If $$g$$ is the inverse function of $$f$$ and $$f'\left( x \right) = \sin \,x$$ then $$g'\left( x \right)$$ is :
A.
$${\text{cosec}}\left\{ {g\left( x \right)} \right\}$$
B.
$${\text{sin}}\left\{ {g\left( x \right)} \right\}$$
C.
$$ - \frac{1}{{{\text{sin}}\left\{ {g\left( x \right)} \right\}}}$$
D.
none of these
Answer :
$${\text{cosec}}\left\{ {g\left( x \right)} \right\}$$
Solution :
$$\eqalign{
& f\left\{ {g\left( x \right)} \right\} \Rightarrow \frac{{df\left\{ {g\left( x \right)} \right\}}}{{dg\left( x \right)}}.g'\left( x \right) = 1 \cr
& \Rightarrow \sin \left\{ {g\left( x \right)} \right\}.g'\left( x \right) = 1 \cr
& \Rightarrow g'\left( x \right) = \frac{1}{{\sin \left\{ {g\left( x \right)} \right\}}} \cr} $$
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-