Gujarati
Hindi
5. Continuity and Differentiation
normal

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq$ 0 , and $f(x)=g(x) \sin x$.

$STATEMENT$ $-1: \lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.and

$STATEMENT$ $-2: f^{\prime}(0)=g(0)$.

A

Statement-$1$ is True, Statement -$2$ is True; Statement-$2$ is a correct explanation for Statement-$1$

B

Statement -$1$ is True, Statement - $2$ is True; Statement-$2$ is $NOT$ a correct explanation for Statement-$1$

C

Statement -$1$ is True, Statement -$2$ is False

D

Statement -$1$ is False, Statement -$2$ is True

(IIT-2008)

Solution

$ f^{\prime}(x)=g(x) \cos x+\sin x \cdot g^{\prime}(x) $

$ \Rightarrow f^{\prime}(0)=g(0) $

$ f^{\prime \prime}(x)=2 g^{\prime}(x) \cos x-g(x) \sin x+\sin x g^{\prime \prime}(x) $

$ \Rightarrow f^{\prime}(0)=2 g^{\prime}(0)=0$

But $\lim _{x \rightarrow 0}[g(x) \cot x-g(0) \operatorname{cosec} x]=\lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x}=\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=g^{\prime}(0)=0=f^{\prime \prime}(0)$.

Standard 12
Mathematics

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