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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]
Ans. 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]
Ans. 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]
Ans. 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)]
Ans. (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 $\mathrm{h}(\mathrm{g}(3))=38$