NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.2) Exercise 6.2

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

Ans

$\begin{array}{l}\mathrm{Function}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}+17\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}‘\left(\mathrm{x}\right)=3>0\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{every}\mathrm{interval}\mathrm{of}\mathrm{R}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{function}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbf{R}\mathrm{.}\end{array}$

Q.2 Show that the function given by f(x) = e^2x is strictly increasing on R

Ans

$\begin{array}{l}\mathrm{Let}{\mathrm{x}}_{\mathrm{1}}\mathrm{and}{\mathrm{x}}_{\mathrm{2}}\mathrm{be}\mathrm{any}\mathrm{two}\mathrm{numbers}\mathrm{in}\mathbf{R}.\\ \mathrm{Then},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{have}:\\ {\mathrm{x}}_{\mathrm{1}}<{\mathrm{x}}_{\mathrm{2}}\Rightarrow 2{\mathrm{x}}_{\mathrm{1}}<\; 2{\mathrm{x}}_{\mathrm{2}}\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{e}}^{2{\mathrm{x}}_{\mathrm{1}}}<{\mathrm{e}}^{2{\mathrm{x}}_{\mathrm{2}}}\\ \Rightarrow \mathrm{f}\left({\mathrm{x}}_1\right)<\mathrm{f}\left({\mathrm{x}}_2\right)\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathbf{R}\mathrm{.}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{f}(\mathrm{x})\; =\sin \mathrm{x}\mathrm{.}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cosx}\\ \left(\mathrm{a}\right)\mathrm{Since}\mathrm{for}\mathrm{each}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\in (0,\frac{\mathrm{\pi }}{2}),\mathrm{\hspace{0.17em}}\mathrm{cosx}>0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{have}\mathrm{f}‘\left(\mathrm{x}\right)>0.\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{\hspace{0.17em}}(0,\frac{\mathrm{\pi }}{2}).\\ \left(\mathrm{b}\right)\mathrm{Since}\mathrm{for}\mathrm{each}\hspace{0.17em}\hspace{0.17em}\mathrm{x}\in (\frac{\mathrm{\pi }}{2},\mathrm{\pi }),\mathrm{\hspace{0.17em}}\mathrm{cosx}<0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{\hspace{0.17em}}\mathrm{have}\mathrm{f}‘\left(\mathrm{x}\right)<0.\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{\hspace{0.17em}}(\frac{\mathrm{\pi }}{2},\mathrm{\pi }).\\ \left(\mathrm{c}\right)\mathrm{From}\mathrm{the}\mathrm{results}\mathrm{obtained}\mathrm{in}\left(\mathrm{a}\right)\mathrm{and}(\mathrm{b}),\mathrm{it}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{f}\mathrm{is}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{neither}\mathrm{increasing}\mathrm{nor}\mathrm{\hspace{0.17em}}\mathrm{decreasing}\mathrm{in}(0,\mathrm{\pi }).\end{array}$

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

(a) strictly increasing (b) strictly decreasing

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^{\mathrm{2}}-3\mathrm{x}\mathrm{.}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}‘\left(\mathrm{x}\right)=4\mathrm{x}-3\\ \mathrm{Therefore},\mathrm{f}‘\left(\mathrm{x}\right)=0\Rightarrow 4\mathrm{x}-3=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=\frac{3}{4}\\ \mathrm{Now}\mathrm{the}\mathrm{point}\mathrm{x}=\frac{3}{4}\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{\hspace{0.17em}}\mathrm{namely},(-\mathrm{\infty },\mathrm{\hspace{0.17em}}\frac{3}{4})\mathrm{and}(\frac{3}{4},\mathrm{\infty }).\end{array}$

\begin{array}{l}\text{In the interval}\left(-\infty ,\frac{3}{4}\right),\\ \text{f’}\left(x\right)=4x-3<0\\ Therefore,\text{f is strictly deacreasing in this interval}\text{. Also, in}\\ \text{the interval}\left(\frac{3}{4},\infty \right)\text{, f’}\left(x\right)>0\text{}\text{and so the function f is strictly}\\ \text{increasing in this interval}\text{.}\end{array}

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

Ans

$\begin{array}{l}\therefore \mathrm{f}‘\left(\mathrm{x}\right)=0\Rightarrow 6\mathrm{\hspace{0.33em}}(\mathrm{x}-3)\mathrm{\hspace{0.33em}}(\mathrm{x}+2)=0\\ \Rightarrow \mathrm{x}=-2,3\\ \mathrm{The}\mathrm{points}\mathrm{x}=-2\mathrm{and}\mathrm{x}=\; 3\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{\hspace{0.17em}}(-\infty ,-2),\mathrm{\hspace{0.33em}}(-2,3)\mathrm{and}(3,\infty ).\end{array}$

\begin{array}{l}\text{In intervals}(-\infty ,-2)\text{and}(3,\infty ),f‘(x)\text{is positive while in interval}\\ (-2,3),f‘(x)\text{is negative}\text{.}\\ \text{Hence, the given function (f) is strictly increasing in intervals}\\ (-\infty ,-2)\text{and}(3,\infty ),\text{while function (f) is strictly decreasing in}\\ \text{interval}\left(-2,3\right).\end{array}

Q.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

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{We}\mathrm{have},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{2}}+2\mathrm{x}-\mathrm{5}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0\Rightarrow 2\mathrm{x}+2=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=-1\\ \mathrm{Point}\mathrm{x}=-1\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-1)\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}(-1,\mathrm{\infty }).\\ \mathrm{In}\mathrm{interval}(-\mathrm{\infty },-1),\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2<0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{interval}\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-1).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x}<-1.\\ \mathrm{In}\mathrm{interval}\hspace{0.17em}\hspace{0.17em}(-1,\mathrm{\infty }),\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+2>0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{\hspace{0.17em}}(-1,\mathrm{\infty }).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}>-1.\\ \left(\mathrm{b}\right)\mathrm{We}\mathrm{have},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{10}-6\mathrm{x}-2{\mathrm{x}}^{\mathrm{2}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{\hspace{0.17em}}6-4\mathrm{x}\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0\Rightarrow -\mathrm{\hspace{0.17em}}6-4\mathrm{x}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=-\frac{3}{2}\\ \mathrm{Point}\mathrm{x}=-\frac{3}{2}\mathrm{divides}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{two}\mathrm{disjoint}\mathrm{intervals}\\ \mathrm{i}.\mathrm{e}.,\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-\frac{3}{2})\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}(-\frac{3}{2},\mathrm{\infty }).\\ \mathrm{In}\mathrm{interval}(-\mathrm{\infty },-\frac{3}{2}),\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{\hspace{0.17em}}6-4\mathrm{x}>0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-\frac{3}{2}).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}\mathrm{x}<-\frac{3}{2}\mathrm{.}\\ \mathrm{In}\mathrm{interval}\hspace{0.17em}\hspace{0.17em}(-\frac{3}{2},\mathrm{\infty }),\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=-6-4\mathrm{x}<0\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{interval}\mathrm{\hspace{0.17em}}(-\frac{3}{2},\mathrm{\infty }).\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{for}\mathrm{x}\hspace{0.17em}>-\frac{3}{2}\mathrm{.}\\ \left(\mathrm{c}\right)\mathrm{We}\mathrm{have},\\ \hspace{0.17em}\hspace{0.17em}\mathrm{f}\left(\mathrm{x}\right)=-2{\mathrm{x}}^{\mathrm{3}}-9{\mathrm{x}}^{\mathrm{2}}-12\mathrm{x}+\mathrm{1}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{\hspace{0.17em}}6{\mathrm{x}}^2-18\mathrm{x}-12=-6({\mathrm{x}}^2+3\mathrm{x}+2)\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=0\Rightarrow -6({\mathrm{x}}^2+3\mathrm{x}+2)=0\\ \Rightarrow (\mathrm{x}+1)(\mathrm{x}+2)=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=-1,-2\\ \mathrm{The}\mathrm{points}\mathrm{x}=-1\mathrm{and}\mathrm{x}=-2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-2),(-2,-1)\mathrm{and}(-1,\mathrm{\infty }).\\ \mathrm{In}\mathrm{intervals}\hspace{0.17em}\hspace{0.17em}(-\mathrm{\infty },-2)\mathrm{and}(-1,\mathrm{\infty }),\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)\mathrm{\hspace{0.17em}}\mathrm{is}\mathrm{negative}.\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\\ (-\mathrm{\infty },-2)\mathrm{and}(-1,\mathrm{\infty }),\\ \mathrm{i}.\mathrm{e}.,\mathrm{\hspace{0.17em}}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{when}\mathrm{x}<-2\mathrm{and}\mathrm{x}>-1.\end{array}$

\begin{array}{l}\text{In intervals}\left(-2,-1\right),f‘\left(x\right)\text{is positive}\text{.}\\ \text{So, function (f) is strictly increasing in interval}\left(-2,-1\right),\\ i.e.,\text{function (f) is strictly increasing}when-2<x<-1.\\ \left(d\right)\text{We have,}\\ \text{f}\left(x\right)=\text{6}-\text{9x}-{\text{x}}^{\text{2}}\\ \text{Differentiating both sides with respect to x, we get}\\ \text{f’}\left(x\right)=-\text{9}-2\text{x}\\ \therefore \text{f’}\left(x\right)=0\Rightarrow -\text{9}-2\text{x}=0\\ \Rightarrow x=-\frac{9}{2}\\ \text{Point x}=-\frac{9}{2}\text{divides the real line into two disjoint intervals}\\ \text{i}\text{.e}\text{.,}\left(-\infty ,-\frac{9}{2}\right)and\left(-\frac{9}{2},\infty \right).\\ \text{In interval}\left(-\infty ,-\frac{9}{2}\right),f‘\left(x\right)=-\text{9}-2\text{x}>0\\ \therefore \text{f is strictly increasing in interval}\left(-\infty ,-\frac{9}{2}\right).\\ \text{Thus, f is strictly increasing for x <}-\frac{9}{2}\text{.}\\ \text{In interval}\left(-\frac{9}{2},\infty \right),f‘\left(x\right)=-\text{9}-2\text{x}<0\\ \therefore \text{f is strictly decreasing in interval}\left(-\frac{9}{2},\infty \right).\\ \text{Thus, f is strictly decreasing for x >}-\frac{9}{2}\text{.}\end{array}

\begin{array}{l}\left(e\right)\text{We have,}\\ \text{f(x)}={\text{(x + 1)}}^{\text{3}}\text{(x}-{\text{3)}}^{\text{3}}\\ Differentiating\text{both sides with respect to x, we get}\\ \text{f’}\left(x\right)={\left(\text{x + 1}\right)}^{\text{3}}\frac{d}{dx}{\left(\text{x}-\text{3}\right)}^{\text{3}}+{\left(\text{x}-\text{3}\right)}^{\text{3}}\frac{d}{dx}{\left(\text{x + 1}\right)}^{\text{3}}\\ \left[By\text{product rule}\right]\\ \Rightarrow \text{f’}\left(x\right)={\left(\text{x + 1}\right)}^{\text{3}}3{\left(\text{x}-\text{3}\right)}^{\text{2}}\frac{d}{dx}\left(\text{x}-\text{3}\right)+{\left(\text{x}-\text{3}\right)}^{\text{3}}3{\left(\text{x + 1}\right)}^{\text{2}}\frac{d}{dx}\left(\text{x + 1}\right)\\ \left[By\text{chain rule}\right]\\ \Rightarrow \text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{3}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(1-0\right)+3{\left(\text{x}-\text{3}\right)}^{\text{3}}{\left(\text{x + 1}\right)}^{\text{2}}\left(1+0\right)\\ \Rightarrow \text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x+1+x-3\right)\\ \Rightarrow \text{f’}\left(x\right)=3{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(2x-2\right)\\ =6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)\\ Now,\text{f’}\left(x\right)=0\Rightarrow x=-1,1,3\\ \text{The points x}=-1,x=1,andx=3\text{divide the real line into}\\ \text{four disjoint intervals i}\text{.e}\text{.,}\left(-\infty ,-1\right),\left(-1,1\right),\left(1,3\right)\text{and}\left(3,\infty \right).\\ \text{In intervals}\left(-\infty ,-1\right)\text{and (}-\text{1, 1),}\end{array}

\begin{array}{l}\text{f’}\left(x\right)=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)<0\\ \therefore \text{f is strictly decreasing in intervals}\left(-\infty ,-1\right)\text{and}\left(-\text{1, 1}\right).\\ \text{In intervals (1, 3) and}\left(3,\infty \right),\text{f’}\left(x\right)=6{\left(\text{x + 1}\right)}^{\text{2}}{\left(\text{x}-\text{3}\right)}^{\text{2}}\left(x-1\right)>0\\ \therefore \text{f is strictly increasing in intervals}\left(\text{1, 3}\right)\text{and}\left(3,\infty \right).\end{array}

Q.7

$\begin{array}{l}\mathrm{Show}that\mathrm{y}=\log \left(1+\mathrm{x}\right)-\frac{2\mathrm{x}}{2+\mathrm{x}},\mathrm{x}>-1,\mathrm{is}\mathrm{an}\mathrm{increasing}\\ \mathrm{function}\mathrm{of}\mathrm{x}\mathrm{throughout}\mathrm{its}domain.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\log (1+\mathrm{x})-\frac{2\mathrm{x}}{2+\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{x},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\log (1+\mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2\mathrm{x}}{2+\mathrm{x}}\right)\\ =\frac{1}{1+\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}(1+\mathrm{x})-\frac{(2+\mathrm{x})\frac{\mathrm{d}}{\mathrm{dx}}2\mathrm{x}-2\mathrm{x}\frac{\mathrm{d}}{\mathrm{dx}}(2+\mathrm{x})}{{(2+\mathrm{x})}^2}\\ =\frac{1}{1+\mathrm{x}}\times (0+1)-\frac{2(2+\mathrm{x})-2\mathrm{x}(0+1)}{{(2+\mathrm{x})}^2}\\ =\frac{1}{1+\mathrm{x}}-\frac{4}{{(2+\mathrm{x})}^2}\\ =\frac{4+4\mathrm{x}+{\mathrm{x}}^2-4-4\mathrm{x}}{(1+\mathrm{x}){(2+\mathrm{x})}^2}\\ =\frac{{\mathrm{x}}^2}{(1+\mathrm{x}){(2+\mathrm{x})}^2}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \Rightarrow \mathrm{\hspace{0.17em}}\frac{{\mathrm{x}}^2}{(1+\mathrm{x}){(2+\mathrm{x})}^2}=0\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{x}}^2=0[(1+\mathrm{x})\ne 0\mathrm{and}{(2+\mathrm{x})}^2\ne 0]\\ \Rightarrow \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}=0\\ \mathrm{Since}\mathrm{x}>-1,\mathrm{point}\mathrm{x}=\; 0\mathrm{divides}\mathrm{the}\mathrm{domain}(-1,\mathrm{\infty })\mathrm{in}\mathrm{two}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,-1<\mathrm{\hspace{0.17em}}\mathrm{x}<\; 0\mathrm{and}\mathrm{x}>\; 0\mathrm{.}\\ \mathrm{When}-1<\mathrm{x}<\; 0,\mathrm{we}\mathrm{have}:\\ \mathrm{x}<\hspace{0.17em}0\Rightarrow {\mathrm{x}}^2>0\\ \mathrm{x}>-1\Rightarrow \mathrm{x}+2>0\Rightarrow {(\mathrm{x}+2)}^2>0\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^2}{{(\mathrm{x}+2)}^2}>0\\ \mathrm{Also},\mathrm{when}\mathrm{x}>\; 0:\\ \mathrm{x}>0\Rightarrow {\mathrm{x}}^2>0,\hspace{0.17em}\hspace{0.17em}(2+{\mathrm{x}}^2)>0\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^2}{{(\mathrm{x}+2)}^2}>0\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{increasing}\mathrm{throughout}\mathrm{this}\mathrm{domain}\mathrm{.}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{for}\mathrm{which}\mathrm{y}={\left[\mathrm{x}\left(\mathrm{x}-2\right)\right]}^2\mathrm{is}\mathrm{an}\mathrm{increasing}\\ \mathrm{function}.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{y}={\left[\mathrm{x}(\mathrm{x}-\mathrm{2})\right]}^2\\ ={\mathrm{x}}^2{(\mathrm{x}-2)}^2\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^2{(\mathrm{x}-2)}^2\\ ={\mathrm{x}}^2\frac{\mathrm{d}}{\mathrm{dx}}{(\mathrm{x}-2)}^2+{(\mathrm{x}-2)}^2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^2\\ ={\mathrm{x}}^2.2(\mathrm{x}-2)\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-2)+{(\mathrm{x}-2)}^2.2\mathrm{x}\\ =2{\mathrm{x}}^2(\mathrm{x}-2)(1-0)+2\mathrm{x}{(\mathrm{x}-2)}^2\\ =2\mathrm{x}(\mathrm{x}-2)(\mathrm{x}+\mathrm{x}-2)\\ =2\mathrm{x}(\mathrm{x}-2)(2\mathrm{x}-2)\\ =4\mathrm{x}(\mathrm{x}-2)(\mathrm{x}-1)\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0\Rightarrow \mathrm{x}=0,1,2\\ \mathrm{The}\mathrm{points}\mathrm{x}=\; 0,\mathrm{x}=\; 1,\mathrm{and}\mathrm{x}=\; 2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{four}\\ \mathrm{disjoint}\mathrm{intervals}\mathrm{i}.\mathrm{e}.,(-\mathrm{\infty },0),(0,1),(1,2),(2,\mathrm{\infty }).\\ \mathrm{In}\mathrm{intervals}(-\mathrm{\infty },0)\mathrm{and}\hspace{0.17em}\hspace{0.17em}(1,2),\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}<0\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}\mathrm{\hspace{0.17em}}(-\mathrm{\infty },0)\mathrm{and}\hspace{0.17em}\hspace{0.17em}(1,2).\\ \mathrm{However},\mathrm{in}\mathrm{intervals}(0,\; 1)\mathrm{and}(2,\mathrm{\infty }),\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}>0\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{intervals}(0,\; 1)\mathrm{and}(2,\infty )\mathrm{.}\\ \therefore \mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{for}0\; <\mathrm{x}<\; 1\mathrm{and}\mathrm{x}>\; 2\mathrm{.}\end{array}$

Q.9

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

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\hspace{0.17em}\hspace{0.17em}\mathrm{y}=\frac{4\mathrm{sin\theta }}{(\mathrm{2}+\mathrm{cos\theta })}-\mathrm{\theta }\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{\theta },\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{\mathrm{d}}{\mathrm{d\theta }}\{\frac{4\mathrm{sin\theta }}{(\mathrm{2}+\mathrm{cos\theta })}-\mathrm{\theta }\}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{(2+\mathrm{cos\theta })\frac{\mathrm{d}}{\mathrm{d\theta }}\left(4\mathrm{sin\theta }\right)-4\mathrm{sin\theta }\frac{\mathrm{d}}{\mathrm{d\theta }}(2+\mathrm{cos\theta })}{{(2+\mathrm{cos\theta })}^2}-\frac{\mathrm{d}}{\mathrm{d\theta }}\mathrm{\theta }\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{4(2+\mathrm{cos\theta })\mathrm{cos\theta }-4\mathrm{sin\theta }(0-\mathrm{sin\theta })}{{(2+\mathrm{cos\theta })}^2}-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{4(2+\mathrm{cos\theta })\mathrm{cos\theta }+4{\sin }^2\mathrm{\theta }}{{(2+\mathrm{cos\theta })}^2}-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{8\mathrm{cos\theta }+4({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta })}{{(2+\mathrm{cos\theta })}^2}-1\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=\frac{8\mathrm{cos\theta }+4}{{(2+\mathrm{cos\theta })}^2}-1\\ \mathrm{Now},\mathrm{}\frac{\mathrm{dy}}{\mathrm{d\theta }}=0\\ \Rightarrow \frac{8\mathrm{cos\theta }+4}{{(2+\mathrm{cos\theta })}^2}-1=0\Rightarrow \frac{8\mathrm{cos\theta }+4}{{(2+\mathrm{cos\theta })}^2}=1\\ \Rightarrow 8\mathrm{cos\theta }+4={(2+\mathrm{cos\theta })}^2\\ \Rightarrow 8\mathrm{cos\theta }+\bcancel{4}=\bcancel{4}+4\mathrm{cos\theta }+{\cos }^2\mathrm{\theta }\\ \Rightarrow {\cos }^2\mathrm{\theta }-4\mathrm{cos\theta }=0\\ \Rightarrow \mathrm{cos\theta }(\mathrm{cos\theta }-4)=0\\ \Rightarrow \mathrm{cos\theta }=0\hspace{0.17em}\hspace{0.17em}\mathrm{or}\mathrm{\hspace{0.17em}}\mathrm{cos\theta }=4\mathrm{\hspace{0.17em}}\\ \mathrm{Since}\cos \mathrm{\theta }\ne \mathrm{4}\\ \therefore \mathrm{cos\theta }=0\Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{2}\\ \mathrm{Now},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{d\theta }}=\frac{8\mathrm{cos\theta }+4}{{(2+\mathrm{cos\theta })}^2}-1\\ =\frac{8\mathrm{cos\theta }+4-{(2+\mathrm{cos\theta })}^2}{{(2+\mathrm{cos\theta })}^2}\\ =\frac{8\mathrm{cos\theta }+4-4-4\mathrm{cos\theta }-{\cos }^2\mathrm{\theta }}{{(2+\mathrm{cos\theta })}^2}\\ =\frac{\mathrm{cos\theta }(4-\mathrm{cos\theta })}{{(2+\mathrm{cos\theta })}^2}\\ \mathrm{In}\mathrm{interval}\hspace{0.17em}\hspace{0.17em}(0,\frac{\mathrm{\pi }}{2}),\hspace{0.17em}\hspace{0.17em}\mathrm{we}\mathrm{have}\mathrm{cos\theta }>\; 0.\mathrm{Also},\; 4\; >\mathrm{cos\theta }\\ \Rightarrow \mathrm{4}-\mathrm{cos\theta }\mathrm{\hspace{0.17em}}>0\mathrm{.}\\ \therefore \mathrm{cos\theta }(4-\mathrm{cos\theta })>0\mathrm{and}\mathrm{also}{(2+\mathrm{cos\theta })}^2>0\\ \Rightarrow \frac{\mathrm{cos\theta }(4-\mathrm{cos\theta })}{{(2+\mathrm{cos\theta })}^2}>0\Rightarrow \frac{\mathrm{dy}}{\mathrm{d\theta }}>0\\ \mathrm{Therefore},\mathrm{y}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\hspace{0.17em}\hspace{0.17em}(0,\frac{\mathrm{\pi }}{2}).\\ \mathrm{Also},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{is}\mathrm{continuous}\mathrm{at}\hspace{0.17em}\hspace{0.17em}\mathrm{x}=0\mathrm{and}\mathrm{x}=\frac{\mathrm{\pi }}{2}.\\ \mathrm{Hence},\mathrm{y}\mathrm{is}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{\hspace{0.17em}}(0,\frac{\mathrm{\pi }}{2}).\end{array}$

Q.10

$\begin{array}{l}\mathrm{Provethatthelogarithmicfunctionisstrictly}\mathrm{increasing}\\ \mathrm{on}\left(0,\mathrm{\infty }\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{function}\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}\\ \therefore \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{for}\mathrm{x}>\; 0,\mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}>0\\ \mathrm{Hence},\mathrm{f}(\mathrm{x})\; =\log \mathrm{x}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}(0,\infty )\mathrm{.}\\ \end{array}$

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

Ans

$\begin{array}{l}{\text{The given function is f(x) = x}}^{\text{2}}-\text{x + 1}\text{.}\\ Differentiating\text{with respect to x, we get}\\ \text{f’}\left(x\right)=2x-1\\ Now,f‘\left(x\right)=0\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}\\ \text{The point}\frac{1}{2}\text{divides the interval}(-\text{1, 1})\text{into two disjoint}\\ \text{intervals i}\text{.e}\text{.,}(-1,\frac{1}{2})and\left(\frac{1}{2}\text{, 1}\right)\text{.}\\ \text{Now, for interval}(-1,\frac{1}{2}),f‘\left(x\right)=2x-1<0\\ \text{Therefore, f is strictly decreasing in interval}(-1,\frac{1}{2}).\\ \text{Now, for interval}\left(\frac{1}{2}\text{, 1}\right),f‘\left(x\right)=2x-1>0\\ \text{So, f is strictly increasing in interval}\left(\frac{1}{2}\text{, 1}\right)\text{.}\\ \text{Hence, f is neither strictly increasing nor decreasing in}\\ \text{interval}(-\text{1, 1}).\end{array}$

Q.12

$\begin{array}{l}\mathrm{Which\; of\; the\; following\; functions\; are}strictly\mathrm{decreasing}\\ \mathrm{on}\left(0,\frac{\mathrm{\pi }}{2}\right)?\end{array}$

$\begin{array}{l}\left(\mathrm{A}\right)\mathrm{cosx}\left(\mathrm{B}\right)\cos 2\mathrm{x}\left(\mathrm{C}\right)\cos 3\mathrm{x}\left(\mathrm{D}\right)\tan x\end{array}$

Ans

\begin{array}{l}\left(A\right)Let{f}_1\left(x\right)=\cos x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{f’}}_{\text{1}}\left(x\right)=\frac{d}{dx}\cos x\\ =-\sin x\\ In\text{interval}\left(0,\frac{\pi }{2}\right),\text{sinx is positive in first quadrant}\text{.}\\ \text{So,}{\text{f’}}_{\text{1}}\left(x\right)<0\\ Therefore,f_1\left(x\right)=\cos x\text{is strictly decreasing in interval}\left(0,\frac{\pi }{2}\right).\\ \left(B\right)Let{f}_2\left(x\right)=\cos 2x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{f’}}_{\text{2}}\left(x\right)=\frac{d}{dx}\cos 2x\\ =-2\sin 2x\\ In\text{interval}\left(0,\frac{\pi }{2}\right),\text{sin2x is positive in first quadrant}\text{.}\\ \text{So,}{\text{f’}}_{\text{2}}\left(x\right)<0\\ Therefore,f_2\left(x\right)=\cos 2x\text{is strictly decreasing in interval}\left(0,\frac{\pi }{2}\right).\\ \left(C\right)Let{f}_3\left(x\right)=\cos 3x\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{f’}}_{\text{3}}\left(x\right)=\frac{d}{dx}\cos 3x\\ =-3\sin 3x\\ Now,{\text{f’}}_{\text{3}}\left(x\right)=0\Rightarrow -3\sin 3x=0\\ \Rightarrow 3x=\pi \Rightarrow x=\frac{\pi }{3}\\ \text{The point}x=\frac{\pi }{3}\text{divides the interval}\left(0,\frac{\pi }{2}\right)\text{into two disjoint}\\ \text{intervals i}\text{.e}\text{.,}\left(0,\frac{\pi }{3}\right)\text{and}\left(\frac{\pi }{3},\frac{\pi }{2}\right).\\ Now,\text{}\text{in interval}\left(0,\frac{\pi }{3}\right),\\ {\text{f’}}_{\text{3}}\left(x\right)=-3\sin 3x<0\left[\because 0<x<\frac{\pi }{3}\Rightarrow 0<3x<\pi \right]\\ \therefore {\text{f}}_{\text{3}}\text{is strictly decreasing in interval}\left(0,\frac{\pi }{3}\right).\\ Now,in\text{interval}\left(\frac{\pi }{3},\frac{\pi }{2}\right),\\ \frac{\pi }{3}<x<\frac{\pi }{2}\Rightarrow \pi <3x<\frac{3\pi }{2}\\ So,\text{sin3x is in third quadrant, where sin3x is negative}\text{.}\\ \therefore {\text{f}}_{\text{3}}\left(x\right)=-3\sin 3x>0\\ Therefore,{\text{f}}_{\text{3}}\text{is strictly increasing in interval}\left(\frac{\pi }{3},\frac{\pi }{2}\right).\\ {\text{Hence, f}}_{\text{3}}\text{is neither increasing nor decreasing in interval}\left(0,\frac{\pi }{2}\right).\\ \left(D\right)Let{f}_4\left(x\right)=\tan x\\ Differentiating\text{w}\text{.r}\text{.t}\text{. x, we get}\\ {\text{f’}}_{\text{4}}\left(x\right)=\frac{d}{dx}\tan x\\ ={\sec }^{2}x\\ In\mathrm{int}erval\left(0,\frac{\pi }{2}\right),\text{}\sec x>0\Rightarrow {\sec }^{2}x>0\\ \therefore f_4\left(x\right)>0\\ \therefore \text{f4 is strictly increasing in interval}\left(0,\frac{\pi }{2}\right).\\ \text{Therefore, functions cos x and cos 2x are strictly decreasing}\\ \text{in}\left(0,\frac{\pi }{2}\right).\\ \text{Hence, the correct answers are A and B}\text{.}\end{array}

Q.13

$\begin{array}{l}\mathrm{On}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{intervals}\mathrm{is}\mathrm{the}\mathrm{function}\mathrm{f}\mathrm{given}\\ \mathrm{by}f\left(\mathrm{x}\right)={\mathrm{x}}^{100}+\mathrm{sinx}–\; 1\mathrm{strictly}\mathrm{decreasing}?\\ \left(\mathrm{A}\right)\left(0,1\right)\left(\mathrm{B}\right)\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\left(\mathrm{C}\right)\left(0,\frac{\mathrm{\pi }}{2}\right)\left(\mathrm{D}\right)\mathrm{None}\mathrm{of}these\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\hspace{0.17em}\hspace{0.17em}\mathrm{have},\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{100}+\mathrm{sinx}-1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=100{\mathrm{x}}^{99}+\mathrm{cosx}\\ \mathrm{f}‘\left(\mathrm{x}\right)>0[\because 100{\mathrm{x}}^{99}>0\mathrm{and}\mathrm{cosx}>0\mathrm{in}\mathrm{interval}(0,1)]\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}(0,\; 1)\mathrm{.}\\ \mathrm{In}\mathrm{interval}(\frac{\mathrm{\pi }}{2},\mathrm{\pi }),\\ \mathrm{cosx}<0\mathrm{and}100{\mathrm{x}}^{\mathrm{99}}>0\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)>0\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}(\frac{\mathrm{\pi }}{2},\mathrm{\pi }).\\ \mathrm{In}\mathrm{interval}(0,\frac{\mathrm{\pi }}{2}),\\ \mathrm{cosx}>0\mathrm{and}100{\mathrm{x}}^{\mathrm{99}}>0\\ \Rightarrow 100{\mathrm{x}}^{99}+\mathrm{cosx}>0\\ \mathrm{So},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)>0\\ \mathrm{Thus},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}(0,\frac{\mathrm{\pi }}{2}).\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{in}\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{intervals}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.14 Find the least value of a such that the function f given f(x) =x²+ax+1 is strictly increasing on (1, 2).

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{ax}+1\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=2\mathrm{x}+\mathrm{a}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{function}\mathrm{f}\mathrm{will}\mathrm{be}\mathrm{increasing}\mathrm{in}\mathrm{\hspace{0.17em}}(1,2),\mathrm{if}\\ \mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)>0\Rightarrow 2\mathrm{x}+\mathrm{a}>0\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}>-\frac{\mathrm{a}}{2}\\ \mathrm{Therefore},\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{such}\mathrm{that}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{x}>-\frac{\mathrm{a}}{2},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{when}\mathrm{\hspace{0.17em}}\mathrm{x}\in (1,2)\\ \Rightarrow \mathrm{\hspace{0.17em}}\mathrm{x}>-\frac{\mathrm{a}}{2},\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}(\mathrm{when}\mathrm{\hspace{0.17em}}1<\mathrm{x}<2)\\ \mathrm{Thus},\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{for}\mathrm{f}\mathrm{to}\mathrm{be}\mathrm{increasing}\mathrm{on}(1,\; 2)\mathrm{is}\\ \mathrm{given}\mathrm{by},\\ \mathrm{–}\frac{\mathrm{a}}{2}=1\Rightarrow \mathrm{a}=-2\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{is}-2.\end{array}$

Q.15

$\begin{array}{l}\mathrm{Let}\mathrm{I}\mathrm{be}\mathrm{any}\mathrm{interval}\mathrm{disjoint}\mathrm{from}\left(-1,1\right).\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{function}\\ \mathrm{fgiven}by\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\frac{1}{\mathrm{x}}\mathrm{is}\mathrm{strictly}\mathrm{increasing}on\mathrm{I}.\end{array}$

Ans

$\begin{array}{l}\mathrm{we}\mathrm{}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\frac{1}{\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^2}\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)=0\Rightarrow 1-\frac{1}{{\mathrm{x}}^2}=0\\ \Rightarrow 1=\frac{1}{{\mathrm{x}}^2}\Rightarrow \mathrm{x}=\pm 1\\ \mathrm{The}\mathrm{points}\mathrm{x}=1\mathrm{and}\mathrm{x}=-1\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{in}\mathrm{three}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-1),(-1,1)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}(1,\mathrm{\infty }).\\ \mathrm{In}\mathrm{interval}(-1,\; 1),\mathrm{it}\mathrm{is}\mathrm{noticed}\mathrm{that}:\\ -1<\mathrm{x}<1\Rightarrow {\mathrm{x}}^2<1\\ \Rightarrow 1<\frac{1}{{\mathrm{x}}^2},\hspace{0.17em}\hspace{0.17em}\mathrm{x}\ne 0\\ \Rightarrow 1-\frac{1}{{\mathrm{x}}^2}<0,\hspace{0.17em}\hspace{0.17em}\mathrm{x}\ne 0\\ \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^2}<0\hspace{0.17em}\hspace{0.17em}\mathrm{on}\mathrm{}(-1,1)~\left\{0\right\}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{on}\hspace{0.17em}\hspace{0.17em}(-1,1)~\left\{0\right\}\\ \mathrm{In}\mathrm{intervals}\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-1)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}(1,\mathrm{\infty }),\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\\ \mathrm{x}<-\mathrm{1\hspace{0.17em}}\mathrm{or}\mathrm{x}>1\\ \Rightarrow {\mathrm{x}}^{\mathrm{2}}>\mathrm{1}\Rightarrow 1>\frac{1}{{\mathrm{x}}^2}\\ \Rightarrow 1-\frac{1}{{\mathrm{x}}^2}>0\\ \therefore \mathrm{f}‘\left(\mathrm{x}\right)=1-\frac{1}{{\mathrm{x}}^2}>0\mathrm{on}\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-1)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}(1,\mathrm{\infty }).\\ \therefore \mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{on}\mathrm{\hspace{0.17em}}(-\mathrm{\infty },-1)\hspace{0.17em}\hspace{0.17em}\mathrm{and}\hspace{0.17em}\hspace{0.17em}(1,\mathrm{\infty }).\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}\mathrm{I}\mathrm{disjoint}\mathrm{from}(-1,\; 1)\mathrm{.}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Provethatthefunctionfgivenbyf}(\mathrm{x})\; =\mathrm{logsinxisstrictly}\\ \mathrm{increasingon}\left(0,\frac{\mathrm{\pi }}{2}\right)\mathrm{andstrictlydecreasing}on\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{logsinx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{sinx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{sinx}[\mathrm{By}\mathrm{\hspace{0.17em}}\mathrm{chain}\mathrm{rule}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{\mathrm{sinx}}\mathrm{cosx}=\mathrm{cotx}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cotx}>0\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{interval}(0,\frac{\mathrm{\pi }}{2})\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{function}\mathrm{in}\mathrm{interval}(0,\frac{\mathrm{\pi }}{2}).\\ \mathrm{f}‘\left(\mathrm{x}\right)=\mathrm{cotx}<0\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{interval}(\frac{\mathrm{\pi }}{2},\mathrm{\pi })\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{function}\mathrm{in}\mathrm{interval}(\frac{\mathrm{\pi }}{2},\mathrm{\pi }).\end{array}$

Q.17 Prove that the function f given by f(x) = logcos x is strictly decreasing on ( 0 , π 2 ) and strictly increasing on ( π 2 , π ) .

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{logcosx}\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{\hspace{0.17em}}\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=\frac{1}{\mathrm{cosx}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{cosx}[\mathrm{By}\mathrm{\hspace{0.17em}}\mathrm{chain}\mathrm{rule}]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=\frac{1}{\mathrm{cosx}}\times -\mathrm{sinx}=-\mathrm{tanx}\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{tanx}<0\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{interval}(0,\frac{\mathrm{\pi }}{2})\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{decreasing}\mathrm{function}\mathrm{in}\mathrm{interval}(0,\frac{\mathrm{\pi }}{2}).\\ \mathrm{f}‘\left(\mathrm{x}\right)=-\mathrm{tanx}>0\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{in}\mathrm{interval}(\frac{\mathrm{\pi }}{2},\mathrm{\pi })\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{function}\mathrm{in}\mathrm{interval}(\frac{\mathrm{\pi }}{2},\mathrm{\pi }).\end{array}$

Q.18

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}\mathrm{given}\mathrm{byf}(\mathrm{x})\; ={\mathrm{x}}^2-3{\mathrm{x}}^2+3\mathrm{x}-100\\ \mathrm{is}\mathrm{increasingin}R.\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-3{\mathrm{x}}^2+3\mathrm{x}-100\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{f}‘\left(\mathrm{x}\right)=3{\mathrm{x}}^2-6\mathrm{x}+3\\ =3({\mathrm{x}}^2-2\mathrm{x}+1)\\ =3{(\mathrm{x}-1)}^2\\ \mathrm{For}\mathrm{any}\mathrm{x}\in \mathbf{R},\hspace{0.17em}{(\mathrm{x}-1)}^2>0.\\ \mathrm{Thus},\mathrm{f}‘\left(\mathrm{x}\right)\hspace{0.17em}\hspace{0.17em}\mathrm{is}\mathrm{always}\mathrm{positive}\mathrm{in}\mathbf{R}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{increasing}\mathrm{in}\mathbf{R}\mathrm{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{The}\mathrm{interval}\mathrm{in}\mathrm{which}\mathrm{y}={\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}\mathrm{is}\mathrm{increasing}\mathrm{is}\\ \left(\mathrm{A}\right)\left(-\mathrm{\infty },\mathrm{\infty }\right)(\mathrm{B})\left(-2,0\right)\left(\mathrm{C}\right)\left(2,\mathrm{\infty }\right)\left(\mathrm{D}\right)\left(0,2\right)\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{have},\\ \mathrm{y}={\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^2\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}=-{\mathrm{x}}^2{\mathrm{e}}^{-\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}\left(2\mathrm{x}\right)\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}={\mathrm{e}}^{-\mathrm{x}}(-{\mathrm{x}}^2+2\mathrm{x})\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \Rightarrow {\mathrm{e}}^{-\mathrm{x}}(-{\mathrm{x}}^2+2\mathrm{x})=0\\ \Rightarrow \mathrm{x}=0,2[\because {\mathrm{e}}^{-\mathrm{x}}\ne 0]\\ \mathrm{The}\mathrm{points}\mathrm{x}=0\mathrm{and}\mathrm{x}=2\mathrm{divide}\mathrm{the}\mathrm{real}\mathrm{line}\mathrm{into}\mathrm{three}\mathrm{disjoint}\\ \mathrm{intervals}\mathrm{i}.\mathrm{e}.,(-\mathrm{\infty },0),(0,2)\mathrm{\hspace{0.17em}}\mathrm{and}\hspace{0.17em}\hspace{0.17em}(2,\mathrm{\infty }).\\ \mathrm{f}‘\left(\mathrm{x}\right)<0\mathrm{in}\mathrm{intervals}(-\mathrm{\infty },0)\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}(2,\mathrm{\infty }).\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{is}\mathrm{decreasing}\mathrm{in}\mathrm{intervals}(-\mathrm{\infty },0)\mathrm{\hspace{0.17em}}\mathrm{and}\mathrm{\hspace{0.17em}}(2,\mathrm{\infty }).\\ \mathrm{Now},\mathrm{\hspace{0.17em}}\mathrm{f}‘\left(\mathrm{x}\right)>0\mathrm{in}\mathrm{interval}(0,2)\\ \mathrm{So},\mathrm{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{is}\mathrm{increasing}\mathrm{in}\mathrm{intervals}(0,2)\mathrm{\hspace{0.17em}}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{strictly}\mathrm{increasing}\mathrm{in}\mathrm{interval}(0,\; 2)\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$