Let a, b in R and f: R rightarrow R be define | vedclass

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(C) Given $f(x) = a \cos (|x^3-x|) + b|x| \sin (|x^3+x|)$.Since $\cos(	heta) = \cos(-	heta)$,we have $\cos(|x^3-x|) = \cos(x^3-x)$.Also,$|x| \sin(|x

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$A, B$

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(C) Given $f(x) = a \cos (|x^3-x|) + b|x| \sin (|x^3+x|)$.Since $\cos(	heta) = \cos(-	heta)$,we have $\cos(|x^3-x|) = \cos(x^3-x)$.Also,$|x| \sin(|x^3+x|) = x \sin(x^3+x)$ for all $x \in R$ because if $x \ge 0$,$|x|=x$ and $|x^3+x|=x^3+x$,and if $x Thus,$f(x) = a \cos(x^3-x) + b x \sin(x^3+x)$ for all $x \in R$.Since $\cos(x^3-x)$ and $x \sin(x^3+x)$ are differentiable functions on $R$,$f(x)$ is differentiable for all $x \in R$ for any $a, b \in R$.Therefore,$f$ is differentiable at $x=0$ and $x=1$ for any $a, b \in R$.Checking the options:$(A)$ Differentiable at $x=0$ if $a=0, b=1$ (True).$(B)$ Differentiable at $x=1$ if $a=1, b=0$ (True).$(C)$ Not differentiable at $x=0$ if $a=1, b=0$ (False).$(D)$ Not differentiable at $x=1$ if $a=1, b=1$ (False).Thus,options $A$ and $B$ are correct.

Let $a, b \in R$ and $f: R \rightarrow R$ be defined by $f(x)=a \cos (|x^3-x|)+b|x| \sin (|x^3+x|)$. Then $f$ is

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