CUET UG 2025 — Mathematics Calculus

A function decreases when its derivative is negative, so we need to find where $f'(x) < 0$f′(x)<0.

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Given $f(x) = x^x$f(x)=xx, take the natural logarithm of both sides:

$$\ln f(x) = \ln(x^x)$$lnf(x)=ln(xx)

$$\ln f(x) = x \ln x$$lnf(x)=xlnx

Differentiate both sides:

$$\frac{f'(x)}{f(x)} = \frac{d}{dx}[x \ln x]$$f(x)f′(x)​=dxd​[xlnx]

Using the product rule on the right side:

$$\frac{f'(x)}{f(x)} = 1 \cdot \ln x + x \cdot \frac{1}{x}$$f(x)f′(x)​=1⋅lnx+x⋅x1​

$$\frac{f'(x)}{f(x)} = \ln x + 1$$f(x)f′(x)​=lnx+1

$$f'(x) = f(x) \cdot (\ln x + 1)$$f′(x)=f(x)⋅(lnx+1)

$$f'(x) = x^x(\ln x + 1)$$f′(x)=xx(lnx+1)

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For $f'(x) < 0$f′(x)<0:

$$x^x(\ln x + 1) < 0$$xx(lnx+1)<0

Since $x > 0$x>0, we have $x^x > 0$xx>0 always.

Therefore:

$$\ln x + 1 < 0$$lnx+1<0

$$\ln x < -1$$lnx<−1

$$x < e^{-1}$$x<e−1

$$x < \frac{1}{e}$$x<e1​

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Combined with the condition $x > 0$x>0, the function decreases on the interval $(0, \frac{1}{e})$(0,e1​).