Ncert Solutions Maths class 12th

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1-cos2x}{x^2}=\underset{h\to 0}{lim}\frac{1-cos2\left(0-h\right)}{{\left(0-h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{1-cos\left(-2h\right)}{h^2}=\underset{h\to 0}{lim}\frac{1-cos\left(2h\right)}{h^2}\\ =\underset{h\to 0}{lim}\frac{2si{n}^{2}h}{h^2}\left[?1-cos\theta =2si{n}^2\frac{\theta }{2}\right]\\ =\underset{h\to 0}{lim}\frac{2sinh}{h}.\frac{sinh}{h}=2.1.1=2\left[\underset{x\to 0}{lim}\frac{sinx}{x}=1\right]\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{1-cos2x}{x^2}\\ =\underset{h\to 0}{lim}\frac{1-cos2\left(0+h\right)}{{\left(0+h\right)}^2}=\underset{h\to 0}{lim}\frac{1-cos\left(2h\right)}{h^2}\\ =\underset{h\to 0}{lim}\frac{2si{n}^{2}h}{h^2}=\underset{h\to 0}{lim}\frac{2sinh}{h}.\frac{sinh}{h}=2.1.1=2\\ \underset{x\to 0}{lim}f\left(x\right)=5\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\underset{x\to 2^{+}}{lim}f\left(x\right)=3x+5\\ =\underset{h\to 0}{lim}3\left(2+h\right)+5=11\\ \underset{x\to 2}{lim}f\left(x\right)=3x+5=3\left(2\right)+5=11\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=x^2=\underset{h\to 0}{lim}{\left(2-h\right)}^2\\ =\underset{h\to 0}{lim}{\left(2\right)}^2+h^2-4h={\left(2\right)}^2=4\\ Since\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)\ne \underset{x\to 2^{-}}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=2.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthaty=f\left(x\right)willbecontinuousatx=aif\\ \underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)\\ Givenf\left(x\right)=x^3+2x^2-1\\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left(1+h\right)}^3+2{\left(1+h\right)}^2-1=1+2-1=2\\ \underset{x\to 1}{lim}f\left(x\right)={\left(1\right)}^3+2{\left(1\right)}^2-1=1+2-1=2\\ \underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}{\left(1+h\right)}^3+2{\left(1+h\right)}^2-1=1+2-1=2\\ \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=2.\\ Hence,f\left(x\right)iscontinuousatx=1.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatx=sintandy=sinpt\\ Differentiatingbothsidesw.r.t.t\\ \Rightarrow \frac{dx}{dt}=costand\frac{dy}{dt}=cospt.p=p.cospt\\ \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{p.cospt}{cost}\\ \Rightarrow \frac{dy}{dx}=\frac{p.cospt}{cost}\\ Againdifferentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=p.\frac{d}{dx}\left(\frac{cospt}{cost}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\frac{d}{dx}\left(cospt\right)-cospt.\frac{d}{dx}\left(cost\right)}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\left(-sinpt\right).p\frac{dt}{dx}-cospt.\left(-sint\right).\frac{dt}{dx}}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{dt}{dx}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{1}{cost}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\\ Nowwehavetoprovethat\\ \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+ p^{2}y=0\\ L.H.S.=\left(1-x^2\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-x\left(\frac{p.cospt}{cost}\right)+ p^{2}y\\ \Rightarrow =\left(1-si{n}^{2}t\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =co{s}^{2}t\left[\frac{-p^{2}cost.sinpt+pcospt.sint}{co{s}^{3}t}\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint}{cost}-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint-p.sintcospt+p^2.sinptcost}{cost}\\ \Rightarrow =\frac{0}{cost}=0=R.H.S.\\ Hence,proved.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}\left(i\right)Giventhat{x}^m.y^n={\left(x+y\right)}^{m+n}\\ Takinglogonbothsides,weget,\\ \Rightarrow log{x}^m.y^n=log{\left(x+y\right)}^{m+n}\left[?logxy=logx+logy\right]\\ \Rightarrow log{x}^m+log{y}^n=\left(m+n\right)log\left(x+y\right)\\ \Rightarrow mlogx+nlogy=\left(m+n\right)log\left(x+y\right)\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow m.\frac{d}{dx}logx+n.\frac{d}{dx}logy=\left(m+n\right)\frac{d}{dx}log\left(x+y\right)\\ \Rightarrow m.\frac{1}{x}+n.\frac{1}{y}.\frac{dy}{dx}=\left(m+n\right)\frac{1}{x+y}\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}.\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}.\frac{dy}{dx}\\ \Rightarrow \frac{n}{y}.\frac{dy}{dx}-\frac{m+n}{x+y}.\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{n}{y}-\frac{m+n}{x+y}\right).\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{nx+ny-my-ny}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{mx+nx-mx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \left(\frac{nx-my}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{nx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{nx-my}{x\left(x+y\right)}\times \frac{y\left(x+y\right)}{nx-my}=\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}Hence,proved.\end{array}$

$\begin{array}{l}\left(ii\right)Giventhat\frac{dy}{dx}=\frac{y}{x}\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{y}{x}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{x.\frac{dy}{dx}-y.1}{x^2}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{x.\frac{y}{x}-y}{x^2}\left[?\frac{dy}{dx}=\frac{y}{x}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{y-y}{x^2}=\frac{0}{x^2}=0\\ Hence,\frac{d^{2}y}{d{x}^2}=0Hence,proved.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\{\begin{array}{cc}\hfill x^2+3x+p,\hfill & \hfill x\le 1\hfill \\ \hfill qx+2,\hfill & \hfill x>1\hfill \end{array}atx=1.\\ L.H.L.f\text{'}\left(c\right)=\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}\\ f\text{'}\left(1\right)=\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{-}}{lim}\frac{\left(x^2+3x+p\right)-\left(1+3+p\right)}{x-1}\\ =\underset{h\to 0}{lim}\frac{\left[{\left(1-h\right)}^2+3\left(1-h\right)+p\right]-\left(1+3+p\right)}{1-h-1}\\ =\underset{h\to 0}{lim}\frac{\left[1+h^2-2h+3-3h+p\right]-\left(4+p\right)}{-h}\\ =\underset{h\to 0}{lim}\frac{\left[h^2-5h+4+p\right]-\left[4+p\right]}{-h}\\ =\underset{h\to 0}{lim}\frac{h^2-5h+4+p-4-p}{-h}\\ =\underset{h\to 0}{lim}\frac{h^2-5h}{-h}=\underset{h\to 0}{lim}\frac{h\left[h-5\right]}{-h}=5\\ R.H.L.f\text{'}\left(1\right)=\underset{x\to 1^{+}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{+}}{lim}\frac{\left(qx+2\right)-\left(1+3+p\right)}{x-1}\\ =\underset{h\to 0}{lim}\frac{\left[q\left(1+h\right)+2\right]-\left[4+p\right]}{1+h-1}\\ =\underset{h\to 0}{lim}\frac{q+qh+2-4-p}{h}=\underset{h\to 0}{lim}\frac{q+qh-2-p}{h}\\ Forexistingthelimit\\ q-2-p=0\Rightarrow q-p=0\dots \left(1\right)\\ \underset{h\to 0}{\Rightarrow lim}\frac{qh-0}{h}=q\\ IfL.H.L.f\text{'}\left(1\right)=R.H.L.f\text{'}\left(1\right)thenq=5.\\ Nowp\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\left|sinx+cosx\right|atx=\pi \\ Putg\left(x\right)=sinx+cosxandh\left(x\right)=\left|x\right|\\ \therefore h\left[g\left(x\right)\right]=h\left(sinx+cosx\right)=\left|sinx+cosx\right|\\ Now,g\left(x\right)=sinx+cosxisacontinuousfunctionsincesinxandcosxaretwocontinuous\\ functionsatx=\pi .\\ Weknowthateverymodulusfunctionisacontinuousfunctioneverywhere.\\ Hence,f\left(x\right)=\left|sinx+cosx\right|isacontinuousfunctionatx=\pi .\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Wehavef\left(x\right)=\frac{1}{t^2+t-2}\\ \Rightarrow f\left(t\right)=\frac{1}{{\left(\frac{1}{x-1}\right)}^2+\frac{1}{x-1}-2}\left[Puttingt=\frac{1}{x-1}\right]\\ =\frac{1}{\frac{1+x-1-2{\left(x-1\right)}^2}{{\left(x-1\right)}^2}}=\frac{{\left(x-1\right)}^2}{x-2x^2-2+4x}\\ =\frac{{\left(x-1\right)}^2}{-2x^2+5x-2}=\frac{{\left(x-1\right)}^2}{-\left(2x^2-5x+2\right)}\\ =\frac{{\left(x-1\right)}^2}{-\left[2x^2-4x-x+2\right]}=\frac{{\left(x-1\right)}^2}{-\left[2x\left(x-2\right)-1\left(x-2\right)\right]}\\ =\frac{{\left(x-1\right)}^2}{-\left(x-2\right)\left(2x-1\right)}=\frac{{\left(x-1\right)}^2}{\left(2-x\right)\left(2x-1\right)}\\ So,iff\left(t\right)isdiscontinuous,then2-x=0\therefore x=2\\ and2x-1=0\therefore x=\frac{1}{2}\\ Hence,therequiredofdiscontinuityare2and\frac{1}{2}.\end{array}$