NIMCET 2012 Question Paper with solutions

Let y = ${\mathit{x}}^{\mathit{x}}$
Taking log on both sides,
we get log y = x log x
On differentiating
$\frac{1}{\mathit{y}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = x . $\frac{1}{\mathit{x}}$ + log x
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = y (1 + log x) = ${\mathit{x}}^{\mathit{x}}$ . (1 + log x) ... (i)
For decreasing of y,
Here , $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ < 0
${\mathit{x}}^{\mathit{x}}$ . (1 + log x) < 0 (but ${\mathit{x}}^{\mathit{x}}$ ≮ 0 and x > 0)
⇒ 1 + log x < 0
⇒ log x < - 1
⇒ log x < log ${\mathit{e}}^{-1}$
⇒ x < $\frac{1}{\mathit{e}}$ and x > 0
∴ x ∊ $\left(0,\frac{1}{\mathit{e}}\right)$