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

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We have, f\left( x \right) = \left| \cos x \right| `$\Rightarrow$ f(x) ={(cosx , 2npile x < (4n +1)_2^pi),(0, x= (4n +1)_2^pi),(-cos x, (4n +1)_2^pi < x< (4n +3)^_2^pi),(0,x = (4n + 3)_2^pi),(cos x, (4n +3)_2^pi < x le (2n +2)pi):}` \text { When, x is in first quadrant, i . e . 2n}$\pi$ \leq x < \left( 4n + 1 \right)$\pi$/2 , \text { we have} f\left( x \right) = \text { cos x which being a trigonometrical function is continuous and differentiable in} \left( 2n$\pi$, \left( 4n + 1 \right)$\pi$/2 \right) \text { When, x is in second quadrant or in third quadrant, i . e }. , \left( 4n + 1 \right)$\pi$/2 < x < \left( 4n + 3 \right)$\pi$/2 , we have f\left( x \right) = - \text { cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 1 \right)$\pi$/2, \left( 4n + 3 \right)$\pi$/2 \right) \text { When, x is in fourth quadrant, i . e} . , \left( 4n + 3 \right)$\pi$/2 < x \leq \left( 2n + 2 \right)$\pi$ ,\text { we have } f\left( x \right) =\text{cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 3 \right)$\pi$/2, \left( 2n + 2 \right)$\pi$ \right) \text { Thus possible point of non - differentiability of } f\left( x \right)\text { are x} = \left( 4n + 1 \right)$\pi$/2, \left( 4n + 3 \right)$\pi$/2 \text { Now, LHD } \left[ \text { at x }= \left( 4n + 1 \right)$\pi$/2 \right] = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^-} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)$\pi$/2 \right)}{x - \left( 4n + 1 \right)$\pi$/2} = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^-} \frac{\cos x - 0}{x - \left( 4n + 1 \right)$\pi$/2} = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^-} - \sin x/1 - 0 \left[\text { By L'Hospital rule } \right] = - 1 \text { And RHD } \left( \text { at x } = \left( 4n + 1 \right)$\pi$/2 \right) = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^+} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)$\pi$/2 \right)}{x - \left( 4n + 1 \right)$\pi$/2} = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^+} \frac{- \cos x - 0}{x - \left( 4n + 1 \right)$\pi$/2} = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^+} \sin x/1 - 0 \left[ \text { By L'Hospital rule} \right] = 1\ \therefore \lim_{x \to \left( 4n + 1 \right) $\pi$/2^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 1 \right) $\pi$/2^+} f\left( x \right) \text { So }f\left( x \right)\text { is not differentiable at x }= \left( 4n + 1 \right)$\pi$/2 \text { Now, LHD} \left[\text { at x } = \left( 4n + 3 \right)$\pi$/2 \right] = \lim_{x \to \left( 4n + 1 \right) $\pi$/2^-} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)$\pi$/2 \right)}{x - \left( 4n + 3 \right)$\pi$/2} = \lim_{x \to \left( 4n + 3 \right) $\pi$/2^-} \frac{- \cos x - 0}{x - \left( 4n + 3 \right)$\pi$/2} = \lim_{x \to \left( 4n + 3 \right) $\pi$/2^-} \sin x/1 - 0 \left[\text { By L'Hospital rule }\right] = 1 \text { And RHD } \left( \text { at x } = \left( 4n + 3 \right)$\pi$/2 \right) = \lim_{x \to \left( 4n + 3 \right) $\pi$/2^+} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)$\pi$/2 \right)}{x - \left( 4n + 3 \right)$\pi$/2} = \lim_{x \to \left( 4n + 3 \right) $\pi$/2^+} \frac{\cos x - 0}{x - \left( 4n + 3 \right)$\pi$/2} = \lim_{x \to \left( 4n + 3 \right) $\pi$/2^+} - \sin x/1 - 0 \left[\text { By L'Hospital rule} \right] = - 1\ \therefore \lim_{x \to \left( 4n + 3 \right) $\pi$/2^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 3 \right) $\pi$/2^+} f\left( x \right) \text { So} f\left( x \right) \text { is not differentiable at x} = \left( 4n + 3 \right)$\pi$/2 \text { Therefore} , f\left( x \right)\text { is neither differentiable at }\left( 4n + 1 \right)$\pi$/2 \text { nor at } \left( 4n + 3 \right)$\pi$/2 \text { i . e } . f\left( x \right) \text { is not differentiable at odd multiples of } $\pi$/2 \text { i . e .} f\left( x \right) \text { is not differentiable at x }= \left( 2n + 1 \right)$\pi$/2 \text { Therefore, f(x) is everywhere continuous but not differentiable at } \left( 2n + 1 \right)$\pi$/2 .

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