SSC CHSL Model Paper 19

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Question : 70 of 99

Marks: +1, -0

If

a^2 = b + c, b^2 = a + c, c^2 = b + a,

then what will be the value of

1/{a+1}+1/{b+1}+1/{c+1}\ ?

-1
2
1
0

Solution:

a^2 = b + c

⇒

a^2 + a= a+ b + c

⇒

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

⇒

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

Similarly,

b^2 = a + c

⇒

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

and

c^2 = b+ a

⇒

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

∴

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

=

a/{a+ b + c}+b/{a+ b + c}+c/{a+ b + c}

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

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