Class 12th

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

Sol:

$\begin{array}{l}Giventhat{\left(x^2+y^2\right)}^2=xy\\ \Rightarrow x^4+y^4+2x^2{y}^2=xy\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(y^4\right)+2.\frac{d}{dx}\left(x^2{y}^2\right)=\frac{d}{dx}\left(xy\right)\\ \Rightarrow 4x^3+4y^3.\frac{dy}{dx}+2\left[x^2.2y.\frac{dy}{dx}+y^2.2x\right]=x.\frac{dy}{dx}+y.1\\ \Rightarrow 4x^3+4y^3.\frac{dy}{dx}+4x^{2}y.\frac{dy}{dx}+4x{y}^2=x.\frac{dy}{dx}+y\\ \Rightarrow 4y^3.\frac{dy}{dx}+4x^{2}y.\frac{dy}{dx}-x.\frac{dy}{dx}=y-4x^3-4x{y}^2\\ \Rightarrow \left[4y^3+4x^{2}y-x\right]\frac{dy}{dx}=y-4x^3-4x{y}^2\\ \Rightarrow \frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4y^3+4x^{2}y-x}\\ Hence,\frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4y^3+4x^{2}y-x}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatta{n}^{-1}\left(x^2+y^2\right)=a\\ \Rightarrow x^2+y^2=tana\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(tana\right)\\ \Rightarrow 2x+2y.\frac{dy}{dx}=0\Rightarrow 2y.\frac{dy}{dx}=-2x\\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}\\ Hence,\frac{dy}{dx}=\frac{-x}{y}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatsec\left(x+y\right)=xy\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}sec\left(x+y\right)=\frac{d}{dx}\left(xy\right)\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\frac{d}{dx}\left(x+y\right)=x.\frac{dy}{dx}+y.1\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\left(1+\frac{dy}{dx}\right)=x.\frac{dy}{dx}+y\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right)+sec\left(x+y\right).tan\left(x+y\right).\frac{dy}{dx}=x.\frac{dy}{dx}+y\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\frac{dy}{dx}-x.\frac{dy}{dx}=y-sec\left(x+y\right).tan\left(x+y\right)\\ \Rightarrow \left[sec\left(x+y\right).tan\left(x+y\right)-x\right]\frac{dy}{dx}=y-sec\left(x+y\right).tan\left(x+y\right)\\ \Rightarrow \frac{dy}{dx}=\frac{y-sec\left(x+y\right).tan\left(x+y\right)}{sec\left(x+y\right).tan\left(x+y\right)-x}\\ Hence,\frac{dy}{dx}=\frac{y-sec\left(x+y\right).tan\left(x+y\right)}{sec\left(x+y\right).tan\left(x+y\right)-x}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatsinxy+\frac{x}{y}=x^2-y\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}sin\left(xy\right)+\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(y\right)\\ \Rightarrow cosxy.\frac{d}{dx}\left(xy\right)+\frac{y.\frac{d}{dx}\left(x\right)-x.\frac{dy}{dx}}{y^2}=2x-\frac{dy}{dx}\\ \Rightarrow cosxy\left[x.\frac{dy}{dx}+y.1\right]+\frac{y.1}{y^2}-\frac{x}{y^2}\frac{dy}{dx}=2x-\frac{dy}{dx}\\ \Rightarrow xcosxy.\frac{dy}{dx}+ycosxy+\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}=2x-\frac{dy}{dx}\\ \Rightarrow xcosxy.\frac{dy}{dx}-\frac{x}{y^2}\frac{dy}{dx}+\frac{dy}{dx}=2x-ycosxy-\frac{1}{y}\\ \Rightarrow \left[xcosxy-\frac{x}{y^2}+1\right]\frac{dy}{dx}=2x-ycosxy-\frac{1}{y}\\ \Rightarrow \left[\frac{x{y}^{2}cosxy-x+y^2}{y^2}\right]\frac{dy}{dx}=\frac{2xy-y^{2}cosxy-1}{y}\\ \Rightarrow \frac{dy}{dx}=\frac{2xy-y^{2}cosxy-1}{y}*\frac{y^2}{x{y}^{2}cosxy-x+y^2}\\ =\frac{2x{y}^2-y^{3}cos\left(xy\right)-y}{x{y}^{2}cos\left(xy\right)-x+y^2}\\ Hence,\frac{dy}{dx}=\frac{2x{y}^2-y^{3}cos\left(xy\right)-y}{x{y}^{2}cos\left(xy\right)-x+y^2}\end{array}$

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:

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

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

Sol:

Sample question papers help you know the difficulty level and type of questions asked for the Class 12 board exam. We cannot predict that questions will not be repeated from the sample paper. Students can expect a few questions or similar questions from the sample paper. Questions in the Class 12 board examination may be from any part of the syllabus. So, prepare thoroughly from the entire syllabus.

No, examiners will not deduct marks for exceeding the word limit. Marks will be deducted for spelling mistakes in the Language Papers. Students are advised to be careful with the spellings, so that they won't loose any marks. They are also advised not to waste time by writing much more than the word limit. 

Students have to follow the Class 12 syllabus applicable for the year for which they are appearing the Class 12 board exam. They can check the Class 12th syllabus on the official website of the board. They can also take help from their teachers for the Class 12th  syllabus.