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Question

f(x)=k=1⎜ ⎜ ⎜ ⎜1+2cos(2x3k)3⎟ ⎟ ⎟ ⎟, then the number of points where [xf(x)]+|xf(x)|+(x1)|x23x+2| is non-differentiable in x(0,3π) is equal to (where [.] denotes the greatest integer function)

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Solution

f(x)=k=11+2 cos(2x3k)3=k=113(1+24sin2x3k)=k=113 sin(x3k){3 sin(x3k)4 sin3(x3k)}=k=1sin(x3k1)3 sin(x3k)=limk13ksin xsinx3×sinx3sinx9×.....×sin(x3k1)sin(x3k)=limksin xxsin(x3k)(x3k)f(x)=sin xxxf(x)=sin x
So [sin x]+|sin x|+(x1)|(x1)(x2)|
is non-differentiable at 5 (π2,π,2π,5π2,2) points

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