SSC CGL 4 Sep 2016 Paper

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Question : 65 of 100

Marks: +1, -0

If

a^2/{b+c}

=

b^2/{c+a}

=

c^2/{a+b}

= 1 then find the value of

2/{1+a}+2/{1+b}+2/{1+c}

0
1
2
3

Solution:

a^2/{b+c}

=

b^2/{c+a}

=

c^2/{a+b}

= 1
⇒

a^2/{b+c}

= 1 ⇒

a^2

= b + c
⇒

a^2

+ a = a + b + c
⇒ a (a + 1) = a + b + c
⇒ a + 1 =

{a+b+c}/1

⇒

1/{a+1}

=

a/{a+b+c}

Similarly,

b^2/{c+a}

= 1
⇒

1/{b+1}

=

b/{a+b+c}

and ,

c^2/{a+b}

= 1
⇒

1/{c+1}

=

c/{a+b+c}

∴

2/{1+a}+2/{1+b}+2/{1+c}

= 2

(a/{a+b+c}

+

b/{a+b+c}

+

c/{a+b+c})

= 2

({a+b+c}/{a+b+c})

= 2

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