NIMCET 2017 Question Paper with solutions

For critical points
$\frac{\mathit{\delta }\mathit{f}}{\mathit{\delta }\mathit{x}}$ = 0 and $\frac{\mathit{\delta }\mathit{f}}{\mathit{\delta }\mathit{y}}$ = 0
∴ 2x -2 = 0 and 4y + 4 = 0
⇒ x = land y = -1
Hence, critical point of the given function is (1, -1).
The given function is maximum at (1, -1), if
rt - ${\mathit{s}}^2$ > 0 and r < 0
And minimum at (1, -1), if rt - ${\mathit{s}}^2$ > 0 and r > 0
For maximum and minimum
r = $\frac{{\mathit{\delta }}^2\mathit{f}}{\mathit{\delta }{\mathit{x}}^2}$ = 2 ,
t = $\frac{{\mathit{\delta }}^2\mathit{f}}{\mathit{\delta }{\mathit{y}}^2}$ = 4
and s = $\frac{{\mathit{\delta }}^2\mathit{f}}{\mathit{\delta }\mathit{x}\mathit{\delta }\mathit{y}}$ = 0
Clearly, by putting the value of r, t and a, we get r > 0 and rt - ${\mathit{s}}^2$ > 0
Hence, the given function is minimum at (1, - 1).