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UPSC NDA Math Model Paper 5

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Question : 46 of 120

Marks: +1, -0

The function

f(x)=\log (x+\sqrt{x^{2}+1}).

is ______ function.

even
odd
periodic
none of these

Solution:

Given:

f(x)=\log (x+\sqrt{x^{2}+1}).

Replace x by -x,

\Rightarrow f(-x)=\log (-x+\sqrt{(-x)^{2}+1})

=\log (-x+\sqrt{x^{2}+1})

=\log (\sqrt{x^{2}+1}-x)

=\log \left[(\sqrt{x^{2}+1}-x) \times \frac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}-x}\right]

=\log \left[\frac{(\sqrt{x^{2}+1})^{2}-x^{2}}{\sqrt{x^{2}+1}-x}\right]

(\because (a+b)(a-b)=a^{2}-b^{2}

=\log \left[\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}-x}\right]=\log \left(\frac{1}{\sqrt{x^{2}+1}-x}\right)

=\log 1-\log (\sqrt{x^{2}+1}-x)

=-\log (\sqrt{x^{2}+1}-x)=-f(x)

Hence function is odd.

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