Q34 / 69 · JEE Advanced 2008 (Paper 1) · Mathematics
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 $1 \lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.
Statement $2 f^{\prime}(0)=g(0)$.
- AStatement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
- Correct answerBStatement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
- CStatement 1 is true, Statement 2 is false.
- DStatement 1 is false, Statement 2 is true
Solution
We have,
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \\
& =\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0
\end{aligned}
$$
Since, $f(x)=g(x) \sin x$
$$
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x \\
& \Rightarrow \quad f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x \\
& \Rightarrow \quad f^{\prime \prime}(0)=0
\end{aligned}
$$
Thus, $\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)$
$\Rightarrow$ Statement 1 is true.
Statement 2. $f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$
$$
\Rightarrow \quad f^{\prime}(0)=g(0)
$$
Statement 2 is not a correct explanation of Statement 1.
Hence, the correct answer is option: 'B'.