Let f : [–1, 1] → R be a function defined by \f(x)=\\begin{array}{c}x^2|\cos (\pi/x)|\end{array},\for x\neq 0;0,\for x= 0The set of points where f is not differentiable is

Correct Answer is : {x ∈ [–1, 1] : x =22n+1, n ∈ Z}

{x ∈ [–1, 1] : x = 0 or x = 2/{2n=1}, n ∈ Z}

{x ∈ [–1, 1] : x =2/{2n+1}, n ∈ Z}

KVPY SX SB Exam 2018 Question Paper

Question : 11 of 120

Marks: +1, -0

Let f : [–1, 1] → R be a function defined by

\f(x)=\{\table x^2{|\cos (π/x)|},\for x≠ 0;0,\for x= 0

The set of points where f is not differentiable is

Solution:

Modulus changes it’s definition at x

=2/{2n+1}

\f'(x)=\{\table -\sin π/x+2x\cos (π/x),x ∈ U(2/{4n-1},2/{4n+1});\sin π/x-2x\cos(π/x), x∈U(2/{4n+1},2/{4n+3})

LHD ≠ RHD for

x=2/(2n+1)

And at x = 0, LHD

=\lim↙{h→0}{h^2{|\cosπ/h|}-0}/{-h}=0

RHD =

=\lim↙{h→0}{h^2{|\cosπ/h|}-0}/{ h}=0

LHD = RHD, hence differentiable at zero.

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