Question

The function f(x) = x^x decreases on the interval

विकल्प

• (0, e)

• (0, 1)

• (0, 1/e)

• none of these

Answer 1

 (0, 1/e)

$$\text{ Given }:f\left(x\right)=x^x$$

$$\text{ Applying log with base e on both sides, we get }$$

$$\log \left(f\left(x\right)\right)=x{\log }_{e}x$$

$$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=1+{\log }_{e}x$$

$$f^{\prime }\left(x\right)=f\left(x\right)(1+{\log }_{e}x)=x^x(1+{\log }_{e}x)$$

$$\text{ For f(x) to be decreasing, we must have }$$

$$f^{\prime }\left(x\right)<0$$

$$\Rightarrow x^x(1+{\log }_{e}x)<0$$

$$\text{ Here, logaritmic function is defined for positive values of x }.$$

$$\Rightarrow x^x>0$$

$$\Rightarrow 1+{\log }_{e}x<0[\text{ Since }{x}^x>0,x^x(1+{\log }_{e}x)<0\Rightarrow 1+{\log }_{e}x<0]$$

$$\Rightarrow {\log }_{e}x<-1$$

$$\Rightarrow x<e^{-1}[\because l{og}_{a}x<N\Rightarrow x<a^N\text{ for }a>1]$$

$$\text{ Here },$$

$$e>1$$

$$\Rightarrow {\log }_{e}x<-1\Rightarrow x<e^{-1}$$

$$\Rightarrow x\in (0,e^{-1})$$

$$\text{ So,f(x) is decreasing on }(0,\frac{1}{e}).$$