Mathematics - Continuity - Differentiability Question with Solution | TestHub

We have, $$\begin{array}{rl}& \underset{x\to 0}{\lim }\frac{g(x)\cos x-g(0)}{\sin x}\\ & =\underset{x\to 0}{\lim }\frac{g^{\prime }(x)\cos x-g(x)\sin x}{\cos x}=0\end{array}$$ Since, $f(x)=g(x)\sin x$ $$\begin{array}{rl}& \Rightarrow f^{\prime }(x)=g^{\prime }(x)\sin x+g(x)\cos x\\ & \Rightarrow f^{\prime \prime }(x)=g^{\prime \prime }(x)\sin x+2g^{\prime }(x)\cos x-g(x)\sin x\\ & \Rightarrow f^{\prime \prime }(0)=0\end{array}$$ Thus, ${\lim }_{x\to 0}[g(x)\cos x-g(0)\mathrm{cosec}x]=0=f^{\prime \prime }(0)$ $\Rightarrow$ Statement 1 is true. Statement 2. $f^{\prime }(x)=g^{\prime }(x)\sin x+g(x)\cos x$ $$\Rightarrow f^{\prime }(0)=g(0)$$ Statement 2 is not a correct explanation of Statement 1.