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:

$\begin{array}{l}Lety=\frac{x}{sinx}andz=sinx\\ Differentiatingboththeparametricfunctionsw.r.t.x\\ \frac{dy}{dx}=\frac{sinx.\frac{d}{dx}\left(x\right)-x.\frac{d}{dx}\left(sinx\right)}{{\left(sinx\right)}^2}\\ =\frac{sinx.1-x.cosx}{\left(si{n}^{2}x\right)}=\frac{sinx-x.cosx}{si{n}^{2}x}\\ \frac{dz}{dx}=cosx\\ \therefore \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{\frac{sinx-x.cosx}{si{n}^{2}x}}{cosx}=\frac{sinx-xcosx}{si{n}^{2}xcosx}\\ =\frac{sinx}{si{n}^{2}xcosx}-\frac{xcosx}{si{n}^{2}xcosx}\\ =\frac{tanx}{si{n}^{2}x}-\frac{x}{si{n}^{2}x}=\frac{tanx-x}{si{n}^{2}x}\\ Hence,\frac{dy}{dz}=\frac{tanx-x}{si{n}^{2}x}\end{array}$

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:

$\begin{array}{l}Giventhatx=asin2t\left(1+cos2t\right),y=bcos2t\left(1-cos2t\right)\\ Differentiatingboththeparametricfunctionsw.r.t.t,weget\\ \Rightarrow \frac{dx}{dt}=a\left[sin2t.\frac{d}{dt}\left(1+cos2t\right)+\left(1+cos2t\right).\frac{d}{dt}\left(sin2t\right)\right]\\ \Rightarrow =a\left[sin2t.\left(-sin2t\right).2+\left(1+cos2t\right)\left(cos2t\right).2\right]\\ \Rightarrow =a\left[-2si{n}^{2}2t+2cos2t+2co{s}^{2}2t\right]\\ \Rightarrow =a\left[2\left(co{s}^{2}2t-si{n}^{2}2t\right)+2cos2t\right]\\ \Rightarrow =a\left[2cos4t+2cos2t\right]\left[?cos2x=co{s}^{2}x-si{n}^{2}x\right]\\ \Rightarrow =2a\left[cos4t+cos2t\right]\\ Now,y=bcos2t\left(1-cos2t\right)\\ \Rightarrow \frac{dy}{dt}=b\left[cos2t.\frac{d}{dt}\left(1-cos2t\right)+\left(1-cos2t\right).\frac{d}{dt}\left(cos2t\right)\right]\\ \Rightarrow =b\left[cos2t.\left(sin2t\right).2+\left(1-cos2t\right)\left(-sin2t\right).2\right]\\ \Rightarrow =b\left[2sin2tcos2t-2sin2t+2sin2tcos2t\right]\\ \Rightarrow =b\left[sin4t-2sin2t+sin4t\right]\left[?sin2x=2sinxcosx\right]\\ \Rightarrow =b\left[2sin4t-2sin2t\right]=2b\left(sin4t-sin2t\right)\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2b\left[sin4t-sin2t\right]}{2a\left[cos4t+cos2t\right]}=\frac{b}{a}\left[\frac{sin4t-sin2t}{cos4t+cos2t}\right]\\ Putt=\frac{\pi }{4}\\ \therefore {\left(\frac{dy}{dx}\right)}_{att=\frac{\pi }{4}}=\frac{b}{a}\left[\frac{sin4\left(\frac{\pi }{4}\right)-sin2\left(\frac{\pi }{4}\right)}{cos4\left(\frac{\pi }{4}\right)+cos2\left(\frac{\pi }{4}\right)}\right]=\frac{b}{a}\left[\frac{sin\pi -sin\frac{\pi }{2}}{cos\pi +cos\frac{\pi }{2}}\right]\\ \text{\&thi}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatx=e^{cos2t},y=e^{sin2t}\\ \Rightarrow cos2t=logxandsin2t=logy\\ Differentiatingboththeparametricfunctionsw.r.t.t,weget\\ \Rightarrow \frac{dx}{dt}=e^{cos2t}.\frac{d}{dt}\left(cos2t\right)\\ \Rightarrow =e^{cos2t}\left(-sin2t\right).\frac{d}{dt}\left(2t\right)\\ \Rightarrow =e^{cos2t}\left(-sin2t\right).2=-2e^{cos2t}sin2t\\ Now,y=e^{sin2t}\\ \Rightarrow \frac{dy}{dt}=e^{sin2t}.\frac{d}{dt}\left(sin2t\right)\\ \Rightarrow =e^{sin2t}\left(cos2t\right).\frac{d}{dt}\left(2t\right)\\ \Rightarrow =e^{sin2t}\left(cos2t\right).2\\ \Rightarrow =2e^{sin2t}cos2t\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2e^{sin2t}cos2t}{-2e^{cos2t}sin2t}=\frac{e^{sin2t}cos2t}{-e^{cos2t}sin2t}\\ =\frac{ycos2t}{-xsin2t}=\frac{ylogx}{-xlogy}\left[\begin{array}{l}?cos2t=logx\\ sin2t=logy\end{array}\right]\\ Hence,\frac{dy}{dx}=\frac{ylogx}{-xlogy}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatx=\frac{1+logt}{t^2},y=\frac{3+2logt}{t}\\ Differentiatingboththeparametricfunctionsw.r.t.t,weget\\ \Rightarrow \frac{dx}{dt}=\frac{t^2.\frac{d}{dt}\left(1+logt\right)-\left(1+logt\right).\frac{d}{dt}\left(t^2\right)}{t^4}\\ \Rightarrow =\frac{t^2.\left(\frac{1}{t}\right)-\left(1+logt\right).2t}{t^4}\\ \Rightarrow =\frac{t-\left(1+logt\right).2t}{t^4}\\ \Rightarrow =\frac{t\left[1-2-2logt\right]}{t^4}=\frac{\left[-1-2logt\right]}{t^3}\\ \Rightarrow \frac{dx}{dt}=\frac{-\left(1+2logt\right)}{t^3}\\ Now,y=\frac{3+2logt}{t}\\ \Rightarrow \frac{dy}{dt}=\frac{t.\frac{d}{dt}\left(3+2logt\right)-\left(3+2logt\right).\frac{d}{dt}\left(t\right)}{t^2}\\ \Rightarrow =\frac{t.\left(\frac{2}{t}\right)-\left(3+2logt\right).1}{t^2}\\ \Rightarrow =\frac{2-3-2logt}{t^2}\\ \Rightarrow =\frac{-\left(1+2logt\right)}{t^2}\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-\left(1+2logt\right)}{t^2}}{\frac{-\left(1+2logt\right)}{t^3}}=\frac{t^3}{t^2}=t\\ Hence,\frac{dy}{dx}=t\end{array}$

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

Sol:

$\begin{array}{l}Now,takingtany=\frac{2t}{1-t^2}\\ Differentiatingbothsidesw.r.t.t,weget\\ \frac{d}{dt}\left(tany\right)=\frac{d}{dt}\left(\frac{2t}{1-t^2}\right)\\ \Rightarrow se{c}^{2}y.\frac{dy}{dt}=\frac{\left(1-t^2\right).\frac{d}{dt}\left(2t\right)-2t.\frac{d}{dt}\left(1-t^2\right)}{{\left(1-t^2\right)}^2}\\ \Rightarrow se{c}^{2}y.\frac{dy}{dt}=\frac{2.\left(1-t^2\right)-2t.\left(-2t\right)}{{\left(1-t^2\right)}^2}\\ \Rightarrow se{c}^{2}y.\frac{dy}{dt}=\frac{2-2t^2+4t^2}{{\left(1-t^2\right)}^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2+2t^2}{{\left(1-t^2\right)}^2}\times \frac{1}{se{c}^{2}y}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{1}{1+ta{n}^{2}y}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{1}{1+{\left(\frac{2t}{1-t^2}\right)}^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{1}{\frac{{\left(1-t^2\right)}^2+4t^2}{{\left(1-t^2\right)}^2}}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{{\left(1-t^2\right)}^2}{1+t^4-2t^2+4t^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{{\left(1-t^2\right)}^2}{1+t^4+2t^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2\left(1+t^2\right)}{{\left(1-t^2\right)}^2}\times \frac{{\left(1-t^2\right)}^2}{{\left(1+t^2\right)}^2}\\ \Rightarrow \frac{dy}{dt}=\frac{2}{1+t^2}\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{1+t^2}}{\frac{2}{1+t^2}}=1\\ Hence,\frac{dy}{dx}=1\end{array}$

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

Sol:

$\begin{array}{l}Giventhatx=3cos\theta -2co{s}^3\theta ,y=3sin\theta -2si{n}^3\theta \\ Differentiatingboththegivenparametricfunctionsw.r.t.\theta \\ \frac{dx}{d\theta }=-3sin\theta -6co{s}^2\theta .\frac{d}{d\theta }\left(cos\theta \right)\\ =-3sin\theta -6co{s}^2\theta .\left(-sin\theta \right)\\ =-3sin\theta +6co{s}^2\theta .sin\theta \\ \frac{dy}{d\theta }=3cos\theta -6si{n}^2\theta .\frac{d}{d\theta }\left(sin\theta \right)\\ =3cos\theta -6si{n}^2\theta .cos\theta \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3cos\theta -6si{n}^2\theta .cos\theta }{-3sin\theta +6co{s}^2\theta .sin\theta }\\ =\frac{cos\theta \left(3-6si{n}^2\theta \right)}{sin\theta \left(-3+6co{s}^2\theta \right)}=\frac{cos\theta \left[3-6\left(1-co{s}^2\theta \right)\right]}{sin\theta \left[-3+6co{s}^2\theta \right]}\\ =cot\theta \left(\frac{3-6+6co{s}^2\theta }{-3+6co{s}^2\theta }\right)=cot\theta \left(\frac{-3+6co{s}^2\theta }{-3+6co{s}^2\theta }\right)=cot\theta \\ Hence,\frac{dy}{dx}=cot\theta \end{array}$

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

Sol:

$\begin{array}{l}Giventhatx=e^{\theta }\left(\theta +\frac{1}{\theta }\right),y=e^{-\theta }\left(\theta -\frac{1}{\theta }\right)\\ Differentiatingboththeparametricfunctionsw.r.t.\theta \\ \frac{dx}{d\theta }=e^{\theta }\left(1-\frac{1}{{\theta }^2}\right)+\left(\theta +\frac{1}{\theta }\right).e^{\theta }\\ \frac{dx}{d\theta }=e^{\theta }\left(1-\frac{1}{{\theta }^2}+\theta +\frac{1}{\theta }\right)\Rightarrow e^{\theta }\left(\frac{{\theta }^2-1+{\theta }^3+\theta }{{\theta }^2}\right)\\ =e^{\theta }\frac{\left({\theta }^3+{\theta }^2+\theta -1\right)}{{\theta }^2}\\ y=e^{-\theta }\left(\theta -\frac{1}{\theta }\right)\\ \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}\right)+\left(\theta -\frac{1}{\theta }\right).\left(-e^{-\theta }\right)\\ \frac{dy}{d\theta }=e^{-\theta }\left(1+\frac{1}{{\theta }^2}-\theta +\frac{1}{\theta }\right)\Rightarrow e^{-\theta }\left(\frac{{\theta }^2+1-{\theta }^3+\theta }{{\theta }^2}\right)\\ =e^{-\theta }\frac{\left(-{\theta }^3+{\theta }^2+\theta +1\right)}{{\theta }^2}\\ \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{e^{-\theta }\frac{\left(-{\theta }^3+{\theta }^2+\theta +1\right)}{{\theta }^2}}{e^{\theta }\frac{\left({\theta }^3+{\theta }^2+\theta -1\right)}{{\theta }^2}}\\ =e^{-2\theta }\left(\frac{-{\theta }^3+{\theta }^2+\theta +1}{{\theta }^3+{\theta }^2+\theta -1}\right)\\ Hence,\frac{dy}{dx}=e^{-2\theta }\left(\frac{-{\theta }^3+{\theta }^2+\theta +1}{{\theta }^3+{\theta }^2+\theta -1}\right)\end{array}$

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

Sol:

$\begin{array}{l}Giventhatx=t+\frac{1}{t},y=t-\frac{1}{t}\\ Differentiatingboththegivenparametricfunctionsw.r.t.t\\ \frac{dx}{dt}=1-\frac{1}{t^2},\frac{dy}{dt}=1+\frac{1}{t^2}\\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1}\\ Hence,\frac{dy}{dx}=\frac{t^2+1}{t^2-1}\end{array}$