SATHEE: Limit Continuity And Differentiability Ques 36

Limit Continuity And Differentiability Ques 36

Assertion and Reason

For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.

(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I

(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I

(d) Statement I is false; Statement II is true

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 I $\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$ and

Statement II $f^{\prime}(0)=g(0)$.

$(2008,3 \mathrm{M})$

Show Answer

Correct Answer: 36.(b)

Solution: (b) Statement II

We have, $\lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \quad $ $ \left[\frac{0}{0} \text { form }\right] $

$ =\quad \lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0 $

Since, $f(x)=g(x) \sin x$

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

and $f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x$ $ \quad \Rightarrow \quad f^{\prime \prime}(0)=0$

Thus, $\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)$

$\Rightarrow \quad$ Statement I is true.

Statement II $f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$ $\quad \Rightarrow \quad f^{\prime}(0)=g(0)$

Statement II is not a correct explanation of Statement I.