Limit Continuity And Differentiability Ques 3
- If $g$ is the inverse of a function $f$ and $f^{\prime}(x)=\frac{1}{1+x^5}$, then $g(x)$ is equal to
(2015)
(a) $1+x^5$
(b) $5 x^4$
(c) $\frac{1}{1+\{g(x)\}^5}$
(d) $1+\{g(x)\}^5$
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Answer:
Correct Answer: 3.(d)
Solution: (d) Here, $g$ is the inverse of $f(x)$.
$\Rightarrow \quad f \circ g(x)=x$
On differentiating w.r.t. $x$, we get
$ \begin{aligned} & f^{\prime}\{g(x)\} \times g^{\prime}(x)=1 \Rightarrow g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))} \\ & =\frac{1}{\frac{1}{1+\{g(x)\}^5}} \quad\left[\because f^{\prime}(x)=\frac{1}{1+x^5}\right] \\ & \Rightarrow \quad g^{\prime}(x)=1+\{g(x)\}^5 \\ \end{aligned} $
BETA
