Let f: R R be a function such that f(0)=1/π and f(x)={x}{e^{π x}-1} for x ≠ 0, then

Correct Answer is : None of the above

f(x) is continuous but not differentiable at x=0

f(x) is differentiable at x=0 and f^{'}(0)=-\;{π}/{2}

Correct Answer is : None of the above

Test Index

NIMCET 2024 Paper

Section: Mathematics

Question : 46 of 120

Marks: +1, -0

Let

f: R \longrightarrow R

f(0)=\frac{1}{\pi}

f(x)=\frac{x}{e^{\pi x}-1}

x \neq 0

Solution:

Let us check the limit of the function at

x=0

\lim\limits_{x \to 0} \frac{x}{e^{\pi x}-1} = \frac{1}{\pi e^{\pi x}} = \frac{1}{\pi}

Hence the function us continuous at

x=0

Let us check derivative of the function at

x=0

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

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

= \lim\limits_{h \to 0} \frac{h\pi - (e^{h\pi}-1)}{h\pi (e^{h\pi}-1)} = \frac{h\pi - \left(h\pi + \frac{h^{2}\pi^{2}}{2!} + \cdots\right)}{h\pi \left(h\pi + \frac{h^{2}\pi^{2}}{2!} + \cdots\right)}

= \frac{-\frac{h^{2}\pi^{2}}{2!} + \cdots}{h^{2}\pi^{2} + \frac{h^{3}\pi^{3}}{2!} + \cdots} = -\frac{1}{2}

Similarly, it can be shown that

f'(0^{-}) = \frac{1}{2}

Correct answer is (4).

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