Question

Let f (x) = |sin x|. Then,

विकल्प

• f (x) is everywhere differentiable.

• f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z

• f (x) is everywhere continuous but not differentiable at  \[x = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z\]

• none of these

Answer 1

(b) f (x) is everywhere continuous but not differentiable at x = nπ, n ∈ Z We have,

\[f\left( x \right) = \left| \sin x \right|\]

`⇒ f(x) = {(0, x = 2npi),(sin x ,2npi<x< (2n +1)pi),(0, x = (2n +1)pi),(-sin x, (2n  +1)pi <x <(2n + 2)pi):}`

\[\text{When, x is in first or second quadrant}, i . e . , 2n\pi < x < \left( 2n + 1 \right)\pi , we have\]

\[ f\left( x \right) = \text{sin x which being a trigonometrical function is continuous and differentiable in} \left( 2n\pi, \left( 2n + 1 \right)\pi \right)\]

\[\text{When, x is in third or fourth quadrant}, i . e . , \left( 2n + 1 \right)\pi < x < \left( 2n + 2 \right)\pi , we have\]

\[ f\left( x \right) = - \text{sin x which being a trigonometrical function is continuous and differentiable in} \left( \left( 2n + 1 \right)\pi, \left( 2n + 2 \right)\pi \right)\]

\[\text{Thus possible point of non - differentiability of } f\left( x \right) \text { are x = 2n}\pi \text { and } \left( 2n + 1 \right)\pi\]

\[\text { Now, LHD } \left[\text {  at x = 2n }\pi \right] = \lim_{x \to 2n \pi^-} \frac{f\left( x \right) - f\left( 2n\pi \right)}{x - 2n\pi}\]

\[ = \lim_{x \to 2n \pi^-} \frac{- \sin x - 0}{x - 2n\pi}\]

\[ = \lim_{x \to 2n \pi^-} \frac{- \cos x}{1 - 0} \left[\text {  By L'Hospital rule } \right]\]

\[ = - 1\]

\[\text { And RHD } \left( at x = 2n\pi \right) = \lim_{x \to 2n \pi^+} \frac{f\left( x \right) - f\left( 2n\pi \right)}{x - 2n\pi}\]

\[ = \lim_{x \to 2n \pi^+} \frac{\sin x - 0}{x - 2n\pi}\]

\[ = \lim_{x \to 2n \pi^+} \frac{\cos x}{1 - 0} \left[ \text { By L'Hospital rule } \right]\]

\[ = 1\]

\[ \therefore \lim_{x \to 2n \pi^-} f\left( x \right) \neq \lim_{x \to 2n \pi^+} f\left( x \right)\]

\[\text { So }f\left( x \right) \text { is not differentiable at x = 2n }\pi\]

\[\text { Now, LHD } \left[ at x = \left( 2n + 1 \right)\pi \right] = \lim_{x \to \left( 2n + 1 \right) \pi^-} \frac{f\left( x \right) - f\left( \left( 2n + 1 \right)\pi \right)}{x - \left( 2n + 1 \right)\pi}\]

\[ = \lim_{x \to \left( 2n + 1 \right) \pi^-} \frac{\sin x - 0}{x - \left( 2n + 1 \right)\pi}\]

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

\[ = - 1\]

\[\text { And RHD } \left( at x = \left( 2n + 1 \right)\pi \right) = \lim_{x \to \left( 2n + 1 \right) \pi^+} \frac{f\left( x \right) - f\left( \left( 2n + 1 \right)\pi \right)}{x - \left( 2n + 1 \right)\pi}\]

\[ = \lim_{x \to \left( 2n + 1 \right) \pi^+} \frac{- \sin x - 0}{x - \left( 2n + 1 \right)\pi}\]

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

\[ = 1\]

\[ \text { therefore } \lim_{x \to \left( 2n + 1 \right) \pi^-} f\left( x \right) \neq \lim_{x \to \left( 2n + 1 \right) \pi^+} f\left( x \right)\]

\[\text { So }f\left( x \right) \text{is not differentiable at} x = \left( 2n + 1 \right)\pi\]

\[\text{Therefore, }f\left( x \right) \text { is neither differentiable at} 2n\pi \text { nor at} \left( 2n + 1 \right)\pi\]

\[i . e . f\left( x \right) \text{is neither differentiable at even multiple of} \pi\text {  nor at odd multiple of} \pi\]

\[i . e . f\left( x \right) \text{is not differentiable at x} = n\pi\]

\[\text{Therefore, f(x) is everywhere continuous but not differentiable at} n\pi .\]