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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 : 60 of 105

Marks: +1, -0

Which of the following statements is/are correct?

g(x)

is differentiable at

x=0

B(x)

is differentiable at

x=2

Select the correct answer using the code given below:

1 only
2 only
Both 1 and 2
Neither 1 nor 2

Solution:

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)|

At

x=0

For

LHD:g(x)=-2+|-2|=-2+2=0\Rightarrow g(x)=0

LHD=\lim\limits_{x\to 0^{-}}\;\frac{g(x)-g(0)}{x-0}=\lim\limits_{h\to 0}\;\frac{g(-h)-g(0)}{-h}

=\lim\limits_{h\to 0}\;\frac{0-0}{-h}=\lim\limits_{h\to 0}0

LHD

=0

For

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

g(x)=x-2-(x-2)

x > 0

(and just greater than zero)

g(x)=x-2-x+2=0

\Rightarrow RHD==\lim\limits_{x\to 0^{+}}\;\frac{g(x)-g(0)}{x-0}=\lim\limits_{h\to 0}\;\frac{g(h)-g(0)}{h}

=\lim\limits_{h\to 0}0=0

⇒ LHD = RHD ⇒ so

g(x)

is differentiable at

x=0

.

At

x=2

For LHD:

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

\therefore LHD=\lim\limits_{x\to 2^{-}}\;\frac{g(x)-g(2)}{x-2}=\lim\limits_{x\to 2}\;\frac{0}{x-2}=0

For RHD:

g(x)-|x|-2+|x-2|-x-2+x-2-2x-4

\Rightarrow RHD=\lim\limits_{x\to 2^{+}}\;\frac{g(x)-g(2)}{x-2}=\lim\limits_{x\to 2}\;\frac{2x-4-2(2)+4}{x-2}

=\lim\limits_{x\to 2}\;\frac{2(x-2)}{x-2}=2

\Rightarrow LHD\neq RHD

Thus g(;x;) is not differentiable at

x=2

.

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