Question

Let f (x) = |cos x|. Then,

विकल्प

• ^{f (x) is everywhere differentable}

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

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

• (d) none of these

Answer 1

(c) f (x) is everywhere continuous but not differentiable at

\[x = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z\]

We have,

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

`⇒ f(x) = {(cos x ,2npi le 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):}`

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

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

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

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

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

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

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

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

\[ = - 1\]

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

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

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

\[ = 1\]

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

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

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

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

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

\[ = - 1\]

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

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

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

\[ = 1\]

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

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

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

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

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

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