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Question

The function f (x) = |cos x| is
(a) differentiable at x = (2n + 1) π/2, n ∈ Z
(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
(c) neither differentiable nor continuous at x = n ∈ Z
(d) none of these

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Solution

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

We have,fx=cos xfx=cos x, 2nπx<4n+1π20, x=4n+1π2 -cos x, 4n+1π2<x<4n+3π2 0 , x=4n+3π2 cos x, 4n+3π2< x2n+2πWhen, x is in first quadrant, i.e. 2nπx<4n+1π2 , we have fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x2n+2π , we have fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx4n+1π2- fx- f4n+1π2x-4n+1π2 =limx4n+1π2- cos x- 0x-4n+1π2 =limx4n+1π2- -sin x1-0 By L'Hospital rule =-1And RHD at x=4n+1π2 =limx4n+1π2+ fx- f4n+1π2x-4n+1π2 =limx4n+1π2+ -cos x- 0x-4n+1π2 =limx4n+1π2+ sin x1-0 By L'Hospital rule =1lim x4n+1π2-fx limx4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx4n+1π2- fx- f4n+3π2x-4n+3π2 =limx4n+3π2- -cos x- 0x-4n+3π2 =limx4n+3π2- sin x1-0 By L'Hospital rule =1And RHD at x=4n+3π2 =limx4n+3π2+ fx- f4n+3π2x-4n+3π2 =limx4n+3π2+ cos x- 0x-4n+3π2 =limx4n+3π2+ -sin x1-0 By L'Hospital rule =-1lim x4n+3π2-fx limx4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .

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