48x + 25\) on \([0, 3]\).

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\[\begin{aligned}f'(x) &= 12x^3 - 24x^2 + 24x - 48 \\ &= 12(x - 2)(x^2 + 2)\end{aligned}\]: critical point \(x = 2\).
\(f(0) = 25\), \(f(2) = 48 - 64 + 48 - 96 + 25 = -39\), \(f(3) = 243 - 216 + 108 - 144 + 25 = 16\).

Answer: Maximum 25 (at 0), minimum \(-39\) (at 2)

Exercise 6.3, Question 8

At what points in \([0, 2\pi]\) does \(\sin 2x\) attain its maximum value?

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\(\sin 2x = 1\) when \(2x = \tfrac{\pi}{2}, \tfrac{5\pi}{2}\) (with \(2x \in [0, 4\pi]\)).

Answer: \(x = \tfrac{\pi}{4}\) and \(x = \tfrac{5\pi}{4}\)

Exercise 6.3, Question 9

What is the maximum value of \(\sin x + \cos x\)?

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\[\sin x + \cos x = \sqrt2\sin\left(x + \tfrac{\pi}{4}\right)\], at most \(\sqrt2\) (reached at \(x = \tfrac{\pi}{4}\)).

Answer: \(\sqrt2\)

Exercise 6.3, Question 10

(a) Find the maximum of \(2x^3 - 24x + 107\) on \([1, 3]\).

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\(f'(x) = 6x^2 - 24 = 0\) at \(x = \pm2\); in \([1, 3]\) only 2.
\(f(1) = 85\), \(f(2) = 75\), \(f(3) = 89\).

Answer: 89 (at \(x = 3\))

(b) … and on \([-3, -1]\).

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Critical point \(-2\): \(f(-3) = 125\), \(f(-2) = 139\), \(f(-1) = 129\).

Answer: 139 (at \(x = -2\))

Exercise 6.3, Question 11

\(x^4 - 62x^2 + ax + 9\) attains its maximum on \([0, 2]\) at \(x = 1\). Find a.

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An interior maximum is a critical point: \(f'(1) = 4 - 124 + a = 0\).
Check: with \(a = 120\), \(f''(1) = 12 - 124 < 0\), and \(f(1) = 68\) exceeds \(f(0) = 9\) and \(f(2) = 25\).

Answer: \(a = 120\)

Exercise 6.3, Question 12

Find the maximum and minimum of \(x + \sin 2x\) on \([0, 2\pi]\).

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\[\begin{aligned}&f'(x) = 1 + 2\cos 2x = 0 \\ \Rightarrow\ &\cos 2x = -\tfrac12\end{aligned}\]: \[x = \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, \tfrac{5\pi}{3}\]
Values: \(\tfrac{\pi}{3} + \tfrac{\sqrt3}{2}\), \(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\), \(\tfrac{4\pi}{3} + \tfrac{\sqrt3}{2}\), \(\tfrac{5\pi}{3} - \tfrac{\sqrt3}{2}\); ends \(f(0) = 0\), \(f(2\pi) = 2\pi\).

Answer: Maximum \(2\pi\) (at \(x = 2\pi\)), minimum 0 (at \(x = 0\))

Exercise 6.3, Question 13

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

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\(P = x(24 - x)\), \(P' = 24 - 2x = 0\) at 12; \(P'' = -2 < 0\).

Answer: 12 and 12

Exercise 6.3, Question 14

Find positive x, y with \(x + y = 60\) and \(xy^3\) maximum.

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\(f(y) = (60 - y)y^3\), \(f'(y) = 180y^2 - 4y^3 = 4y^2(45 - y)\).
Positive for \(y < 45\), negative after: maximum at \(y = 45\).

Answer: \(x = 15,\ y = 45\)

Exercise 6.3, Question 15

Find positive x, y with sum 35 and \(x^2y^5\) maximum.

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\(f(x) = x^2(35 - x)^5\): \[\begin{aligned}f'(x) &= x(35 - x)^4[2(35 - x) - 5x] \\ &= x(35 - x)^4(70 - 7x)\end{aligned}\]
Zero at \(x = 10\) inside \((0, 35)\), + to −.

Answer: \(x = 10,\ y = 25\)

Exercise 6.3, Question 16

Find two positive numbers with sum 16 whose cubes have the least sum.

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\(S = x^3 + (16 - x)^3\), \[\begin{aligned}&S' = 3x^2 - 3(16 - x)^2 = 0 \\ \Rightarrow\ &x = 8\end{aligned}\]; \(S'' = 6x + 6(16 - x) = 96 > 0\).

Answer: 8 and 8

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Exercise 6.3, Question 17

An 18 cm square of tin becomes an open box by cutting equal squares from the corners. What side of square should be cut off for maximum volume?

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Cut side x: \(V = x(18 - 2x)^2\), \(0 < x < 9\).
\[\begin{aligned}V' &= (18 - 2x)^2 - 4x(18 - 2x) \\ &= (18 - 2x)(18 - 6x)\end{aligned}\]: zero at \(x = 3\) in range; \(V''(3) = 24(3) - 144 < 0\).

Answer: 3 cm

Exercise 6.3, Question 18

A 45 cm by 24 cm sheet becomes an open box by cutting equal squares from the corners. What side should be cut off for maximum volume?

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\(V = x(45 - 2x)(24 - 2x)\), \(0 < x < 12\).
\[\begin{aligned}V' &= 12x^2 - 276x + 1080 \\ &= 12(x - 5)(x - 18)\end{aligned}\]: only \(x = 5\) is feasible; \(V''(5) = 24(5) - 276 < 0\).

Answer: 5 cm

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Exercise 6.3, Question 19

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

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Radius r: a rectangle with vertices on the circle has diagonal 2r; sides \(2r\cos\theta\), \(2r\sin\theta\), \(0 < \theta < \tfrac{\pi}{2}\).
Area \[\begin{aligned}A &= 4r^2\sin\theta\cos\theta \\ &= 2r^2\sin 2\theta\end{aligned}\], largest when \(2\theta = \tfrac{\pi}{2}\), i.e. \(\theta = \tfrac{\pi}{4}\): equal sides, a square.

Answer: Shown

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Exercise 6.3, Question 20

Show that the right circular cylinder of given surface area and maximum volume has height equal to the diameter of its base.

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\(S = 2\pi r^2 + 2\pi rh\) fixed, so \(h = \dfrac{S - 2\pi r^2}{2\pi r}\) and \(V = \pi r^2h = \tfrac12(Sr - 2\pi r^3)\).
\[\begin{aligned}&V'(r) = \tfrac12(S - 6\pi r^2) = 0 \\ \Rightarrow\ &S = 6\pi r^2\end{aligned}\]; \(V'' = -6\pi r < 0\).
Then \(2\pi rh = S - 2\pi r^2 = 4\pi r^2\), so \(h = 2r\).

Answer: Shown: \(h = 2r\)

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Exercise 6.3, Question 21

Of all closed cylindrical cans of volume 100 cm\(^3\), find the dimensions of the one with least surface area.

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\[\begin{aligned}&\pi r^2h = 100 \\ \Rightarrow\ &h = \dfrac{100}{\pi r^2}\end{aligned}\]; \(S = 2\pi r^2 + \dfrac{200}{r}\).
\[\begin{aligned}&S' = 4\pi r - \dfrac{200}{r^2} = 0 \\ \Rightarrow\ &r^3 = \dfrac{50}{\pi}\end{aligned}\]; \(S'' = 4\pi + \dfrac{400}{r^3} > 0\).
\[\begin{aligned}h &= \dfrac{100}{\pi r^2} \\ &= \dfrac{2 \cdot 50}{\pi r^3} \cdot r \\ &= 2r\end{aligned}\]

Answer: \(r = \left(\dfrac{50}{\pi}\right)^{1/3}\) cm, \(h = 2\left(\dfrac{50}{\pi}\right)^{1/3}\) cm

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Exercise 6.3, Question 22

A 28 m wire is cut in two; one piece bent into a square, the other into a circle. Find the lengths for the least combined area.

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Square piece x: side \(\tfrac{x}{4}\). Circle piece \(28 - x\): radius \(\dfrac{28 - x}{2\pi}\).
\[A = \dfrac{x^2}{16} + \dfrac{(28 - x)^2}{4\pi}\], \[\begin{aligned}&A' = \dfrac{x}{8} - \dfrac{28 - x}{2\pi} = 0 \\ \Rightarrow\ &\pi x = 4(28 - x)\end{aligned}\]; \(A'' > 0\).

Answer: Square: \(\dfrac{112}{\pi + 4}\) m; circle: \(\dfrac{28\pi}{\pi + 4}\) m

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Exercise 6.3, Question 23

Prove the largest cone inscribed in a sphere of radius R has volume \(\tfrac{8}{27}\) of the sphere's.

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Let the cone's base be at distance x below the centre: height \(R + x\), base radius\(^2\) \(R^2 - x^2\).
\(V = \tfrac{\pi}{3}(R^2 - x^2)(R + x)\), \[\begin{aligned}&V' = \tfrac{\pi}{3}(R + x)(R - 3x) = 0 \\ \Rightarrow\ &x = \tfrac{R}{3}\end{aligned}\] (a maximum).
\[\begin{aligned}V &= \tfrac{\pi}{3} \cdot \tfrac{8R^2}{9} \cdot \tfrac{4R}{3} \\ &= \tfrac{32\pi R^3}{81} \\ &= \tfrac{8}{27} \cdot \tfrac43\pi R^3\end{aligned}\]

Answer: Proved

Exercise 6.3, Question 24

Show that the right circular cone of least curved surface area for a given volume has height \(\sqrt2\) times its base radius.

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\(V = \tfrac13\pi r^2h\) fixed, so \(h = \dfrac{3V}{\pi r^2}\). Curved surface \(C = \pi r\sqrt{r^2 + h^2}\); minimise \(C^2 = \pi^2r^4 + \dfrac{9V^2}{r^2}\).
\[\begin{aligned}&\dfrac{d(C^2)}{dr} = 4\pi^2r^3 - \dfrac{18V^2}{r^3} = 0 \\ \Rightarrow\ &2\pi^2r^6 = 9V^2 = \pi^2r^4h^2\end{aligned}\]
So \(h^2 = 2r^2\); \(C^2\) is decreasing then increasing through this point, a minimum.

Answer: Shown: \(h = \sqrt2\,r\)

Exercise 6.3, Question 25

Show the semi-vertical angle of the cone of maximum volume for a given slant height is \(\tan^{-1}\sqrt2\).

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Slant l, semi-vertical angle \(\theta\): \(r = l\sin\theta\), \(h = l\cos\theta\), \[V = \tfrac{\pi}{3}l^3\sin^2\theta\cos\theta\]
\[\begin{aligned}&V'(\theta) = \tfrac{\pi}{3}l^3\sin\theta(2\cos^2\theta - \sin^2\theta) = 0 \\ \Rightarrow\ &\tan^2\theta = 2\end{aligned}\]
V rises then falls through this angle, so it is the maximum.

Answer: Shown: \(\theta = \tan^{-1}\sqrt2\)

Exercise 6.3, Question 26

Show the semi-vertical angle of the cone of given total surface area and maximum volume is \(\sin^{-1}\tfrac13\).

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\(S = \pi r^2 + \pi rl\) fixed: \(l = \dfrac{S - \pi r^2}{\pi r}\).
\[\begin{aligned}V^2 &= \tfrac{\pi^2}{9}r^4(l^2 - r^2) \\ &= \tfrac19(S^2r^2 - 2\pi Sr^4)\end{aligned}\]; \[\begin{aligned}&\dfrac{d(V^2)}{dr} = \tfrac19(2S^2r - 8\pi Sr^3) = 0 \\ \Rightarrow\ &S = 4\pi r^2\end{aligned}\]
Then \(\pi rl = 3\pi r^2\), \(l = 3r\), and \(\sin\theta = \dfrac{r}{l} = \dfrac13\) (a maximum: \(V^2\) rises then falls).

Answer: Shown

Exercise 6.3, Question 27

The point on \(x^2 = 2y\) nearest to \((0, 5)\) is: (A) \((2\sqrt2, 4)\) (B) \((2\sqrt2, 0)\) (C) \((0, 0)\) (D) \((2, 2)\)

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\(D^2 = x^2 + (y - 5)^2 = 2y + (y - 5)^2\); derivative \(2 + 2(y - 5) = 0\) at \(y = 4\), \(x^2 = 8\).

Answer: (A)

Exercise 6.3, Question 28

For real x, the minimum of \(\dfrac{1 - x + x^2}{1 + x + x^2}\) is: (A) 0 (B) 1 (C) 3 (D) \(\tfrac13\)

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\(f(x) = 1 - \dfrac{2x}{1 + x + x^2}\); \[f'(x) = \dfrac{2(x^2 - 1)}{(1 + x + x^2)^2}\]: minimum at \(x = 1\).
\(f(1) = \tfrac13\).

Answer: (D) \(\tfrac13\)

Exercise 6.3, Question 29

The maximum of \([x(x - 1) + 1]^{1/3}\), \(0 \le x \le 1\), is: (A) \(\left(\tfrac13\right)^{1/3}\) (B) \(\tfrac12\) (C) 1 (D) 0

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\(x^2 - x + 1\) ranges over \(\left[\tfrac34, 1\right]\) on \([0, 1]\) (minimum at \(\tfrac12\), value 1 at both ends); the cube root is increasing.

Answer: (C) 1