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UPSC NDA 2 2016 Mathematics Solved Paper

Consider the following for the next two (02) items that follow:

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 : 84 of 120

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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