Let $f (x) = a + b | x | + c {| x |}^{4}$ where $a, b$ and $c$ are real constants. Then $f (x)$ is differentiable at $x = 0$ if-

a = 0

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b = 0

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c = 0

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None of these

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Solution

The correct option is B $b = 0$
$f (x) = a + b | x | + c {| x |}^{4}$
$∴ f (x) = a + b x + c x^{4}$ if $x \geq 0$
and $f (x) = a - b x + c x^{4}$ if $x < 0$
RHD $= f^{'} (x) = + b + 4 c x^{3}$ $x \geq 0$
and LHD $= f^{'} (x) = - b + 4 c x^{3}$ $x < 0$
For $f (x)$ to be differential at $x = 0$ , $L H D = R H D$
$\Rightarrow b = - b \Rightarrow 2 b = 0 \Rightarrow b = 0$

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