Ncert Solutions Maths class 12th

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

Sol:

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

Sol:

$\begin{array}{l}Lety=ta{n}^{-1}\left[\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right]\\ Putx=atan\theta \therefore \theta =ta{n}^{-1}\frac{x}{a}\\ \Rightarrow y=ta{n}^{-1}\left[\frac{3a^{2}atan\theta -a^{3}ta{n}^3\theta }{a^3-3a.a^{2}ta{n}^2\theta }\right]\\ \Rightarrow y=ta{n}^{-1}\left[\frac{3a^{3}tan\theta -a^{3}ta{n}^3\theta }{a^3-3a^{3}ta{n}^2\theta }\right]\\ \Rightarrow y=ta{n}^{-1}\left[\frac{3tan\theta -ta{n}^3\theta }{1-3ta{n}^2\theta }\right]\\ \Rightarrow y=ta{n}^{-1}\left(tan3\theta \right)\left[?tan3\theta =\frac{3tan\theta -ta{n}^3\theta }{1-3ta{n}^2\theta }\right]\\ \Rightarrow y=3\theta \Rightarrow y=3ta{n}^{-1}\frac{x}{a}\\ Differentiatingbothsidesw.r.t.x\\ \frac{dy}{dx}=3.\frac{d}{dx}\left(ta{n}^{-1}\frac{x}{a}\right)=3.\frac{1}{1+\frac{x^2}{a^2}}.\frac{d}{dx}.\left(\frac{x}{a}\right)=3.\frac{a^2}{a^2+x^2}.\frac{1}{a}=\frac{3a}{a^2+x^2}\\ Hence,\frac{dy}{dx}=\frac{3a}{a^2+x^2}\end{array}$

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

Sol:

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

Sol:

$\begin{array}{l}Lety=ta{n}^{-1}\left(\frac{acosx-bsinx}{bcosx+asinx}\right)\\ \Rightarrow y=ta{n}^{-1}\left[\frac{\frac{acosx}{bcosx}-\frac{bsinx}{bcosx}}{\frac{bcosx}{bcosx}+\frac{asinx}{bcosx}}\right]\\ \Rightarrow y=ta{n}^{-1}\left[\frac{\frac{a}{b}-tanx}{1+\frac{a}{b}tanx}\right]\\ \Rightarrow y=ta{n}^{-1}\frac{a}{b}-ta{n}^{-1}\left(tanx\right)\left[?ta{n}^{-1}\left(\frac{x-y}{1+xy}\right)=ta{n}^{-1}x-ta{n}^{-1}y\right]\\ \Rightarrow y=ta{n}^{-1}\frac{a}{b}-x\\ Differentiatingbothsidesw.r.t.x\\ \frac{dy}{dx}=\frac{d}{dx}\left[ta{n}^{-1}\frac{a}{b}\right]-\frac{d}{dx}\left(x\right)=0-1=-1\\ Hence,\frac{dy}{dx}=-1\end{array}$

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

Sol:

$\begin{array}{l}Lety=ta{n}^{-1}\left(secx+tanx\right)\\ Differentiatingbothsidesw.r.t.x\\ \frac{dy}{dx}=\frac{d}{dx}\left[ta{n}^{-1}\left(secx+tanx\right)\right]\\ =\frac{1}{1+{\left(secx+tanx\right)}^2}.\frac{d}{dx}\left(secx+tanx\right)\\ =\frac{1}{1+se{c}^{2}x+ta{n}^{2}x+2secxtanx}.\left(secxtanx+se{c}^{2}x\right)\\ =\frac{1}{\left(1+ta{n}^{2}x\right)+se{c}^{2}x+2secxtanx}.secx\left(tanx+secx\right)\\ =\frac{1}{se{c}^{2}x+se{c}^{2}x+2secxtanx}.secx\left(tanx+secx\right)\\ =\frac{1}{2se{c}^{2}x+2secxtanx}.secx\left(tanx+secx\right)\\ =\frac{1}{2secx\left(secx+tanx\right)}.secx\left(tanx+secx\right)=\frac{1}{2}\\ Hence,\frac{dy}{dx}=\frac{1}{2}\end{array}$

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

Sol:

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

Sol:

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

Sol:

$\begin{array}{l}Lety={\left(x+1\right)}^2{\left(x+2\right)}^3{\left(x+3\right)}^4\\ Takinglogonbothsides,\\ logy=log\left[{\left(x+1\right)}^2.{\left(x+2\right)}^3.{\left(x+3\right)}^4\right]\\ \Rightarrow logy=log{\left(x+1\right)}^2+log{\left(x+2\right)}^3+log{\left(x+3\right)}^4\left[?logxy=logx+logy\right]\\ \Rightarrow logy=2log\left(x+1\right)+3log\left(x+2\right)+4log\left(x+3\right)\left[?log{x}^y=ylogx\right]\\ Differentiatingbothsidesw.r.t.x\\ \frac{1}{y}.\frac{dy}{dx}=2.\frac{d}{dx}log\left(x+1\right)+3.\frac{d}{dx}log\left(x+2\right)+4.\frac{d}{dx}log\left(x+3\right)\\ \frac{1}{y}.\frac{dy}{dx}=2.\frac{1}{x+1}+3.\frac{1}{x+2}+4.\frac{1}{x+3}\\ \frac{dy}{dx}=y\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right]\\ ={\left(x+1\right)}^2{\left(x+2\right)}^3{\left(x+3\right)}^4\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right]\\ ={\left(x+1\right)}^2{\left(x+2\right)}^3{\left(x+3\right)}^4\left[\frac{2\left(x+2\right)\left(x+3\right)+3\left(x+1\right)\left(x+3\right)+4\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right]\\ =\left(x+1\right){\left(x+2\right)}^2{\left(x+3\right)}^3\left[2x^2+10x+12+3x^2+12x+9+4x^2+12x+8\right]\\ =\left(x+1\right){\left(x+2\right)}^2{\left(x+3\right)}^3\left[9x^2+34x+29\right]\\ Hence,\frac{dy}{dx}=\left(x+1\right){\left(x+2\right)}^2{\left(x+3\right)}^3\left[9x^2+34x+29\right]\end{array}$

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

Sol:

$\begin{array}{l}Lety=si{n}^{m}xco{s}^{n}x\\ Differentiatingbothsidesw.r.t.x\\ \frac{dy}{dx}=\frac{d}{dx}\left(si{n}^{m}xco{s}^{n}x\right)\\ =si{n}^{m}x.\frac{d}{dx}\left(co{s}^{n}x\right)+co{s}^{n}x.\frac{d}{dx}\left(si{n}^{m}x\right)\\ =si{n}^{m}x.n.co{s}^{n-1}x\frac{d}{dx}\left(cosx\right)+co{s}^{n}x.m.si{n}^{m-1}x\frac{d}{dx}\left(sinx\right)\\ =n.si{n}^{m}x.co{s}^{n-1}x.\left(-sinx\right)+m.co{s}^{n}x.si{n}^{m-1}x.cosx\\ =-n.si{n}^{m+1}x.co{s}^{n-1}x+m.co{s}^{n+1}x.si{n}^{m-1}x\\ =si{n}^{m}x.co{s}^{n}x\left[-n\frac{sinx}{cosx}+m.\frac{cosx}{sinx}\right]\\ Hence,\frac{dy}{dx}=si{n}^{m}x.co{s}^{n}x\left[-ntanx+m.cotx\right]\end{array}$

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

Sol:

$\begin{array}{l}Lety={\left(sinx\right)}^{cosx}\\ Takinglogonbothsides,\\ logy=log{\left(sinx\right)}^{cosx}\\ logy=cosx.log\left(sinx\right)\\ Differentiatingbothsidesw.r.t.x\\ \frac{1}{y}.\frac{dy}{dx}=\frac{d}{dx}cosx.log\left(sinx\right)\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=cosx.\frac{d}{dx}log\left(sinx\right)+log\left(sinx\right).\frac{d}{dx}cosx\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=cosx.\frac{1}{sinx}.\frac{d}{dx}\left(sinx\right)+log\left(sinx\right).\left(-sinx\right)\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=cotx.cosx-sinx.log\left(sinx\right)\\ \frac{dy}{dx}=y\left[cotx.cosx-sinx.log\left(sinx\right)\right]\\ Hence,\frac{dy}{dx}={\left(sinx\right)}^{cosx}\left[\frac{co{s}^{2}x}{sinx}-sinx.log\left(sinx\right)\right]\end{array}$