Let f(x)={cases} x² 1/x, & x≠ 0 \\ 0, & x=0 {cases} Then which of the following is true:

Correct Answer is : f′(x)f'(x)f′(x) is not continuous at x=0x=0x=0

f(x) is not differentiable at x=0

Correct Answer is : f′(x)f'(x)f′(x) is not continuous at x=0x=0x=0

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NIMCET 2024 Paper

Section: Mathematics

Question : 32 of 120

Marks: +1, -0

Let

f(x)=\begin{cases} x^2 \sin\frac{1}{x}, & x\neq 0 \\ 0, & x=0 \end{cases}

Solution:

We know that for any value of

x, \sin x

is finite and its values remains between -1 and 1 .

\;f'(0^+) = \frac{h^2 \sin(1/h) - 0}{h} = h \sin(1/h) = 0

\;f'(0^-) = \frac{h^2 \sin(-1/h) - 0}{-h} = h \sin(1/h) = 0

Hence function is differentiable at

x=0

\;f'(x) = 2x \sin(1/x) + x^2 \cos(1/x) \left(-\frac{1}{x^2}\right)

\Rightarrow f'(x) = 2x \sin(1/x) - \cos(1/x)

When

x \to 0, \cos(1/x)

does not give a definite value, hence

f'(x)

is not continuous at

x=0

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