SSC CHSL 6 Dec 2015 Shift 1 Paper

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Question : 120 of 200

Marks: +1, -0

The value of

1+1/√2+√3+1-1/√2+√3

is:

1
$4( √3 + √2)$
$√3$
$√2$

Solution:

Expression :

1+1/√2+√3+1-1/√2+√3

Rationalizing the denominator, we get :
=

( 1/{1+ √3+√ 2} ×{1+√ 3−√ 2}/{1+ √3−√ 2})

+(1/{1+ √3-√ 2} ×{1+√ 3+√ 2}/{1+√ 3+√ 2})

=

[{1+ √3-√ 2}/{(1+ √3)^2-(√2)^2}]

+[{1+ √3+√ 2}/{(1+ √3)^2-(√2)^2}]

=

{(1+√ 3− √2)+(1+ √3+√ 2)}/{(1+√3+2 √3)−(2)}

=

{2+2√ 3}/{2+2√ 3}=1

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