SSC CGL 26 Oct 2014 Shift 1 Paper

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Question : 132 of 200

Marks: +1, -0

If

[1/2(a-b)]^2

+ ab = p

(a+b)^2

, then the value of p is (assume that a ≠ – b)

$1/8$
1
$1/2$
$1/4$

Solution:

[1/2(a-b)]^2

+ ab = p

(a+b)^2

⇒

1/4 (a^2-2ab+b^2)

+ ab = p

(a+b)^2

⇒

1/4 (a^2-2ab+b^2+4ab)

= p

(a+b)^2

⇒

1/4 (a^2+2ab+b^2)

= p

(a+b)^2

⇒

1/4 (a+b)^2

= p

(a+b)^2

⇒ p =

1/4

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