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Question
The function f (x) = |cos x| is
Options
everywhere continuous and differentiable
everywhere continuous but not differentiable at (2n + 1) π/2, n ∈ Z
neither continuous nor differentiable at (2n + 1) π/2, n ∈ Z
none of these
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Solution
We have,
\[f\left( x \right) = \left| \cos x \right|\]
`⇒ 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):}`
\[\text { When, x is in first quadrant, i . e . 2n}\pi \leq x < \left( 4n + 1 \right)\frac{\pi}{2} , \text { we have} \]
\[ f\left( x \right) = \text { cos x which being a trigonometrical function is continuous and differentiable in} \left( 2n\pi, \left( 4n + 1 \right)\frac{\pi}{2} \right)\]
\[\text { When, x is in second quadrant or in third quadrant, i . e }. , \left( 4n + 1 \right)\frac{\pi}{2} < x < \left( 4n + 3 \right)\frac{\pi}{2} , we have\]
\[ f\left( x \right) = - \text { cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2} \right)\]
\[\text { When, x is in fourth quadrant, i . e} . , \left( 4n + 3 \right)\frac{\pi}{2} < x \leq \left( 2n + 2 \right)\pi ,\text { we have }\]
\[ f\left( x \right) =\text{cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 3 \right)\frac{\pi}{2}, \left( 2n + 2 \right)\pi \right)\]
\[\text { Thus possible point of non - differentiability of } f\left( x \right)\text { are x} = \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text { Now, LHD } \left[ \text { at x }= \left( 4n + 1 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{\cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{- \sin x}{1 - 0} \left[\text { By L'Hospital rule } \right]\]
\[ = - 1\]
\[\text { And RHD } \left( \text { at x } = \left( 4n + 1 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{- \cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{\sin x}{1 - 0} \left[ \text { By L'Hospital rule} \right]\]
\[ = 1\
\[ \therefore \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} f\left( x \right)\]
\[\text { So }f\left( x \right)\text { is not differentiable at x }= \left( 4n + 1 \right)\frac{\pi}{2}\]
\[\text { Now, LHD} \left[\text { at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{- \cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{\sin x}{1 - 0} \left[\text { By L'Hospital rule }\right]\]
\[ = 1\]
\[\text { And RHD } \left( \text { at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{\cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{- \sin x}{1 - 0} \left[\text { By L'Hospital rule} \right]\]
\[ = - 1\
\[ \therefore \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} f\left( x \right)\]
\[\text { So} f\left( x \right) \text { is not differentiable at x} = \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text { Therefore} , f\left( x \right)\text { is neither differentiable at }\left( 4n + 1 \right)\frac{\pi}{2} \text { nor at } \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text { i . e } . f\left( x \right) \text { is not differentiable at odd multiples of } \frac{\pi}{2}\]
\[\text { i . e .} f\left( x \right) \text { is not differentiable at x }= \left( 2n + 1 \right)\frac{\pi}{2}\]
\[\text { Therefore, f(x) is everywhere continuous but not differentiable at } \left( 2n + 1 \right)\frac{\pi}{2} .\]
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