Question

If $\operatorname{f}(x)={{x}^{x}}$ , then $\operatorname{f}(x)$ is increasing in the interval
(A) $\left[ 0,e \right]$
(B) $\left( \dfrac{1}{e},\infty \right)$
(C) $\left[ 0,1 \right]$
(D) None of these

Answer 1

Hint: Assume $y=\operatorname{f}(x)={{x}^{x}}$ . We know that a function is increasing, if and only if the slope of the function is greater than zero. So, the slope of the function $\operatorname{f}(x)$ should be greater than 0. Now, take ln in both LHS and RHS of the equation $y={{x}^{x}}$ . Now, to obtain the slope, differentiate the LHS and RHS of the equation, $\ln y=\ln \left( {{x}^{x}} \right)$ . Now, use the formula, $\ln \left( {{m}^{n}} \right)=n\ln m$ , $\dfrac{d\left( uv \right)}{dx}=v\left( \dfrac{du}{dx} \right)+u\left( \dfrac{dv}{dx} \right)$ , and $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$, and then, simply the equation \dfrac{d\left( \ln y \right)}{dx}=\dfrac{d\left\\{ \ln \left( {{x}^{x}} \right) \right\\}}{dx} . For increasing function, make $\dfrac{dy}{dx}>0$ . Since ${{x}^{x}}\left( \ln x+1 \right)$ is greater than zero, so either both ${{x}^{x}}$ and $\ln \left( x+1 \right)$ should be greater than zero or both ${{x}^{x}}$ and $\left( \ln x+1 \right)$ should be less than zero. In case ${{1}^{st}}$ , we have both ${{x}^{x}}$ and $\left( \ln x+1 \right)$ greater than zero. The expression ${{x}^{x}}$ can be greater than 0, if $x>0$ . Now, solve $\ln x+1>0$ and get the range of values of x. In case ${{2}^{nd}}$ , we have both ${{x}^{x}}$ and $\left( \ln x+1 \right)$ less than zero. The expression ${{x}^{x}}$ can be less than 0, if x is less than zero. But the term $\ln x$ is undefined when x is less than zero. So, we can say that the case ${{2}^{nd}}$ is not possible. Now, take the intersection of the range of values of x and conclude the range of values of x.

Complete step-by-step solution-
According to the question, we have a function, $\operatorname{f}(x)={{x}^{x}}$ and we have to get the interval in which the function $\operatorname{f}(x)$ is increasing.
We know that a function is increasing, if and only if the slope of the function is greater than zero. To obtain the slope of the function we have to differentiate the given function.
Let us assume,
$y=\operatorname{f}(x)={{x}^{x}}$ ………………………….(1)
Now, we have to differentiate the function $\operatorname{f}(x)$ to obtain the slope of the function.
Slope of the function= $\dfrac{dy}{dx}=\dfrac{d\operatorname{f}(x)}{dx}$ ……………………………..(2)
Taking ln in LHS and RHS in equation (1), we get
$\Rightarrow \ln y=\ln \left( {{x}^{x}} \right)$ ……………………………(3)
We know the formula, $\ln \left( {{m}^{n}} \right)=n\ln m$ ………………………………..(4)
Now, from equation (3) and equation (4), we get
$\Rightarrow \ln y=\ln \left( {{x}^{x}} \right)$
$\Rightarrow \ln y=x\ln \left( x \right)$ ………………………………..(5)
Now, differentiating equation (5) with respect to x, we get
\Rightarrow \dfrac{d\left( \ln y \right)}{dx}=\dfrac{d\left\\{ x\ln \left( x \right) \right\\}}{dx} ……………………………….(6)
We have a formula, $\dfrac{d\left( uv \right)}{dx}=v\left( \dfrac{du}{dx} \right)+u\left( \dfrac{dv}{dx} \right)$ …………………………………..(7)
Now, from equation (6) an equation (7), we get
\Rightarrow \dfrac{d\left( \ln y \right)}{dx}=\dfrac{d\left\\{ x\ln \left( x \right) \right\\}}{dx}
$\Rightarrow \dfrac{d\left( \ln y \right)}{dx}=\ln x\left( \dfrac{dx}{dx} \right)+x\left( \dfrac{\ln x}{dx} \right)$ …………………………………..(8)
We know the formula, $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$ ………………………………….(9)
From equation (8) and equation (9), we have
$\Rightarrow \dfrac{d\left( \ln y \right)}{dx}=\ln x\left( \dfrac{dx}{dx} \right)+x\left( \dfrac{\ln x}{dx} \right)$
$\Rightarrow \dfrac{1}{y}.\dfrac{dy}{dx}=\ln x+x.\dfrac{1}{x}$
$\Rightarrow \dfrac{dy}{dx}=y\left( \ln x+1 \right)$ ………………………………(10)
Now, putting the value of y from equation (1) in equation (10), we get
$\Rightarrow \dfrac{dy}{dx}={{x}^{x}}\left( \ln x+1 \right)$ ……………………………..(11)
From equation (2) and equation (11), we can say that
Slope of the function = $\dfrac{dy}{dx}={{x}^{x}}\left( \ln x+1 \right)$ ………………………(12)
If a function is increasing then its slope is greater than zero in that interval. Mathematically, we can say that,
Slope of the function > 0 ………………………(13)
From equation (12) and equation (13), we get
${{x}^{x}}\left( \ln x+1 \right)>0$
Since ${{x}^{x}}\left( \ln x+1 \right)$ is greater than zero, so either both ${{x}^{x}}$ and $\ln \left( x+1 \right)$ should be greater than zero or both ${{x}^{x}}$ and $\left( \ln x+1 \right)$ should be less than zero.
Now, we have two cases.
In case ${{1}^{st}}$ , we have both ${{x}^{x}}$ and $\left( \ln x+1 \right)$ greater than zero.
${{x}^{x}}>0$
The expression ${{x}^{x}}$ can be greater than 0, if x is greater than zero. So, $x>0$ ………………..(14)
Now, we have,

& \left( \ln x+1 \right)>0 \\\ & \ln x>-1 \\\ & x>{{e}^{-1}} \\\ & x>\dfrac{1}{e} \\\ \end{aligned}$$ So, $$x\in \left( \dfrac{1}{e},\infty \right)$$ …………………………(15) In case $${{2}^{nd}}$$ , we have both $${{x}^{x}}$$ and $$\left( \ln x+1 \right)$$ less than zero. $${{x}^{x}}<0$$ The expression $${{x}^{x}}$$ can be less than 0, if x is less than zero. But the term $$\ln x$$ is undefined when x is less than zero. So, we can say that the case $${{2}^{nd}}$$ is not possible. From equation (14) and equation (15), we have the range of values of x. For the common value of x we have to take the intersection of values of x.  Now, on taking the intersection, we get $$x\in \left( \dfrac{1}{e},\infty \right)$$ . Hence, option (B) is the correct one. Note: In this question, while differentiating $${{x}^{x}}$$ one might think to use the formula, $$\dfrac{d\left( {{a}^{x}} \right)}{dx}=x{{a}^{x-1}}$$ . This is wrong. We cannot apply this formula here because in the formula $$\dfrac{d\left( {{a}^{x}} \right)}{dx}=x{{a}^{x-1}}$$ , a is the constant term. But in the expression $${{x}^{x}}$$ , we have x in place of a and x is variable.