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NDA Math Limit, Continuity and Differentiability

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Consider the following for the next three (03) items that follow:

Let

f(x)=\begin{cases} -2, & -3 \le x \le 0 \\ x-2, & 0 < x \le 3 \end{cases}

and

g(x)=f(|x|)+|f(x)|

[NDA II 2016]

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Question : 61 of 105

Marks: +1, -0

What is the value of the differential coefficient of g(x) at

x=-2

$-1$
$0$
$1$
$2$

Solution:

For

x=-2

g(x)=-2+|-2|=-2+2

\Rightarrow g(x)=0

⇒ differential coefficient at

x=-2

is given as:

g'(x)=\lim\limits_{h\rightarrow 0}\;\;\frac{g(x+h)-g(x)}{h}=\lim\limits_{h\rightarrow 0}\;\;\frac{0-0}{4}=0

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