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

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(b) continuous but not differentiable at x = (2n + 1) $\pi$/2, n ∈ Z We have, `$\Rightarrow$f(x) = {(cosx , 2npile x<(4n +1)pi/2),(0, x = (4n + 1)pi/2),(-cos x , (4n+1)pi/2 < x<(4n + 3)pi/2),(0, x = (4n +3)pi/2),(cos x , (4n + 3)pi/2 < xle (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 , \text { 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[ 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( 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 i . e . f\left( x \right) \text{is not differentiable at odd multiples of} $\pi$/2 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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