# The set of points where the function f (x) given by

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) $(0, \infty)$

(d) none of these

Solution:

(b) $R-(3)$

$(\mathrm{LHD}$ at $x=3)=\lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}$

$(\mathrm{LHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{3-h-3}$

$(\mathrm{LHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h}$

$(\mathrm{LHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{|3-h-3| \cos (3-h)-f(3)}{-h}$

$(\mathrm{LHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{h \cos (3-h)-0}{-h}=-\cos 3$

$(\mathrm{RHD}$ at $x=3)=\lim _{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}$

$(\mathrm{RHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{3+h-3}$

$(\mathrm{RHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}$

$(\mathrm{RHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{|3+h-3| \cos (3+h)-f(3)}{h}$

$(\mathrm{RHD}$ at $x=3)=\lim _{h \rightarrow 0} \frac{h \cos (3+h)-0}{h}=\cos 3$

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.