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:

Solution:

“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)