Question : If $x$ and $y$ are real numbers, then the least possible value of $4 (x -2)^2+ (y-3)^2-2 (x-3)^2$ is:
Option 1: 3
Option 2: –4
Option 3: 1
Option 4: –8
Correct Answer: –4
Solution :
Given: $x$ and $y$ are real numbers.
The expression is $4 (x -2)^2+ (y -3)^2-2 (x-3)^2$
Use the algebraic identity, $(a-b)^2=a^2+b^2-2ab$
For the least possible value, substitute $(y-3)^2=0$
⇒ $4 (x-2)^2+0-2 (x-3)^2$
$=4(x^2+4-2x)-2(x^2+9-6x)$
$=4x^2+16-8x-2x^2-18+12x$
$=2(x^2-1-2x)$
$=2(x^2+1-2x-2)$
$=2[(x-1)^2-2]$
For the least possible value, substitute $(x-1)^2=0$.
⇒ $2\times -2=-4$
Hence, the correct answer is –4.
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