Mathematics - Quadratic Equation Question with Solution | TestHub

$|x|$ is non-differentiable function at $\begin{array}{rl}& x=0\text{ as L.H.D }=-1\text{ and R.H. }\mathrm{D}=1\\ & \because |x|=\{\begin{array}{cl}x,& x\ge 0\\ -x,& x<0\end{array}\end{array}$ But $\cos |h|$ is differentiable $\therefore$ Any combination of two such functions will be non-differentiable. Hence option (a) and (b) are ruled out. Now, consider $\sin |x|+|x|$ $\begin{array}{rl}& L^{\prime }=\underset{h\to 0}{\lim }\frac{\sin |-h|+|-h|}{-h}=\underset{h\to 0}{\lim }\frac{\sin h}{-h}-1=-1-1=-2\\ & R^{\prime }=\underset{h\to 0}{\lim }\frac{\sin |h|+|h|}{h}=\underset{h\to 0}{\lim }\frac{\sin h}{h}+1=1+1=2\\ & \text{ Consider }\sin |x|-|x|\\ & L^{\prime }=\underset{h\to 0}{\lim }\frac{\sin |-h|-|-h|}{-h}=\underset{h\to 0}{\lim }\frac{\sin h}{-h}+1=0\\ & \end{array}$