Mathematics - Application of Derivative Question with Solution | TestHub

(A) RHD at $x=0:{\lim }_{h\to 0}\frac{\left(2-2h^2-h^2\sin \frac{1}{h}\right)-2}{h}=0$ Similarly LHD at $\mathrm{x}=0$ is also equal to 0 . $\therefore$ Differentiable at $\mathrm{x}=0$ (B) $f^{\prime }(x)=-4x-2x\sin \frac{1}{x}-x^2\left(\cos \frac{1}{x}\right)\left(\frac{-1}{x^2}\right)$ $\begin{array}{rl}& f^{\prime }(x)=-\left(4x+2x\sin \frac{1}{x}\right)+\cos \frac{1}{x}\\ & f^{\prime }(x)=-\left(2x\left(4-\sin \frac{1}{x}\right)\right)+\cos \frac{1}{x}\end{array}$ for $\mathrm{x}\in (0,\delta )$ $\Rightarrow$ We can't say $f(x)$ is decreasing on $(0,\delta )$ as $\cos \frac{1}{x}$ oscillates. (C) for $x\in (-\delta ,0)$, for any $\delta >0$ $\Rightarrow f(x)$ is not increasing on $(-\delta ,0)$ as $\cos \frac{1}{x}$ oscillates from -1 to 1. (D) $f(0)=2$ $\begin{array}{rl}& f(0+h)<2\\ & f(0-h)<2\end{array}$ $\therefore \mathrm{x}=0$ is local maxima