A) ?1
B) 0
C) 1
D) 2
Correct Answer: D
Solution :
Since function \[|x|\] is not differentiable at \[x=0\] \[\therefore \,|{{x}^{2}}-3x+2|=|(x-1)(x-2)|\] Hence is not differentiable at \[x=1\] and 2 Now\[f(x)=({{x}^{2}}-1)|{{x}^{2}}-3x+2|\cos (|x|)\] is not differentiable at \[x=2\] For \[1<x<2\], \[f(x)=-({{x}^{2}}-1)({{x}^{2}}-3x+2)+\cos x\] For \[2<x<3\], \[f(x)=+({{x}^{2}}-1)({{x}^{2}}-3x+2)+\cos x\] \[Lf'(x)=-({{x}^{2}}-1)(2x-3)-2x({{x}^{2}}-3x+2)-\sin x\] \[Lf'(2)=-3-\sin 2\] \[Rf'(x)=({{x}^{2}}-1)(2x-3)+2x({{x}^{2}}-3x+2)-\sin x\] \[Rf'(2)=(4-1)(4-3)+0-\sin 2=3-\sin 2\] Hence \[Lf'(2)\ne Rf'(2)\].
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