# The function f (x) = |cos x| is

Question:

The function $f(x)=|\cos x|$ is

(a) everywhere continuous and differentiable

(b) everywhere continuous but not differentiable at $(2 n+1) \pi / 2, n \in Z$

(c) neither continuous nor differentiable at $(2 n+1) \pi / 2, n \in Z$

(d) none of these

Solution:

(b) everywhere continuous but not differentiable at $(2 n+1) \pi / 2, n \in Z$

We have,

$f(x)=|\cos x|$

$\Rightarrow f(x)=\left\{\begin{array}{rc}\cos x, & 2 n \pi \leq x<(4 n+1) \frac{\pi}{2} \\ 0, & x=(4 n+1) \frac{\pi}{2} \\ -\cos x, & (4 n+1) \frac{\pi}{2} When,$x$is in first quadrant, i.e.$2 n \pi \leq x<(4 n+1) \frac{\pi}{2}$, we have$f(x)=\cos x$which being a trigonometrical function is continuous and differentiable in$\left(2 n \pi,(4 n+1) \frac{\pi}{2}\right)$When,$x$is in second quadrant or in third quadrant, i.e.,$(4 n+1) \frac{\pi}{2}

$f(x)=-\cos x$ which being a trigonometrical function is continuous and differentiable in $\left((4 n+1) \frac{\pi}{2},(4 n+3) \frac{\pi}{2}\right)$

When, $x$ is in fourth quadrant, i.e., $(4 n+3) \frac{\pi}{2}$f(x)=\cos x$which being a trigonometrical function is continuous and differentiable in$\left((4 n+3) \frac{\pi}{2},(2 n+2) \pi\right)$Thus possible point of non-differentiability of$f(x)$are$x=(4 n+1) \frac{\pi}{2},(4 n+3) \frac{\pi}{2}$Now, LHD [at$\left.x=(4 n+1) \frac{\pi}{2}\right]=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{f(x)-f\left((4 n+1) \frac{\pi}{2}\right)}{x-(4 n+1) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{\cos x-0}{x-(4 n+1) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{-\sin x}{1-0} \quad$[By L' Hospital rule]$=-1$And$\operatorname{RHD}\left(\right.$at$\left.x=(4 n+1) \frac{\pi}{2}\right)=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} \frac{f(x)-f\left((4 n+1) \frac{\pi}{2}\right)}{x-(4 n+1) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} \frac{-\cos x-0}{x-(4 n+1) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} \frac{\sin x}{1-0} \quad$[By L'Hospital rule]$=1\therefore \lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} f(x) \neq \lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{+}} f(x)$So$f(x)$is not differentiable at$x=(4 n+1) \frac{\pi}{2}$Now, LHD [at$\left.x=(4 n+3) \frac{\pi}{2}\right]\lim _{x \rightarrow(4 n+1) \frac{\pi}{2}^{-}} \frac{f(x)-f\left((4 n+3) \frac{\pi}{2}\right)}{x-(4 n+3) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{-}} \frac{-\cos x-0}{x-(4 n+3) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{-}} \frac{\sin x}{1-0} \quad$[By L' Hospital rule]$=1$And$\operatorname{RHD}\left(\right.$at$\left.x=(4 n+3) \frac{\pi}{2}\right)=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} \frac{f(x)-f\left((4 n+3) \frac{\pi}{2}\right)}{x-(4 n+3) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} \frac{\cos x-0}{x-(4 n+3) \frac{\pi}{2}}=\lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} \frac{-\sin x}{1-0} \quad$[By L' Hospital rule]$=-1\therefore \lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{-}} f(x) \neq \lim _{x \rightarrow(4 n+3) \frac{\pi}{2}^{+}} f(x)$So$f(x)$is not differentiable at$x=(4 n+3) \frac{\pi}{2}$Therefore,$f(x)$is neither differentiable at$(4 n+1) \frac{\pi}{2}$nor at$(4 n+3) \frac{\pi}{2}$i.e.$f(x)$is not differentiable at odd multiples of$\frac{\pi}{2}$i. e.$f(x)$is not differentiable at$x=(2 n+1) \frac{\pi}{2}$Therefore,$f(x)$is everywhere continuous but not differentiable at$(2 n+1) \frac{\pi}{2}\$.