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SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 120 of 200

Marks: +1, -0

The value of

1+\frac{1}{\sqrt{2}}+\sqrt{3}+1-\frac{1}{\sqrt{2}}+\sqrt{3}

is:

1
$4(\sqrt{3}+\sqrt{2})$
$\sqrt{3}$
$\sqrt{2}$

Solution:

Expression :

1+\frac{1}{\sqrt{2}}+\sqrt{3}+1-\frac{1}{\sqrt{2}}+\sqrt{3}

Rationalizing the denominator, we get :

=

\left( \frac{1}{1+\sqrt{3}+\sqrt{2}} \times \frac{1+\sqrt{3}-\sqrt{2}}{1+\sqrt{3}-\sqrt{2}} \right)

+\left( \frac{1}{1+\sqrt{3}-\sqrt{2}} \times \frac{1+\sqrt{3}+\sqrt{2}}{1+\sqrt{3}+\sqrt{2}} \right)

=

\left[ \frac{1+\sqrt{3}-\sqrt{2}}{(1+\sqrt{3})^{2}-(\sqrt{2})^{2}} \right]

+\left[ \frac{1+\sqrt{3}+\sqrt{2}}{(1+\sqrt{3})^{2}-(\sqrt{2})^{2}} \right]

=

\frac{(1+\sqrt{3}-\sqrt{2})+(1+\sqrt{3}+\sqrt{2})}{(1+\sqrt{3}+2\sqrt{3})-(2)}

=

\frac{2+2\sqrt{3}}{2+2\sqrt{3}}=1

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