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Application of Derivatives question

2024 · 6 Apr · Shift 1 · Q49

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Application of Derivatives
2024 · 6 Apr · Shift 1 · Q49

Application of Derivatives question

2024 · 6 Apr · Shift 1 · Q49

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1

The interval in which the function

f(x)=x^x, x\gt 0

, is strictly increasing is

A
$(0, \infty)$
B
$\left(0, \frac{1}{e}\right]$
C
$\left[\frac{1}{e^2}, 1\right)$
D
$\left[\frac{1}{e}, \infty\right)$

View written solutionFree

Correct answer: D

We need the interval where

\\text{for }x>0$$ is strictly increasing. 2. Rewrite the function in exponential form: $$x^x=e^{x\ln x}.$$ Now differentiate: $$f'(x)=e^{x\ln x}\cdot \frac{d}{dx}(x\ln x).$$ 3. Compute the derivative of $x\ln x$: $$\frac{d}{dx}(x\ln x)=\ln x+1.$$ Hence, $$f'(x)=x^x(\ln x+1).$$ 4. Since $x>0$, we have $$x^x>0.$$ So the sign of $f'(x)$ depends only on $$\ln x+1.$$ 5. For the function to be strictly increasing, we need $$f'(x)>0 \iff \ln x+1>0.$$ That gives $$\ln x>-1 \iff x>e^{-1}=\frac{1}{e}.$$ 6. At $x=\frac{1}{e}$, $$f'\left(\frac{1}{e}\right)=0.$$ For $x<\frac{1}{e}$, $f'(x)<0$, so the function decreases there. For $x>\frac{1}{e}$, $f'(x)>0$, so the function increases there. Thus the function is increasing on $$\left[\frac{1}{e},\infty\right).$$ (Equivalently, it is strictly increasing for $x>\frac{1}{e}$, and among the given options this corresponds to option D.) 7. Check options: - A: $(0,\infty)$ — false, since the function decreases on $(0,1/e)$. - B: $\left(0,\frac{1}{e}\right]$ — false, this is the decreasing part. - C: $\left[\frac{1}{e^2},1\right)$ — false, it contains values less than $1/e$ where the function decreases. - D: $\left[\frac{1}{e},\infty\right)$ — correct.

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