Question

The function f (x) =  |cos x| is

विकल्प

• differentiable at x = (2n + 1) π/2, n ∈ Z

•  continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z

• neither differentiable nor continuous at x = n ∈ Z

•  none of these

Answer 1

(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z 

We have, 

`⇒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):}`\[\text{When, x is in first quadrant}, i . e . 2n\pi \leq x < \left( 4n + 1 \right)\frac{\pi}{2} , \text { we have }\]
\[ f\left( x \right) = \text{cos x which being a trigonometrical function is continuous and differentiable in} \left( 2n\pi, \left( 4n + 1 \right)\frac{\pi}{2} \right)\]
\[\text{When, x is in second quadrant or in third quadrant}, i . e . , \left( 4n + 1 \right)\frac{\pi}{2} < x < \left( 4n + 3 \right)\frac{\pi}{2} , \text { we have }\]
\[ f\left( x \right) = - \text{cos x which being a trigonometrical function is continuous and differentiable in} \left( \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2} \right)\]
\[\text{When, x is in fourth quadrant}, i . e . , \left( 4n + 3 \right)\frac{\pi}{2} < x \leq \left( 2n + 2 \right)\pi , \text { we have }\]

\[ f\left( x \right) = \text { cos x which being a trigonometrical function is continuous and differentiable in } \left( \left( 4n + 3 \right)\frac{\pi}{2}, \left( 2n + 2 \right)\pi \right)\]
\[\text { Thus possible point of non - differentiability of } f\left( x \right)\text {  are x } = \left( 4n + 1 \right)\frac{\pi}{2}, \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text { Now, LHD } \left[ at x = \left( 4n + 1 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{\cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{- \sin x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]
\[ = - 1\]
\[\text { And RHD } \left( at x = \left( 4n + 1 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 1 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{- \cos x - 0}{x - \left( 4n + 1 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} \frac{\sin x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]
\[ = 1\]
\[ \therefore \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^+} f\left( x \right)\]
\[\text { So } f\left( x \right)\text { is not differentiable at x }= \left( 4n + 1 \right)\frac{\pi}{2}\]
\[\text { Now, LHD }\left[\text {  at x } = \left( 4n + 3 \right)\frac{\pi}{2} \right] = \lim_{x \to \left( 4n + 1 \right) \frac{\pi}{2}^-} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{- \cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} \frac{\sin x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]
\[ = 1\]

\[ \text { And RHD } \left(\text {  at x }  = \left( 4n + 3 \right)\frac{\pi}{2} \right) = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{f\left( x \right) - f\left( \left( 4n + 3 \right)\frac{\pi}{2} \right)}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{\cos x - 0}{x - \left( 4n + 3 \right)\frac{\pi}{2}}\]
\[ = \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} \frac{- \sin x}{1 - 0} \left[ \text{By L'Hospital rule} \right]\]
\[ = - 1\]

\[ \therefore \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^-} f\left( x \right) \neq \lim_{x \to \left( 4n + 3 \right) \frac{\pi}{2}^+} f\left( x \right)\]
\[\text { So } f\left( x \right) \text{is not differentiable at x} = \left( 4n + 3 \right)\frac{\pi}{2}\]
\[\text{Therefore}, f\left( x \right)\text {  is neither differentiable at} \left( 4n + 1 \right)\frac{\pi}{2}\text {  nor at } \left( 4n + 3 \right)\frac{\pi}{2}\]
\[i . e . f\left( x \right) \text{is not differentiable at odd multiples of} \frac{\pi}{2}\]
\[i . e . f\left( x \right) \text{is not differentiable at x} = \left( 2n + 1 \right)\frac{\pi}{2}\]
\[\text{Therefore, f(x) is everywhere continuous but not differentiable at} \left( 2n + 1 \right)\frac{\pi}{2} .\]