SSC CHSL 24 Mar 2018 Shift 2 Paper

Section: English Language

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Question : 14 of 100

Marks: +1, -0

If

x=[1/(√5+√3)],y=[1/(√7+√5)]

and

z=[1/(√7+√3)]

, then what is the value of (x+y+z) ?

${3(√ 7−√ 3)}/4$
$√5 −√ 3$
$√7 + √5$
$1/{2(√7+√3)}$

Solution:

Given :

x=[1/(√5+√3)]

Rationalizing the denominator, we get :
⇒

x=[1/(√5+√3)]

×[(√5-√3)/(√5-√3)]

⇒

x =( √5−√ 3)/{ ( √5)^2 −( √3)^2

⇒

x =( √5− √3)/ 2

Similarly,

y=(√7-√5)/2

and

z=(√7-√3)/4

To find :

x + y + z

=

( √5− √3)/ 2+(√7-√5)/2

+ (√7-√3)/4

=

√7/2+√7/4-√3/2-√3/4

=

{3√ 7}/4 − { 3 √3 }/4 ={3( √7−√3)}/4

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