Question

Assertion :Let $f$ and $g$ be real-valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime }\left(x\right)$ is countinuous, $g\left(0\right)\ne 0,g^{\prime }\left(0\right)=0,g^{\prime \prime }\left(0\right)\ne 0$ and $f\left(x\right)=g\left(x\right).\sin x.$

$\underset{x\to 0}{lim}\{g\left(x\right)\cdot \cot x-g\left(0\right)\cdot co\sec x=f^{\prime \prime }\left(0\right)\},$ and

Reason: $f^{\prime }\left(0\right)=g\left(0\right).$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

As $f\left(x\right)=g\left(x\right).\sin x.$

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

Putting $x=0$

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

Now, $f^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)-f^{\prime }\left(0\right)}{x}$

$=\underset{x\to 0}{lim}\frac{g\left(x\right)\cos x+g^{\prime }\left(x\right)\sin x-g\left(0\right)}{x}$

$=\underset{x\to 0}{lim}\frac{g\left(x\right)\cos x-g\left(0\right)}{x}+\underset{x\to 0}{lim}\frac{g^{\prime }\left(x\right)\sin x}{x}$

$=\underset{x\to 0}{lim}\frac{g\left(x\right)\cos x-g\left(0\right)}{x}$

$=\underset{x\to 0}{lim}\frac{g\left(x\right)\cos x-g\left(0\right)}{\sin x}$

$={lim}_{x\to 0}\left(g\left(x\right)\cot x-g\left(0\right)\csc x\right)$