MCQ
Suppose for a differentiable function $h, h(0)=0$, $\mathrm{h}(1)=1$ and $\mathrm{h}^{\prime}(0)=\mathrm{h}^{\prime}(1)=2$. If $\mathrm{g}(\mathrm{x})=\mathrm{h}\left(\mathrm{e}^{\mathrm{x}}\right) \mathrm{e}^{\mathrm{h}(\mathrm{x})}$, then $g^{\prime}(0)$ is equal to:
  • A
    $5$
  • B
    $3$
  • C
    $8$
  • $4$

Answer

Correct option: D.
$4$
d
$ g(x)=h\left(e^x\right) \cdot e^{h(x)} $

$ g^{\prime}(x)=h\left(e^x\right) \cdot e^{h(x)} \cdot h^{\prime}(x)+e^{h(x)} h^{\prime}\left(e^x\right) \cdot e^x $

$ g^{\prime}(0)=h(1) e^{h(0)} h^{\prime}(0)+e^{h(0)} h^{\prime}(1) $

$ =2+2=4$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $n$ be a positive integer. Let  $A =\sum_{ k =0}^{ n }(-1)^{ k } n _{ C _{ k }}\left[\left(\frac{1}{2}\right)^{ k }+\left(\frac{3}{4}\right)^{ k }+\left(\frac{7}{8}\right)^{ k }+\left(\frac{15}{16}\right)^{ k }+\left(\frac{31}{32}\right)^{ k }\right]$ . If $63 A =1-\frac{1}{2^{30}},$ then $n$ is equal to ...... .
If ${S_n} = \frac{{n(n + 1)\left( {n + 2} \right)}}{6}$ then $\sum\limits_{n = 1}^\infty  {\frac{1}{{{t_n}}}}  = $
The number of ways that $8$ beads of different colours be string as a necklace is
If  $x = \sin \left( {2{{\tan }^{ - 1}}2} \right),\,y = \sin \left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{4}{3}} \right),$ then -
A curve is given by the equations $x = a\cos \theta + {1 \over 2}b\cos 2\theta ,$ $y = a\sin \theta + {1 \over 2}b\,\sin \,2\theta $, then the points for which ${{{d^2}y} \over {d{x^2}}} = 0,$ is given by
Let $H$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is
$\int_0^{\pi /2} {} \log \sin x\,dx = $
The solution set of $x<5$ and $x \geq 2$ is...
Let $f(\mathrm{x})=\left(\sin \left(\tan ^{-1} \mathrm{x}\right)+\sin \left(\cot ^{-1} \mathrm{x}\right)\right)^{2}-1,|\mathrm{x}|>1$ If $\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1}(f(x))\right) $ and $ y(\sqrt{3})=\frac{\pi}{6}$ then $y(-\sqrt{3})$ is equal to
If $A = \left[ {\begin{array}{*{20}{c}}{1/3}&2\\0&{2x - 3}\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}3&6\\0&{ - 1}\end{array}} \right]$and $AB = I$, then $x =$