SSC CGL 30 Aug 2015 Paper

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Question : 105 of 200

Marks: +1, -0

If (m + 1) =

√n

+ 3 , the value of

1/2 ({m^3-6m^2+12m-8}/√n-n)

is

0
1
2
3

Solution:

m + 1 =

√n

+ 3 (Given)
⇒ m + 1 - 3 =

√n

⇒ m - 2 =

√n

On cubing both sides,

(m-2)^3

=

(√n)^3

⇒

m^3-3m^2

× 2 + 3 m

(2)^2 - 2^3

= n

√n

⇒

m^3-6m^2

+ 12m - 8 = n

√n

⇒

{m^3-6m^2+12m-8}/{√n}

= n
⇒

{m^3-6m^2+12m-8}/√n

- n = 0
⇒

1/2 [{m^3-6m^2+12m-8}/√n - n]

= 0

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