NIMCET 2014 Question Paper with solutions

Let two numbers be x and y, where x > 0, y > 0.
Given, x + y = 9 ... (i)
and z = $\mathit{x}-{\mathit{y}}^2$ ... (ii)
⇒ z = ${\left(9-\mathit{x}\right)}^2$
= x (81 + ${\mathit{x}}^2$ - 18x) = ${\mathit{x}}^2-18{\mathit{x}}^2$ + 81x
∴ $\frac{\mathit{d}\mathit{z}}{\mathit{d}\mathit{x}}$ = $3{\mathit{x}}^2$ - 36x + 81
= 3 (${\mathit{x}}^2$ - 12x + 27)
$\frac{\mathit{d}\mathit{z}}{\mathit{d}\mathit{x}}$ = 0
⇒ ${\mathit{x}}^2$ - 9x - 3x + 27 = 0
⇒ (x - 9) (x - 3) = 0
⇒ x = 3 , x = 9
x = 3 [ x = 9 not possible]
and y = 6
∴ So, numbers are 3 and 6.