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NDA Math Application of Derivatives

Let

f(x)=\begin{cases} \frac{e^x-1}{x} & x>0 \\ 0 & x=0 \end{cases}

be a real valued function.

[NDA II 2016]

Question : 50 of 73

Marks: +1, -0

Solution:

f(x)=\;\;\frac{e^{x}-1}{x}>0

\Rightarrow f'(x)=\;\;\frac{x e^{x}-(e^{x}-1)}{x^{2}}=\frac{(x+1)-e^{x}}{x^{2}}

=\left(\;\;\frac{1}{x}+\;\;\frac{1}{x^{2}}\right)-e^{x}

which is a strictly decreasing function

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