Saturday, 29 October 2011

Algebra-14 (XAT-2010)

If x and y are real numbers, then the minimum value of x2+4xy+6y2-4y+4 is:
(A)-4      (B) 0        (C) 2        (D) 4   (E) None of these
Answer follows here:

“The funda is minimum value of square of a number is zero as it should not be negative”
So we will try to convert the given expression into sum of squares.
x2+4xy+6y2-4y+4 = x2+4xy+4y2+2y2-4y+4
= (x2+4xy+(2y)2)+ ½ (4y2-8y+8)
= (x+2y)2 + ½ {(2y)2-2*2y*2+22+4}
= (x+2y)2 + ½ {(2y-2)2+4}
= (x+2y)2 + ½ (2y-2)2 + 2
Minimum value of (x+2y)2 is zero
Minimum value of (2y-2)2 is zero
The remaining part is ‘2’ which is the minimum value of the whole expression
Answer (C)

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