Exercise 6.2

class 12 maths ex 6.2 ncert solution

Question 1: Show that the function given by $f(x)=3 x+17$ is strictly increasing on $\mathbf{R}$ .

Solution: $f(x)=3 x+17$

$\Rightarrow f'(x)=3>0, \quad \text { For all } x \in \mathbf{R}$

Therefore, the function is strictly increasing on $\mathbf{R}$ .

Question 2: Show that the function given by $f(x)=e^{2 x}$ is strictly increasing on $\mathbf{R}$ .

Solution: $f(x)=e^{2 x}$

$\Rightarrow f^{\prime}(x)=2 e^{2 x}>0$ , For all $x \in \mathbf{R}$

Therefore, the function is strictly increasing on $\mathbf{R}$ .

Question 3: Show that the function given by $f(x)=\sin x$ is

(a) strictly increasing in $\left(0, \frac{\pi}{2}\right)$

(b) strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$

(c) neither increasing nor decreasing in $(0, \pi)$ .

Solution : $f(x)=\sin x$

$\Rightarrow f'(x)=\cos x$

(a) $f'(x)=\cos x>0, \quad$ For all $x \in\left(0, \frac{\pi}{2}\right)$

Therefore, the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ .

(b) $f^{\prime}(x)=\cos x<0$ , For all $x \in\left(\frac{\pi}{2}, \pi\right)$

Therefore, the function is strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$ .

(c) $f^{\prime}(x)=\cos x>0, \quad$ For all $x \in\left(0, \frac{\pi}{2}\right)$

and $f^{\prime}(x)=\cos x<0$ , For all $x \in\left(\frac{\pi}{2}, \pi\right)$

Therefore, the function is neither increasing nor decreasing on $(0, \pi)$ .

Question 4: Find the intervals in which the function $f$ given by $f(x)=2 x^2-3 x$ is

(a) strictly increasing

(b) strictly decreasing

Solution : $f(x)=2 x^2-3 x$

$\Rightarrow f^{\prime}(x)=4 x-3$

If $f'(x)=0$

$\Rightarrow 4 x-3=0 \Rightarrow x=\frac{3}{4}$

(a) $f^{\prime}(x)=4 x-3>0$ , For all $x \in\left(\frac{3}{4}, \infty\right)$ ,

therefore, the function is strictly increasing on $\left(\frac{3}{4}, \infty\right)$ .

(b) $f^{\prime}(x)=4x-3<0$ , For all $x \in\left(-\infty, \frac{3}{4}\right)$ ,

therefore, the function is strictly decreasing on $\left(-\infty, \frac{3}{4}\right)$ .

Question 5: Find the intervals in which the function $f$ given by $f(x)=2 x^3-3 x^2-36 x+7$ is

(a) strictly increasing

(b) strictly decreasing

Solution: $f(x)=2 x^3-3 x^2-36 x+7$

$\Rightarrow f'(x)=6 x^2-6 x-36$

$=6\left(x^2-x-6\right)$

$=6(x-3)(x+2)$

Now $f'(x)=0 \Rightarrow 6(x-3)(x+2)=0$

$\Rightarrow x=-2,3$

$x=-2$ and $x=3$ divides the real number line into three intervals $(-\infty,-2),(-2,3)$ and $(3, \infty)$ .

$\begin{tabular}{|c|c|c|} \hline Intervals & Sign of $f'(x)$ & Nature of the function $f$ \\ \hline$(-\infty,-2)$ & $(-)(-)>0$ & $f$ is increasing \\ \hline$(-2, 3)$ & $(-)(+)<0$ & $f$ is decreasing \\ \hline$(3, \infty)$ & $(+)(+)>0$ & $f$ is increasing \\ \hline \end{tabular}$

(a) Function $f$ is increasing on $(-\infty,-2)$ and $(3, \infty)$ .

(b) Function $f$ is decreasing on $(-2,3)$ .

Question 6: Find the intervals in which the following functions are strictly increasing or decreasing:

(a) $f(x)=x^2+2 x+5$

(b) $f(x)=10-6 x-2 x^2$

(c) $f(x)=-2 x^3-9 x^2-12 x+1$

(d) $f(x)=6-9 x-x^2$

(e) $f(x)=(x+1)^3(x-3)^3$

Solution: (a) $f(x)=x^2+2 x+5$

$\Rightarrow f'(x)=2 x+2$

Now $f'(x)=0 \Rightarrow 2 x+2=0$

$\Rightarrow x=-1$

$f'(x)=2 x+2<0$ , For all $x \in(-\infty,-1)$ ,

therefore, the function is strictly decreasing on $(-\infty,-1)$ .

$f'(x)=2 x+2>0$ , For all $x \in(-1, \infty)$ ,

therefore, the function is strictly increasing on $(-1, \infty)$ .

(b) $f(x)=10-6 x-2 x^2$

$\Rightarrow f'(x)=-6-4 x$

Now $f'(x)=0\Rightarrow-6-4 x=0$

$\Rightarrow x=-\frac{3}{2}$

$f'(x)=-6-4 x>0$ , For all $x \in\left(-\infty,-\frac{3}{2}\right)$ ,

therefore, the function is strictly increasing on $\left(-\infty,-\frac{3}{2}\right)$ .

$f'(x)=-6-4 x<0$ , For all $x \in\left(-\frac{3}{2}, \infty\right)$ ,

therefore, the function is strictly decreasing on $\left(-\frac{3}{2}, \infty\right)$ .

(c) $f(x)=-2 x^3-9 x^2-12 x+1$

$\Rightarrow f'(x)=-6 x^2-18 x-12$

$=-6\left(x^2+3 x+2\right)$

$=-6(x+2)(x+1)$

If $f'(x)=0 \Rightarrow-6(x+2)(x+1)=0$

$\Rightarrow x=-2,-1$

$x=-2$ and $x=-1$ divides the number line into three intervals $(-\infty,-2),(-2,-1)$ and $(-1, \infty)$ .

$\begin{tabular}{|c|c|c|} \hline Interval & Sign of $f'(x)$ & Nature of $f$ \\ \hline$(-\infty,-2)$ & $(-)(-)>0$ & $f$ is increasing \\ \hline$(-2,-1)$ & $(+)(-)<0$ & $f$ is decreasing \\ \hline$(-1, \infty)$ & $(+)(+)>0$ & $f$ is increasing \\ \hline \end{tabular}$

Function $f$ is increasing on $(-\infty,-2) \cup(-1, \infty)$ and decreasing on $(-2,-1)$ .

(d) $f(x)=6-9 x-x^2$

$\Rightarrow f'(x)=-9-2 x$

Now $f'(x)=0$

$\Rightarrow-9-2 x=0$

$\Rightarrow x=-\frac{9}{2}$

$f'(x)=-9-2 x>0$ , For all $x \in\left(-\infty,-\frac{9}{2}\right)$ ,

therefore, the function is strictly increasing on $\left(-\infty,-\frac{9}{2}\right)$ .

$f'(x)=-9-2 x<0$ , For all $x \in\left(-\frac{9}{2}, \infty\right)$ ,

therefore, the function is strictly decreasing on $\left(-\frac{9}{2}, \infty\right)$ .

(e) $f(x)=(x+1)^3(x-3)^3$

$\Rightarrow f'(x)=3(x+1)^2(x-3)^3+(x+1)^3 \cdot 3(x-3)^2$

$=3(x+1)^2(x-3)^2(x-3+x+1)$

$=3(x+1)^2(x-3)^2(2 x-2)$

$=6(x+1)^2(x-3)^2(x-1)$

Now $f'(x)=0$

$\Rightarrow 6(x+1)^2(x-3)^2(x-1)=0$

$\Rightarrow x=-1,1,3$

$x=-1, x=1$ and $x=3$ divides the number lines into four intervals $(-\infty,-1),(-1,1),(1,3)$ and $(3, \infty)$ .

$\begin{tabular}{|c|c|c|} \hline Intervals & Sign of $f'(x)$ & Nature of $f$ \\ \hline$(-\infty,-1)$ & $(+)(+)(-)<0$ & $f$ is decreasing \\ \hline$(-1,1)$ & $(+)(+)(-)<0$ & $f$ is decreasing \\ \hline$(1,3)$ & $(+)(+)(+)>0$ & $f$ is increasing \\ \hline$(3, \infty)$ & $(+)(+)(+)>0$ & $f$ is increasing \\ \hline \end{tabular}$

Function $f$ is decreasing on $(-\infty, 1)$ and increasing on $(1, \infty)$ .

Question 7: Show that $y=\log (1+x)-\frac{2 x}{2+x}, x>-1$ , is an increasing function of $x$ throughout its domain.

Solution : $y=\log (1+x)-\frac{2 x}{2+x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1+x}-\frac{(2+x) \cdot 2-2 x(1)}{(2+x)^2}$

$=\frac{1}{1+x}-\frac{4}{(2+x)^2}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}$

$\Rightarrow \frac{d y}{d x}=\frac{4+x^2+4 x-4-4 x}{(1+x)(2+x)^2}$

$=\frac{x^2}{(1+x)(2+x)^2}$

$x^2>0$ and $(2+x)^2>0$ , as these are perfect square and

$(1+x)>0$ as $x>-1$ .

Therefore, $\frac{d y}{d x}>0$ , if $x>-1$ .

Hence, the function is increasing throughout its domain.

Question 8: Find the values of $x$ for which $y=[x(x-2)]^2$ is an increasing function.

Solution: $y=[x(x-2)]^2 \Rightarrow \frac{d y}{d x}=2[x(x-2)]\cdot \frac{d}{d x}[x(x-2)]$

$=2[x(x-2)]\frac{d}{dx}(x^2-2x) =2[x(x-2)](2x-2)$

$=4[x(x-2)(x-1)]$

Now $f'(x)=0 \Rightarrow 4[x(x-2)(x-1)]=0$

$\Rightarrow x=0, x=1$ and $x=2$ divides the number line into four intervals $(-\infty, 0),(0,1),(1,2)$ and $(2, \infty)$ .

$\begin{tabular}{|c|c|c|} \hline Intervals & Sign of $y'(x)$ & Nature of $y$ \\ \hline$(-\infty, 0)$ & $(-)(-)(-)<0$ & $f$ is decreasing \\ \hline$(0,1)$ & $(+)(-)(-)>0$ & $f$ is increasing \\ \hline$(1,2)$ & $(+)(-)(+)<0$ & $f$ is decreasing \\ \hline$(2, \infty)$ & $(+)(+)(+)>0$ & $f$ is increasing \\ \hline \end{tabular}$

Function $y$ is increasing on $(0,1) \cup(2, \infty)$ .

Question 9: Prove that $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ , is an increasing function of $\theta$ in $\left[0, \frac{\pi}{2}\right]$ .

Solution : $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$

$\Rightarrow \frac{d y}{d \theta}=\frac{(2+\cos \theta) \cdot 4 \cos \theta-4 \sin \theta(0-\sin \theta)}{(2+\cos \theta)^2}-1$

$=\frac{8 \cos \theta+4 \cos ^2 \theta+4 \sin ^2 \theta-(2+\cos \theta)^2}{(2+\cos \theta)^2}$

$\Rightarrow \frac{d y}{d \theta}=\frac{8 \cos \theta+4-4-\cos ^2 \theta-4 \cos \theta}{(2+\cos \theta)^2}$

$=\frac{4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2}$

$=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}$

$(2+\cos \theta)^2>0, \text { as it is a perfect square. } \cos \theta>0 \text { as } \theta \in\left[0, \frac{\pi}{2}\right] .$

$(4-\cos \theta)>0, \text { because } 0 \leq \cos \theta \leq 1, \text { for all } \theta \in\left[0, \frac{\pi}{2}\right] .$

Therefore, $\frac{d y}{d \theta}>0$ , if $\theta \in\left[0, \frac{\pi}{2}\right]$ .

Hence, $y$ is an increasing function of $\theta$ in $\left[0, \frac{\pi}{2}\right]$ .

Question 10: Prove that the logarithmic function is strictly increasing on $(0, \infty)$ .

Solution: $f(x)=\log x$

$\Rightarrow f^{\prime}(x)=\frac{1}{x}>0, \quad \text { For all } x \in(0, \infty) .$

Therefore, the function is strictly increasing on $(0, \infty)$ .

Question 11: Prove that the function $f$ given by $f(x)=x 2-x+1$ is neither strictly increasing nor strictly decreasing on $(-1,1)$ .

Solution: $f(x)=x^2-x+1$

$\Rightarrow f'(x)=2 x-1$

If $f'(x)=0 \Rightarrow 2 x-1=0$

$\Rightarrow x=\frac{1}{2}$

$x=\frac{1}{2}$ divides the interval $(-1,1)$ into two intervals $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$ .

$f'(x)=2 x-1<0$ , For all $x \in\left(-1, \frac{1}{2}\right)$ ,

therefore, the function is strictly decreasing on $\left(-1, \frac{1}{2}\right)$ .

$f'(x)=\frac{1}{x}>0$ , For all $x \in\left(\frac{1}{2}, 1\right)$ , therefore, the function is strictly increasing on $\left(\frac{1}{2}, 1\right)$ .

Therefore, the function is neither increasing nor decreasing on $(-1,1)$ .

Question 12: Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ ?

(A) $\cos x$

(B) $\cos 2 \mathrm{x}$

(D) $\tan \mathrm{x}$

Solution: (A) $f(x)=\cos x \Rightarrow f'(x)=-\sin x$

$f'(x)=-\sin x<0$ , For all $x \in\left(0, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ .

(B) $f(x)=\cos 2 x \Rightarrow f'(x)=-2 \sin 2 x$

If $x \in\left(0, \frac{\pi}{2}\right)$ , then $2 x \in(0, \pi)$

$\Rightarrow \sin 2 x>0$ if $2 x \in(0, \pi)$ .

Therefore, $-\sin 2 x<0$ if $2 x \in(0, \pi)$ .

$f'(x)=-2 \sin 2 x<0$ , For all $x \in\left(0, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ .

(C) $f(x)=\cos 3 x \Rightarrow f'(x)=-3 \sin 3 x$

If $x \in\left(0, \frac{\pi}{2}\right)$ , then $3 x \in\left(0, \frac{3 \pi}{2}\right)$ ,

$\Rightarrow \sin 3 x>0$ if $3 x \in(0, \pi)$ .

$\Rightarrow -\sin 3 x<0$ if $3 x \in(0, \pi)$ or $x \in\left(0, \frac{\pi}{3}\right)$

$f'(x)=-2 \sin 3 x<0$ , For all $x \in\left(0, \frac{\pi}{3}\right)$ ,

therefore, the function is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ .

$\sin 3 x<0$ if $3 x \in\left(\pi, \frac{3 \pi}{2}\right)$ ,

therefore, $-\sin 3 x>0$ if $3 x \in\left(\pi, \frac{3 \pi}{2}\right)$ or $x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$

$f^{\prime}(x)=-2 \sin 3 x>0$ , For all $x \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ .

Hence, the function is neither increasing nor decreasing on $\left(0, \frac{\pi}{2}\right)$ .

(D) $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^2 x$

$f^{\prime}(x)=\sec ^2 x>0$ , For all $x \in\left(0, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ .

Therefore, $\cos x$ and $\cos 2 x$ are decreasing on $\left(0, \frac{\pi}{2}\right)$ .

Hence, the option (A) and (B) are correct.

Question 13: On which of the following intervals is the function $f$ given by $f(x)=x^{100}+\sin x-1$ strictly decreasing?

(A) $(0,1)$

(B) $\left(\frac{\pi}{2}, \pi\right)$

(D) None of these

Solution: $f(x)=x^{100}+\sin x-1$

$\Rightarrow f'(x)=100 x^{99}+\cos x$

(A) $100 x^{99}>0$ and $\cos x>0$ , For all $x \in(0,1)$ ,

therefore, the function is strictly increasing on $(0,1)$ .

(B) $100 x^{99}>0$ and $\cos x<0$ , For all $x \in\left(\frac{\pi}{2}, \pi\right)$ ,

$\Rightarrow 100 x^{99}>\cos x$ , for all $x \in\left(\frac{\pi}{2}\right.$ , $\left.\pi\right)$ .

Therefore, $f^{\prime}(x)=100 x^{99}+\cos x>0$ ,

therefore, the function is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$ .

(C) $100 x^{99}>0$ and $\cos x>0$ , For all $x \in\left(0, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ .

Hence, is function is not decreasing in the given intervals. Hence, the option (D) is correct.

Question 14: Find the least value of $a$ such that the function $f$ given by $f(x)=x^2+a x+1$ is strictly increasing on $(1,2)$ .

Solution : $f(x)=x^2+a x+1$

$\Rightarrow f'(x)=2 x+a$

But $f$ is increasing on $(1,2)$

$\Rightarrow f'(x)>0$

$\Rightarrow 2 x+a>0$

$\Rightarrow x>-\frac{a}{2}$

Here, $x>-\frac{a}{2}$ when $x \in(1,2)$ or $1<x<2$ ,

Therefore, for the least value of $a$ , when the function is increasing on $(1,2)$ , we have

$-\frac{a}{2}=1$

$\Rightarrow a=-2$

Question 15: Let I be any interval disjoint from $(-1,1)$ . Prove that the function $f$ given by $f(x)=x+\frac{1}{x}$ is strictly increasing on I.

Solution: $f(x)=x+\frac{1}{x}$

$\Rightarrow f'(x)=x-\frac{1}{x^2}$

$=\frac{x^2-1}{x^2}$

$=\frac{(x-1)(x+1)}{x^2}$

If $f'(x)=0$

$\Rightarrow(x-1)(x+1)=0$

$\Rightarrow x=-1,1$

$x=-1$ and $x=1$ divides the real value into intervals $(-\infty,-1),(-1,1)$ and $(1, \infty)$ .

Only $(-\infty,-1)$ and $(1, \infty)$ are disjoint from $(-1,1)$ .

$\begin{tabular}{|c|c|c|} \hline Intervals & Sign of $f'(x)$ & Nature of $\boldsymbol{f}$ \\ \hline $(-\infty,-1)$ & $(-)(-)>0$ & $f$ is strictly increasing \\ \hline $(1, \infty)$ & $(+)(+)(+)>0$ & $f$ is strictly increasing\\ \hline\end{tabular}$

Function $f$ is increasing on $(-\infty,-1) \cup(1, \infty)$ .

Hence, the function $f$ is strictly increasing on interval I.

Question 16: Prove that the function $f$ given by $f(x)=\log \sin x$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ and strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$ .

Solution: $f(x)=\log \sin x$

$\Rightarrow f^{\prime}(x)=\frac{1}{\sin x} \cdot \cos x=\tan x$

$f^{\prime}(x)=\tan x>0$ , for all $x \in\left(0, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly increasing on $\left(0, \frac{\pi}{2}\right)$ .

$f^{\prime}(x)=\tan x<0$ , for all $x \in\left(\frac{\pi}{2}, \pi\right)$ ,

therefore, the function is strictly decreasing on $\left(\frac{\pi}{2}, \pi\right)$ .

Question 17: Prove that the function $f$ given by $f(x)=\log \cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ and strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$ .

Solution: $f(x)=\log |\cos x|$

$\Rightarrow f^{\prime}(x)=\frac{1}{\sin x} \cdot \cos x=-\cot x$

$f^{\prime}(x)=-\cot x<0$ , for all $x \in\left(0, \frac{\pi}{2}\right)$ ,

therefore, the function is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ .

$f^{\prime}(x)=-\cot x>0$ , for all $x \in\left(\frac{\pi}{2}, \pi\right)$ ,

therefore, the function is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$ .

Question 18: Prove that the function given by $f(x)=x^3-3 x^2+3 x-100$ is increasing in $\mathbf{R}$ .

Solution : $f(x)=x^3-3 x^2+3 x-100$

$\Rightarrow f^{\prime}(x)=3 x^2-6 x+3$

$=3\left(x^2-2 x+1\right)=3(x-1)^2$

$3(x-1)^2 \geq 0$ , as it is perfect square.

Therefore, the function $f(x)=x^3-3 x^2+3 x-100$ is increasing in $\mathbf{R}$ .

Question 19: The interval in which $y=x^2 e^{-x}$ is increasing is

(A) $(-\infty, \infty)$

(B) $(-2,0)$

(D) $(0,2)$

Solution: $y=x^2 e^{-x}$

$\Rightarrow \frac{d y}{d x}=2 x e^{-x}-x^2 e^{-x}=x e^{-x}(2-x)$

Now $f'(x)=0$

$\Rightarrow x e^{-x}(2-x)=0$

$\Rightarrow x=0,2$

$x=0$ and $x=2$ divides the real values into three intervals $(-\infty, 0),(0,2)$ and $(2, \infty)$ .

$\begin{tabular}{|c|c|c|} \hline Intervals & Sign of $y'(x)$ & Nature of function $y$ \\ \hline$(-\infty, 0)$ & $(-)(+)<0$ & $f$ is strictly decreasing \\ \hline$(0,2)$ & $(+)(+)>0$ & $f$ is strictly increasing \\ \hline$(2, \infty)$ & $(+)(-)<0$ & $f$ is strictly decreasing \\ \hline \end{tabular}$

Function $y$ is strictly increasing on $(0,2)$ .

Therefore, the option (D) is correct.

Case study application of derivative 4 — These days chinese and Indian troops are engaged in aggressive melee, face-offs skirmishes

Exercise 6.2

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