Q49 / 160 · WBJEE 2015 · Mathematics

The least value of $2 x^{2}+y^{2}+2 x y+2 x-3 y+8$ for real numbers $x$ and $y,$ is

  • A
    2
  • B
    8
  • C
    3
  • Correct answer
    D
    $-1 / 2$

Solution

$2 x^{2}+y^{2}+2 x y+2 x-3 y+8$ $=\frac{1}{2}\left|4 x^{2}+2 y^{2}+4 x y+4 x-6 y+16\right|$ $=\frac{1}{2}\left[\left(y^{2}-8 y\right)+\left(4 x^{2}+y^{2}+4 x y+4 x+2 y\right)+16\right].$ $=\frac{1}{2}\left|(y-4)^{2}+(2 x+y+1)^{2}-1\right|$ So, least value will be $-\frac{1}{2}$ at $y=4$ and $x=\frac{-5}{2}$

Hence, the correct answer is option: 'D'.

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