Q95 / 180 · KCET 2024 · Mathematics

The function $x^x ; x\gt0$ is strictly increasing at

  • A
    $\forall x \in R$
  • B
    $x \lt \frac{1}{e}$
  • Correct answer
    C
    $x\gt\frac{1}{e}$
  • D
    $x \lt 0$

Solution

Let $y=x^x$ $\frac{d y}{d x}=x^x(1+\log x)$ For increasing function, $\frac{d y}{d x}\gt0$. $\Rightarrow \quad x^x(1+\log x)\gt0$ $\Rightarrow \quad 1+\log x\gt0$ $\Rightarrow \quad \log _e x\gt\log _e \frac{1}{e}$ Hence, $x\gt\frac{1}{e}$

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

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