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# Method Of Differentiation - JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
Q. Let y be an implicit function of x defined by $x^{2 x}-2 x^{x}$ cot $y-1=0 .$ Then $y^{\prime}(1)$ equals : (1) $\log 2$ $(2)-\log 2$ (3) –1 (4) 1 [AIEEE-2009]
Ans. (3) $\left(x^{x}\right)^{2}-2 \cot y x^{x}-1=0$ $\mathrm{x}^{\mathrm{x}}=\frac{2 \cot y \pm \sqrt{4 \cot ^{2} y+4}}{2}\left\{\begin{array}{l}{\text { at } \mathrm{x}=1} \\ {1=\cot y+\csc y} \\ {\Rightarrow y=\frac{\pi}{2}}\end{array}\right.$ $=\cot y \pm \csc y$ $\mathrm{x}^{\mathrm{x}}=\cot \mathrm{y}+\csc \mathrm{y}$ diff. w.r. to $\mathrm{x}$ $\mathrm{x}^{\mathrm{x}}(1+\log \mathrm{x})=\left[-\csc ^{2} \mathrm{y}-\csc \mathrm{y} \cot \mathrm{y}\right] \frac{\mathrm{d} y}{\mathrm{dx}}$ $1=-\csc y[\csc y+\cot y] \frac{d y}{d x}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=-1$
Q. Let $\mathrm{f}:(-1,1) \rightarrow \mathrm{R}$ be a differentiable function with $\mathrm{f}(0)=-1$ and $\mathrm{f}(0)=1 .$ Let $\mathrm{g}(\mathrm{x})=[\mathrm{f}(2 \mathrm{f}(\mathrm{x})+$ $2)]^{2} .$ Then $\mathrm{g}^{\prime}(0):$ (1) 4            (2) –4             (3) 0              (4) –2 [AIEEE-2010]
Ans. (2) $\mathrm{g}(\mathrm{x})=[\mathrm{f}(2 \mathrm{f}(\mathrm{x})+2)]^{2}$ $\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{f}[2 \mathrm{f}(\mathrm{x})+2] \mathrm{f}^{\prime}[2 \mathrm{f}(\mathrm{x})+2] \cdot 2 \mathrm{f}^{\prime}(\mathrm{x})$ Put $\mathrm{x}=0$ $\mathrm{g}^{\prime}(0)=2 \mathrm{f}[2 \mathrm{f}(0)+2] \mathrm{f}^{\prime}[2 \mathrm{f}(0)+2] 2 \mathrm{f}^{\prime}(0)$ $=2 \mathrm{f}(2(-1)+2) \mathrm{f}^{\prime}(2(-1)+2) 2 \mathrm{f}^{\prime}(0)$ $=2 \mathrm{f}(0) \mathrm{f}^{\prime}(0) 2 \mathrm{f}^{\prime}(0)$ $=4(-1)(1)(1)=-4$
Q. $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}$ equals : (1) $\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2}$ (2) $-\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-3}$ (3) $\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{-1}$ $(4)-\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{-1}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-3}$ [AIEEE-2011]
Ans. (2) $\frac{\mathrm{d}}{\mathrm{dy}}\left(\left(\frac{\mathrm{d} y}{\mathrm{dx}}\right)^{-1}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\left(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-1}\right) \cdot \frac{\mathrm{dx}}{\mathrm{dy}}$ $=-\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-2} \cdot \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dx}^{2}} \cdot \frac{\mathrm{dx}}{\mathrm{dy}}=-\left(\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{-3}$
Q. If $\mathrm{y}=\sec \left(\tan ^{-1} \mathrm{x}\right),$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=1$ is equal to : (1) $\frac{1}{\sqrt{2}}$ (2) $\frac{1}{2}$ (3) 1 (4) $\sqrt{2}$ [JEE-MAIN-2013]
Ans. (1) $\mathrm{y}=\sec \left(\tan ^{-1} \mathrm{x}\right)=\sqrt{1+\mathrm{x}^{2}}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$ $\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{x}=1}=\frac{1}{\sqrt{2}}$
Q. If $\mathrm{g}$ is the inverse of a function $f$ and $f^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}^{5}},$ then $\mathrm{g}^{\prime}(\mathrm{x})$ is equal to : (1) $1+x^{5}$ (2) $5 x^{4}$ (3) $\frac{1}{1+\{\mathrm{g}(\mathrm{x})\}^{5}}$ (4) $1+\{\mathrm{g}(\mathrm{x})\}^{5}$ [JEE-MAIN-2014]
Ans. (4) $\mathrm{f}(\mathrm{g}(\mathrm{x}))=\mathrm{x}$ $\Rightarrow \mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x})) \cdot \mathrm{g}^{\prime}(\mathrm{x})=1$ $\therefore \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x}))}=1+\{\mathrm{g}(\mathrm{x})\}^{5}$
Q. If for $x \in\left(0, \frac{1}{4}\right),$ the derivative of $\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^{3}}\right)$ is $\sqrt{x} \cdot g(x),$ then $g(x)$ equals : (1) $\frac{3}{1+9 x^{3}}$ (2) $\frac{9}{1+9 x^{3}}$ (3) $\frac{3 x \sqrt{x}}{1-9 x^{3}}$ (4) $\frac{3 \mathrm{x}}{1-9 \mathrm{x}^{3}}$ [JEE-MAIN-2017]
Ans. (2) Let $\mathrm{y}=\tan ^{-1}\left(\frac{6 \mathrm{x} \sqrt{\mathrm{x}}}{1-9 \mathrm{x}^{3}}\right)$ where $\mathrm{x} \in\left(0, \frac{1}{4}\right)$ $=\tan ^{-1}\left(\frac{2 \cdot\left(3 \mathrm{x}^{3 / 2}\right)}{1-\left(3 \mathrm{x}^{3 / 2}\right)}\right)=2 \tan ^{-1}\left(3 \mathrm{x}^{3 / 2}\right)$ As $3 x^{3 / 2} \in\left(0, \frac{3}{8}\right)$ $\therefore \frac{d y}{d x}=2 \times \frac{1}{1+9 x^{3}} \times 3 \times \frac{3}{2} \times x^{1 / 2}$ \begin{aligned} &=\frac{9}{1+9 \mathrm{x}^{3}} \sqrt{\mathrm{x}} \\ \therefore \mathrm{g}(\mathrm{x}) &=\frac{9}{1+9 \mathrm{x}^{3}} \end{aligned}
Q. If $\mathrm{x}=\sqrt{2^{\cos \mathrm{c}^{-1} \mathrm{t}}}$ and $\mathrm{y}=\sqrt{2^{\mathrm{sc}-\mathrm{t}}}(|\mathrm{t}| \geq 1),$ then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to : (1) $-\frac{y}{x}$ ( 2)$\frac{\mathrm{x}}{\mathrm{y}}$ $(3)-\frac{x}{y}$ (4) $\frac{y}{x}$ [JEE-MAIN-2018]
Ans. (1)
Q. If $\mathrm{f}(\mathrm{x})=\sin ^{-1}\left(\frac{2 \times 3^{\mathrm{x}}}{1+9^{\mathrm{x}}}\right),$ then $\mathrm{f}^{\prime}\left(-\frac{1}{2}\right)$ equals : (1) $\sqrt{3} \log _{e} \sqrt{3}$ (2) $-\sqrt{3} \log _{e} 3$ (3) $-\sqrt{3} \log _{e} \sqrt{3}$ (4) $\sqrt{3} \log _{e} 3$ [JEE-MAIN-2018]
Ans. (1)
Q. If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{cc}{\cos \mathrm{x}} & {\mathrm{x}} \\ {2 \sin \mathrm{x}} & {\mathrm{x}^{2} 2 \mathrm{x}} \\ {\tan \mathrm{x}} & {\mathrm{x}} & {1}\end{array}\right|,$ then $\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{x}}$ (1) exists and is equal to 0 (2) exists and is equal to –2 (3) exists and is equal to 2 (4) does not exist [JEE-MAIN-2018]
Ans. (2)
Q. If $x^{2}+y^{2}+\sin y=4,$ then the value of $\frac{d^{2} y}{d x^{2}}$ at the point $(-2,0)$ is (1) –34             (2) –32            (3) –2               (4) 4
Ans. (1)

mufassira
Aug. 26, 2021, 8:41 p.m.
sir some of these quetsions solutions have not given can you please make sure that solutions should contain
techno
March 14, 2021, 8:31 p.m.
and one more thing add some more questions....../////
techno
March 14, 2021, 8:30 p.m.
can you send me the solution of last question and 7th question.....
lucky gaur
Feb. 23, 2021, 5:04 p.m.
these question answer have not full solution understand for an average student.......
Csct
Dec. 13, 2020, 9:52 a.m.
Good
Allu arjun
Dec. 9, 2020, 7:45 a.m.
Nice stuff amd content add some more questions
NIHARIKA
Sept. 28, 2020, 4:51 p.m.
some more questions
Saketh
Sept. 22, 2020, 9:41 a.m.
There are some typing mistakes I mean one letter is on another letter
Mahesh
Sept. 8, 2020, 7:46 a.m.
Good one
Sakshi
Sept. 6, 2020, 5:16 p.m.
Nice thanks for uploding this quetions on Google
navyug
Sept. 4, 2020, 10:59 p.m.
bakvas
ravi
July 28, 2020, 11:35 a.m.
super
Alekhya
July 15, 2020, 11:49 a.m.
8th question was incorrect its f(1/2) please see it once
Akash
June 27, 2020, 3:48 p.m.
Nice