If f(x)={cases} x, & x≥ 0 \\ -x, & x

Correct Answer is : f(x)f(x)f(x) is continuous but not differentiable

f(x) is continuous but not differentiable

Correct Answer is : f(x)f(x)f(x) is continuous but not differentiable

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CGPET 2021 Solved Paper

Section: Mathematics

Question : 123 of 150

Marks: +1, -0

f(x)=\begin{cases} x, & x\ge 0 \\ -x, & x<0 \end{cases}.

x=0

Solution:

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

Contuinuity at

x=0

f(0)=0

RHL

=f(o^{+}) = \lim\limits_{x\to 0^{+}} f(x) = \lim\limits_{x\to 0^{+}} x

=\lim\limits_{h\to 0} +(0+h) = 0

LHL

= f(0^{-}) = \lim\limits_{x\to 0^{-}} f(x) = \lim\limits_{x\to 0^{-}} -x

=\lim\limits_{h\to 0} -(-h) = \lim\limits_{h\to 0} h = 0

∵

f(0) = f(0^{+}) = f(0^{-}) = 0

∵

f(x)

is continuous at

x=0

Differentiability at

x=0

LHD

= f'(0^{-}) = \lim\limits_{h\to 0} \frac{f(0-h)-f(0)}{-h} = \lim\limits_{h\to 0} \frac{-(-h)-0}{-h}

\lim\limits_{h\to 0} \frac{h}{-h} = \lim\limits_{h\to 0} -1 = 1

RHD

= f'(0^{+}) = \lim\limits_{h\to 0} \frac{f(0+h)-f(0)}{h}

=\lim\limits_{h\to 0} \frac{h-0}{h} = \lim\limits_{h\to 0} 1 = 1

∵

f'(0^{+}) \neq f'(0^{+})

∴

f(x)

is not differentiable at

x=0

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