Method Of Differentiation – JEE Advanced Previous Year Questions with Solutions
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Q. If the function $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{e}^{\frac{\mathrm{x}}{2}}$ and $\mathrm{g}(\mathrm{x})=\mathrm{f}^{-1}(\mathrm{x}),$ then the value of $\mathrm{g}^{\prime}(1)$ is [JEE 2009, 4]

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Sol. 2 $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{e}^{\mathrm{x} / 2}, \mathrm{g}(\mathrm{x})=\mathrm{f}^{-1}(\mathrm{x})$ $\Rightarrow \mathrm{g}^{\prime}(\mathrm{f}(\mathrm{x})) \cdot \mathrm{f}(\mathrm{x})=1$ $\operatorname{Put} \mathrm{f}(\mathrm{x})=1$ $\Rightarrow \mathrm{x}^{3}+\mathrm{e}^{\mathrm{x} / 2}=1$ $\Rightarrow \mathrm{x}=0$ $\Rightarrow \mathrm{g}^{\prime}(1) \cdot \mathrm{f}^{\prime}(0)=1,\left\{\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+\mathrm{e}^{\mathrm{x} / 2} \cdot \frac{1}{2}, \mathrm{f}^{\prime}(0)=\frac{1}{2}\right\}$ $\Rightarrow \mathrm{g}^{\prime}(1)=2$b

Q. Let $f(\theta)=\sin \left(\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right),$ where $-\frac{\pi}{4}<\theta<\frac{\pi}{4} .$ Then the value of $\frac{\mathrm{d}}{\mathrm{d}(\tan \theta)}(f(\theta))$ is [JEE 2011, 4]

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Sol. 1 Let $f(\theta)=\sin \alpha$ where $\alpha=\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)$ $\Rightarrow \quad \tan \alpha=\frac{\sin \theta}{\sqrt{\cos 2 \theta}}$ $\Rightarrow \quad \sin \alpha=\frac{\sin \theta}{\cos \theta}=\tan \theta \quad\left(\because \theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\right)$ $\Rightarrow \quad f(\theta)=\tan \theta$ $\Rightarrow \quad \frac{\mathrm{d}(f(\theta))}{\mathrm{d}(\tan \theta)}=1$

Q. The slope of the tangent to the curve $\left(y-x^{5}\right)^{2}=x\left(1+x^{2}\right)^{2}$ at the point $(1,3)$ is [JEE(Advanced)-2014, 3]

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Sol. 8 $\left(\mathrm{y}-\mathrm{x}^{5}\right)^{2}=\mathrm{x}\left(1+\mathrm{x}^{2}\right)^{2}$ $\Rightarrow 2\left(\mathrm{y}-\mathrm{x}^{5}\right)\left(\frac{\mathrm{dy}}{\mathrm{dx}}-5 \mathrm{x}^{4}\right)=\left(1+\mathrm{x}^{2}\right)^{2}+2 \mathrm{x}\left(1+\mathrm{x}^{2}\right) \cdot 2 \mathrm{x}$ Put $\mathrm{x}=1, \mathrm{y}=3$ $\frac{\mathrm{dy}}{\mathrm{dx}}=8$

Q. Let $f: R \rightarrow R, g: R \rightarrow R$ and $h: R \rightarrow R$ be differentiable functions such that $f(x)=x^{3}+3 x+2, g(f(x))=x$ and $h(g(g(x)))=x$ for all $x \in R .$ Then- (A) $\mathrm{g}^{\prime}(2)=\frac{1}{15}$ (B) h'(1) = 666 (C) h(0) = 16 (D) h(g(3)) = 36 [JEE(Advanced)-2016, 4(-2)]

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Sol. (B,C) (A) $f^{\prime}(x)=3 x^{2}+3$ so, $g^{\prime}(2)=\frac{1}{f^{\prime}(0)}$ (Given $\left.g(x)=f^{-1}(x)\right)$ $\Rightarrow g^{\prime}(2)=\frac{1}{3}$ (B) $\operatorname{h}(\mathrm{g}(\mathrm{g}(\mathrm{x}))=\mathrm{x}$ $\mathrm{h}^{\prime}\left(\mathrm{g}(\mathrm{g}(\mathrm{x}))=\frac{1}{\mathrm{g}^{\prime}(\mathrm{g}(\mathrm{x})) \cdot \mathrm{g}^{\prime}(\mathrm{x})}\right.$ Now, $\mathrm{g}(\mathrm{g}(\mathrm{x}))=1$ $\mathrm{g}(\mathrm{x})=f(1)=6$ $\therefore \mathrm{x}=f(6)=236$ so h'(1) $=\frac{1}{\mathrm{g}^{\prime}(6) \cdot \mathrm{g}^{\prime}(236)}=\frac{1}{\frac{1}{6} \cdot \frac{1}{111}}$ $\Rightarrow \mathrm{h}^{\prime}(1)=666$ (C) $\mathrm{g}(\mathrm{g}(\mathrm{x}))=0$ $\therefore \mathrm{g}(\mathrm{x})=\mathrm{g}^{-1}(0)$ $\Rightarrow \mathrm{g}(\mathrm{x})=f(0)$ $\Rightarrow \mathrm{g}(\mathrm{x})=2$ $\Rightarrow \mathrm{x}=\mathrm{g}^{-1}(2)$ $\Rightarrow \mathrm{x}=f(2)$ $\Rightarrow \mathrm{x}=16 \mathrm{so} \mathrm{h}(0)=16$ (D) $\mathrm{g}(\mathrm{x})=3$ $\Rightarrow \quad \mathrm{x}=\mathrm{g}^{-1}(3)$ $\Rightarrow \quad \mathrm{x}=f(3)$ $\Rightarrow \mathrm{x}=38$ so $\mathrm{h}(\mathrm{g}(3))=38$

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• October 18, 2020 at 5:18 pm

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• September 21, 2020 at 5:52 am

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