Q96 / 200 · Manipal (MET) 2014 · Mathematics
The function $x^x$ is increasing, when
- Correct answerA$x\gt\frac{1}{e}$
- B$x \lt \frac{1}{e}$
- C$x \lt 0$
- Dfor all x
Solution
Let $y=x^x$
$\Rightarrow \frac{d y}{d x}=x^x(1+\log x)$
For increaṣing function, $\frac{d y}{d x}\gt0$ $\begin{array}{cc} \Rightarrow & x^x(1+\log x)\gt0 \\ \Rightarrow & 1+\log x\gt0 \\ \Rightarrow & \log _e x\gt\log _e \frac{1}{e} \\ \Rightarrow & x\gt\frac{1}{e} \end{array}$ $\therefore$ The function is increasing when $x\gt\frac{1}{e}$
For increaṣing function, $\frac{d y}{d x}\gt0$ $\begin{array}{cc} \Rightarrow & x^x(1+\log x)\gt0 \\ \Rightarrow & 1+\log x\gt0 \\ \Rightarrow & \log _e x\gt\log _e \frac{1}{e} \\ \Rightarrow & x\gt\frac{1}{e} \end{array}$ $\therefore$ The function is increasing when $x\gt\frac{1}{e}$
Hence, the correct answer is option: 'A'.