NCERT Solutions for Class 12 Maths Chapter 6 - Application of Derivatives - Exercise 6.5

3. Find the local maxima and local minima, if any, of the following functions. Also, find the local maximum and the local minimum values, as the case may be as follows :

(i) f(x) = x²

(ii) g(x) = x³ – 3x

$$\textbf{(iii) h(x) = sin x + cos x, 0 < x}\lt\frac{\pi}{\textbf{2}}$$

(iv) f(x) = sin x – cos x, 0 < x < 2π

(v) f(x) = x³ – 6x² + 9x + 15

$$\textbf{(vi)}\space \textbf{g(x)}\textbf{=}\frac{\textbf{x}}{\textbf{2}}+\frac{\textbf{2}}{\textbf{x}},\textbf{x}\gt \textbf{0}\\\textbf{(vii)\space g(x)}=\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{+2}} \\\textbf{(viii)\space f(x)}=\textbf{x}\sqrt{\textbf{1-x}}, \textbf{0}\lt \textbf{x}\lt \textbf{1}$$

Sol. (i) Given function is f(x) = x²
∴ Differentiating w.r.t. x, f ′(x) = 2x and again differentiating w.r.t. x, we get f ′′(x) = 2

For maxima or minima, put f ′(x) = 0

∴ 2x = 0

⇒ x = 0

Thus, x = 0 is the only critical point which could be possibly the point of local maxima or local minima of f.

We have f ′′(0) = 2 > 0 (which is positive)

Therefore, by second derivative test, x = 0 is a point of local minima and local minimum value of f at x = 0 is f(0) = x² = 0² = 0.

(ii) Given function is g(x) = x³ – 3x

∴ g′(x) = 3x² – 3

and g′′(x) = 6x

For maxima and minima put g′(x) = 0

∴ 3x² – 3 = 0

⇒ x² = 1

⇒ x = ± 1

Thus, we expect extremum only at two points – 1 and 1.

At x = – 1, g′′(– 1) = 6(– 1) = – 6 < 0

∴ g has a local maxima at x = – 1 and local maximum value = g(– 1)

= (– 1)³ – 3(– 1)

= – 1 + 3 = 2

At x = 1, g′′(1) = 6 × 1 = 6 > 0

∴ g has a local minima at x = 1 and local minimum value = g(1)

= 1³ – 3 × 1 = – 2

(iii) Given function is,

$$\text{h(x) = sin x + cos x, 0}\lt x\lt \frac{\pi}{2}$$

∴ h′(x) = cos x – sin x

and h′′(x) = – sin x – cos x

For maxima and minima put h′(x) = 0

⇒ sin x = cos x

$$\Rarr\space \frac{\text{sin x}}{\text{cos x}}=1\\\Rarr\space \text{tan x}=1\\\Rarr\space x=\frac{\pi}{4}\epsilon\bigg(0,\frac{\pi}{2}\bigg)\\\text{At}\space x=\frac{\pi}{4},h''\bigg(\frac{\pi}{4}\bigg)\\=-\text{sin}\frac{\pi}{4}-\text{cos}\frac{\pi}{4}\\=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\=-\frac{2}{\sqrt{2}}=-\sqrt{2}\lt 0$$

Therefore, by second derivative test

$$x=\frac{\pi}{4}\space\text{is a point of local maxima and}\\\text{ the local maximum value of h at x}=\frac{\pi}{4}\space\text{is}\\h\bigg(\frac{\pi}{4}\bigg)=\text{sin}\frac{\pi}{4}+\text{cos}\frac{\pi}{4}\\=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$$

(iv) Given function is f(x) = sin x – cos x, 0 < x < 2π

∴ f ′(x) = cos x + sin x

and f ′′(x) = – sin x + cos x

For maxima or minima put f ′(x) = 0

⇒ cos x + sin x = 0

⇒ sin x = – cos x

$$\Rarr\space\frac{\text{sin x}}{\text{cos x}}=-1$$

⇒ tan x = – 1

$$\Rarr\space x=\pi-\frac{\pi}{4},2\pi-\frac{\pi}{4}\\\Rarr\space x=\frac{3\pi}{7},\frac{7\pi}{4},\space x\epsilon(0,2\pi)$$

∴ The points at which extremum may occur are

$$\frac{3\pi}{4}\space\text{and}\frac{7\pi}{4}.\\\text{At x=}\frac{3\pi}{4},\\f''\bigg(\frac{3\pi}{4}\bigg)=-\text{sin}\frac{3\pi}{4}+\text{cos}\frac{3\pi}{4}\\=-\text{sin}\bigg(\pi-\frac{\pi}{4}\bigg) + \text{cos}\bigg(\pi-\frac{\pi}{4}\bigg)$$

[∵ sin (π – θ) = sin θ and cos(π – θ) = – cos θ]

$$=-\text{sin}\frac{\pi}{4}-\text{cos}\frac{\pi}{4}\\=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\=-\sqrt{2}\lt0\\\therefore\space x=\frac{3\pi}{4}\space\text{is a point of maxima.}$$

$$\text{Maxima value = }f\bigg(\frac{3\pi}{4}\bigg)\\=\text{sin}\frac{3\pi}{4}-\text{cos}\space\frac{3\pi}{4}\\=\text{sin}\bigg(\pi-\frac{\pi}{4}\bigg)-\text{cos}\bigg(\pi-\frac{\pi}{4}\bigg)\\=\text{sin}\frac{\pi}{4}-\text{cos}\frac{\pi}{4}\\=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\=\frac{2}{\sqrt{2}}=\sqrt{2}\\\text{At x=}\frac{7\pi}{4},$$

$$f''\bigg(\frac{7\pi}{4}\bigg)=-\text{sin}\frac{7\pi}{4}+\text{cos}\frac{7\pi}{4}\\=-\text{sin}\bigg(2\pi-\frac{\pi}{4}\bigg) + \text{cos}\bigg(2\pi-\frac{\pi}{4}\bigg)$$

[∵ sin(2π – θ) = – sin θ and cos(2π – θ) = cos θ]

$$=\text{sin}\frac{\pi}{4}+\text{cos}\frac{\pi}{4}\\=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\gt 0\\\therefore\space x=\frac{7\pi}{4}\space\text{is a point of minima.}\\\text{Minimum value}\\ f\bigg(\frac{7\pi}{4}\bigg)=\text{sin}\frac{7\pi}{4}-\text{cos}\frac{7\pi}{4}\\=\text{sin}\bigg(2\pi-\frac{\pi}{4}\bigg)-\text{cos}\bigg(2\pi-\frac{\pi}{4}\bigg)\\=-\text{sin}\frac{\pi}{4}-\text{cos}\frac{\pi}{4}\\=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}$$

(v) Given function is

f(x) = x³ – 6x² + 9x + 15

⇒ f ′(x) = 3x² – 12x + 9

and f ′′(x) = 6x – 12

For maxima or minima put f ′(x) = 0

⇒ 3x² – 12x + 9 = 0

⇒ 3(x² – 4x + 3) = 0

⇒ 3(x² – 3x –x + 3) = 0

⇒ 3[x(x – 3) – 1(x – 3)] = 0

⇒ 3(x – 1)(x – 3) = 0

⇒ x = 1 or x = 3

At x = 1, f ′′(1) = 6 × 1 – 12

= 6 – 12 = – 6 < 0

∴ x = 1 is a point of maxima.

Maxima value = f(1) = (1)³ – 6(1)² + 9(1) + 15

= 1 – 6 + 9 + 15 = 19

At x = 3, f ′′(3) = 6 × 3 – 12 = 18 – 12 = 6 > 0

x = 3 is a point of minima

∴ Minimum value = f(3) = 3³ – 6 × 3² + 9 × 3 + 15

= 27 – 54 + 27 + 15 = 15

$$\textbf{(vi)}\space\text{Given function of}\\ \text{g(x) =}\frac{x}{2}+\frac{2}{x},x\gt0\\\Rarr\space g'(x)=\frac{1}{2}+2\bigg(-\frac{1}{x^2}\bigg)\\=\frac{1}{2}-\frac{2}{x^2}\\\text{and}\space g''(x)=-2(-2)^{x^{\normalsize -3}}=\frac{4}{x^3}$$

For maxima or minima put, g′(x) = 0

$$\Rarr\space \frac{1}{2}-\frac{2}{x^2}=0\\\Rarr\space \frac{1}{2}=\frac{2}{x^2}$$

⇒ x² = 4

⇒ x = 2, – 2       [∵ x > 0 (given)]

At x = 2,
$$g''(2)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}\gt 0$$

∴ x = 2 is a point of minima.

Minimum value,

$$\text{g(2)}=\frac{2}{2}+\frac{2}{2}=1+1=2\\\textbf{(vii)}\space\text{Given function is g(x) =}\frac{1}{x^2+2}$$

Now, g(x) = (x² + 2)^{– 1}

$$⇒ g′(x) = – 1(x^2 + 2)^{\normalsize– 1– 1} × 2x=\frac{-2x}{(x^2+2)^2}$$

$$g''(x)=\frac{(x^2+2)^2.(-2)-(-2x).2(x^2+2).2x}{(x^2+2)^4}\\=\frac{-2(x^2+2)^2+8x^2(x^2+2)}{(x^2+2)^4}\\=\frac{(x^2+2)[-2x^2-4+8x^2]}{(x^2+2)^4}\\=\frac{6x^2-4}{(x^2+2)^3}=\frac{2(3x^2-2)}{(x^2+2)^3}$$

For maxima or minima put g′(x) = 0

$$\Rarr\space\frac{-2x}{(x^2+2)^2}=0$$

⇒ – 2x = 0

⇒ x = 0

At x = 0,

$$g''(0)=\frac{2[3(0)^2-2]}{[(0)^2+2]^3}=\frac{-4}{8}\\=-\frac{1}{2}\lt 0$$

∴ x = 0 is a point of maxima.

Maximum value,

$$\text{g(0)}=\frac{1}{(0)^2+2}=\frac{1}{2}\\\textbf{(viii)\space}\text{Given function is f(x) =} x\sqrt{1-x}\\\Rarr\space f'(x)=\frac{x(-1)}{2\sqrt{1-x}}+\sqrt{1-x}\\=\frac{-x+2(1-x)}{2\sqrt{1-x}}\\=\frac{2-3x}{2\sqrt{1-x}}\\f'(x)=\frac{1}{2}\bigg[\frac{\sqrt{1-x}(-3)-(2-3x)\bigg(\frac{1}{2\sqrt{1-x}}\bigg)}{(1-x)}\bigg]$$

$$=\frac{\sqrt{1-x}(-3)-(2-3x)\bigg(\frac{-1}{2\sqrt{1-x}}\bigg)}{2(1-x)}\\=\frac{-6(1-x) + (2-3x)}{4(1-x)^{3/2}}\\=\frac{3x-4}{4(1-x)^{3/2}}\\\text{For maxima or minima put f ′(x) = 0}\\\therefore\space \frac{2-3x}{2\sqrt{1-x}}=0$$

⇒ 2 – 3x = 0

$$\Rarr\space x=\frac{2}{3}\\\text{At x=}\frac{2}{3},\space f''\bigg(\frac{2}{3}\bigg)=\frac{3\bigg(\frac{2}{3}\bigg)-4}{4\bigg(1-\frac{2}{3}\bigg)^{3/2}}\\=\frac{2-4}{4\bigg(\frac{1}{3}\bigg)^{3/2}}\\=\frac{-1}{2\bigg(\frac{1}{3}\bigg)^{3/2}}\lt0$$

$$\therefore\space x=\frac{2}{3}\space\text{is a point of maxima.}\\\text{Maximum value = f}\bigg(\frac{2}{3}\bigg)\\=\frac{2}{3}\sqrt{1-\frac{2}{3}}\\=\frac{2}{3} ×\sqrt{\frac{1}{3}}\\=\frac{2}{3\sqrt{3}}\\=\frac{2\sqrt{3}}{9}$$