If f(x) = a|sin x| + be^{|x|} + c|x|^3,where | vedclass

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(D) The function $f(x) = a|\sin x| + be^{|x|} + c|x|^3$ is differentiable at $x = 0$ if and only if each component is differentiable or their non-diff

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$a = 0, b = 0, c$ is any real number

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(D) The function $f(x) = a|\sin x| + be^{|x|} + c|x|^3$ is differentiable at $x = 0$ if and only if each component is differentiable or their non-differentiable parts cancel out.First,consider $|\sin x|$. The left-hand derivative at $x = 0$ is $\lim_{h 	o 0^-} \frac{|\sin h| - 0}{h} = \lim_{h 	o 0^-} \frac{-\sin h}{h} = -1$. The right-hand derivative is $\lim_{h 	o 0^+} \frac{\sin h}{h} = 1$. Since $-1 
eq 1$,$|\sin x|$ is not differentiable at $x = 0$.Second,consider $e^{|x|}$. The left-hand derivative at $x = 0$ is $\lim_{h 	o 0^-} \frac{e^{-h} - 1}{h} = -1$. The right-hand derivative is $\lim_{h 	o 0^+} \frac{e^h - 1}{h} = 1$. Since $-1 
eq 1$,$e^{|x|}$ is not differentiable at $x = 0$.Third,consider $|x|^3$. Since $|x|^3 = x^3$ for $x \ge 0$ and $-x^3$ for $x For $f(x)$ to be differentiable at $x = 0$,the non-differentiable parts must vanish. This requires $a = 0$ and $b = 0$. The constant $c$ can be any real number because $|x|^3$ is already differentiable.Therefore,the correct option is $D$.

If $f(x) = a|\sin x| + be^{|x|} + c|x|^3$,where $a, b, c \in \mathbb{R}$,is differentiable at $x = 0$,then:

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