Maths NCERT Exemplar Solutions Class 12th Chapter Six

L : y = mx + c, m > 0

y = m (x – 1)

$\frac{x^2}{4}-\frac{y^2}{4}=1$

$y=mx\pm \sqrt[]{4m^2-4}$

$\pm \sqrt[]{4m^2-4}=-m,m>0$

$\Rightarrow \sqrt[]{4m^2-4}=m$

$M\left(\frac{5+\sqrt[]{21}}{2},\sqrt[]{3}+\sqrt[]{7}\right),N\left(\frac{5-\sqrt[]{21}}{2},\sqrt[]{3}-\sqrt[]{7}\right)$

$=$$\text{exception 25:}$$\left(2\text{exception 25:}\right)$$=$$2$

$1+\left(1-2^2.1\right)+\left(1-4^2.3\right)+\dots \dots .+\left(1-{20}^2.19\right)$

$=\alpha -220\beta$

$=11-\left(2^2\cdot 1+4^2\cdot 3+\dots ..+{20}^2\cdot 19\right)$

$=11-2^2\cdot {\sum }_{r=1}^{10}?r^2(2r-1)=11-4\left(\frac{{110}^2}{2}-35*11\right)$

$=11-220\left(103\right)$

$\Rightarrow \alpha =11,\beta =103$

$?y\left(x\right)={\left(x^x\right)}^x$

$\therefore y=x^{x^2}$

$\therefore \frac{dy}{dx}=x^2.x^{x^2-1}-x^{x^2}\mathcal{l}nx.2x$

$\therefore \frac{dx}{dy}=\frac{1}{x^{x^2+1}\left(1+2lnx\right)}$ ….(i)

$\frac{d^{2}x}{dx}=\frac{d}{dx}\left({\left(x^{x^2+1}\left(1+2lnx\right)\right)}^{-1}\right).\frac{dx}{dy}$

$\begin{array}{l}{\left(\frac{d^{2}x}{d{y}^2}\right)}_{x=1}=-4\\ {\left(\frac{d^{2}x}{d{y}^2}\right)}_{x=1}+20=16\end{array}$

${\displaystyle {\int }_{cosx}^1{t}^{2}f\left(t\right)dt=si{n}^{3}x+cosx}$

$\Rightarrow sinxco{s}^{2}xf\left(cosx\right)=3si{n}^{2}xcosx-sinx$

$\Rightarrow f\text{'}\left(cosx\right)\left(-sinx\right)=3se{c}^{2}x-2se{c}^{2}tanx$

$cosx=\frac{1}{\sqrt[]{3}}$

$\therefore f\text{'}\left(\frac{1}{\sqrt[]{3}}\right)\left(-\frac{\sqrt[]{2}}{\sqrt[]{3}}\right)=9-6\sqrt[]{2}$

$\frac{1}{\text{exception 25:}}f\text{'}\left(\frac{1}{\text{exception 25:}}\right)=6-$$\frac{9}{}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Wehavef\left(x\right)=\frac{2x^2-1}{x^4}\\ \Rightarrow f^{\text{'}}\left(x\right)=\frac{x^4\left(4x\right)-\left(2x^2-1\right).4x^3}{x^8}\\ \Rightarrow f^{\text{'}}\left(x\right)=\frac{4x^5-\left(2x^2-1\right).4x^3}{x^8}=\frac{4x^3\left[x^2-2x^2+1\right]}{x^8}=\frac{4\left[-x^2+1\right]}{x^5}\\ Fordecreasingthefunction{f}^{\text{'}}\left(x\right)<0\\ \therefore \frac{4\left(-x^2+1\right)}{x^5}<0\Rightarrow -x^2+1<0\Rightarrow x^2>1\\ \therefore x>\pm 1\Rightarrow x\in \left(1,\infty \right)\\ Hence,therequiredintervalis\left(1,\infty \right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l}Wehavef\left(x\right)=sinx-ax+b\Rightarrow f^{\text{'}}\left(x\right)=cosx-a\\ Forincreasingthefunction{f}^{\text{'}}\left(x\right)>0\\ \therefore cosx-a>0\\ Sincecosx\in \left[-1,1\right]\\ \therefore a<-1\Rightarrow a\in \left(-\infty ,-1\right)\\ Hence,thevalueofa\in \left(-\infty ,-1\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l}Wehavey=tanx.So,\frac{dy}{dx}=se{c}^{2}x\\ \therefore Slopeofthenormal=\frac{-1}{se{c}^{2}x}=-co{s}^{2}x\\ atthepoint\left(0,0\right)theslope=-co{s}^2\left(0\right)=-1\\ So,theequationofnormalat\left(0,0\right)isy-0=-1\left(x-0\right)\\ \Rightarrow y=-x\Rightarrow y+x=0\\ Hence,therequiredequationisy+x=0.\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Wehavey=4x^2+2x-8\dots \left(i\right)\\ andy=x^3-x+13\dots \left(ii\right)\\ Differentiatingeq.\left(i\right)w.r.t.x,wehave\\ \frac{dy}{dx}=8x+2\Rightarrow m_1=8x+2\left[m_{1}isslopeofcurve\left(i\right)\right]\\ Differentiatingeq.\left(ii\right)w.r.t.x,weget\\ \frac{dy}{dx}=3x^2-1\Rightarrow m_2=3x^2-1\left[m_{2}isslopeofcurve\left(ii\right)\right]\\ Ifthetwocurvestoucheachother,then{m}_1=m_2\\ \therefore 8x+2=3x^2-1\\ \Rightarrow 3x^2-8x-3=0\Rightarrow 3x^2-9x+x-3=0\\ \Rightarrow 3x\left(x-3\right)+1\left(x-3\right)=0\Rightarrow \left(x-3\right)\left(3x+1\right)=0\\ \therefore x=3,\frac{-1}{3}\\ Puttingx=3ineq\left(i\right),weget\\ y=4{\left(3\right)}^2+2\left(3\right)-8=36+6-8=34\\ So,therequiredpointis\left(3,34\right)\\ Nowforx=\frac{-1}{3}\\ y=4{\left(\frac{-1}{3}\right)}^2+2\left(\frac{-1}{3}\right)-8=4*\frac{1}{9}-\frac{2}{3}-8\\ =\frac{4}{9}-\frac{2}{3}-8=\frac{4-6-72}{9}=\frac{-74}{9}\\ \therefore Otherrequiredpointis\left(\frac{-1}{3},\frac{-74}{9}\right).\\ Hence,there\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Wehavef\left(x\right)={\left(\frac{1}{x}\right)}^x\\ Takinglogofbothsides,wehave\\ logf\left(x\right)=xlog\frac{1}{x}\\ logf\left(x\right)=xlog{x}^{-1}\Rightarrow log\left[f\left(x\right)\right]=-\left[xlogx\right]\\ Differentiatingbothsidesw.r.t.x,weget\\ \frac{1}{f\left(x\right)}{f}^{\text{'}}\left(x\right)=-\left[x.\frac{1}{x}+logx.1\right]\\ \Rightarrow f^{\text{'}}\left(x\right)=-f\left(x\right)\left[1+logx\right]=-{\left(\frac{1}{x}\right)}^x\left[1+logx\right]\\ Forlocalmaximaandlocalminima{f}^{\text{'}}\left(x\right)=0\\ -{\left(\frac{1}{x}\right)}^x\left[1+logx\right]=0\Rightarrow {\left(\frac{1}{x}\right)}^x\left[1+logx\right]=0\\ {\left(\frac{1}{x}\right)}^x\ne 0\therefore 1+logx=0\\ \Rightarrow logx=-1\Rightarrow x=e^{-1}\Rightarrow x=\frac{1}{e}\\ So,x=\frac{1}{e}isthestationarypoint.\\ Now,f^{\text{'}}\left(x\right)=-{\left(\frac{1}{x}\right)}^x\left[1+logx\right]\\ f^{\text{'}\text{'}}\left(x\right)=-\left[{\left(\frac{1}{x}\right)}^x\left(\frac{1}{x}\right)+\left(1+logx\right).\frac{d}{dx}{\left(x\right)}^x\right]\\ f^{\text{'}\text{'}}\left(x\right)=-\left[{\left(e\right)}^{1/e}\left(e\right)+\left(1+log\frac{1}{e}\right)\frac{d}{dx}{\left(\frac{1}{e}\right)}^{1/e}\right]\\ x=\frac{1}{e}=-e^{\frac{1}{e}+1}<0maxima\\ \therefore Maximumvalueofthefunctionatx=\frac{1}{e}is\\ f\left(\frac{1}{e}\right)={\left(\frac{1}{1/e}\right)}^{1/e}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$