Question

The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
(a) R
(b) R − {3}
(c) (0, ∞)
(d) none of these

Answer 1

(b) $R-\left(3\right)$

$$\begin{gathered} \left(\mathrm{LHD}\mathrm{at}x=3\right)=\underset{x\to 3^{-}}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3} \\ \left(\mathrm{LHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{f\left(3-h\right)-f\left(3\right)}{3-h-3} \\ \left(\mathrm{LHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{f\left(3-h\right)-f\left(3\right)}{-h} \\ \left(\mathrm{LHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{\left|3-h-3\right|\cos \left(3-h\right)-f\left(3\right)}{-h} \\ \left(\mathrm{LHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{h\cos \left(3-h\right)-0}{-h}=-\cos 3 \\ \left(\mathrm{RHD}\mathrm{at}x=3\right)=\underset{x\to 3^{+}}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3} \\ \left(\mathrm{RHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{3+h-3} \\ \left(\mathrm{RHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{h} \\ \left(\mathrm{RHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{\left|3+h-3\right|\cos \left(3+h\right)-f\left(3\right)}{h} \\ \left(\mathrm{RHD}\mathrm{at}x=3\right)=\underset{h\to 0}{\lim }\frac{h\cos \left(3+h\right)-0}{h}=\cos 3 \end{gathered}$$

So, f(x) is not differentiable at x = 3.

Also, f(x) is differentiable at all other points because both modulus and cosine functions are differentiable and the product of two differentiable function is differentiable.