JEE Advanced · Mathematics · 23. C&D
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 \(2 f^{\prime}(0)=g(0)\).
- 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
Answer & Solution
Correct Answer
(B)
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
Step-by-step Solution
Detailed explanation
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.
\[
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \\
& =\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0
\end{aligned}
\]
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.
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