CUET UG 2025 — Mathematics Algebra

A function is increasing when its derivative is positive.

Given: $f(x) = 2\log_e(x-2) - x^2 + 4x + 1$f(x)=2loge​(x−2)−x2+4x+1

Differentiating term by term:

$$\frac{d}{dx}[2\log_e(x-2)] = \frac{2}{x-2}$$dxd​[2loge​(x−2)]=x−22​

$$\frac{d}{dx}[-x^2] = -2x$$dxd​[−x2]=−2x

$$\frac{d}{dx}[4x] = 4$$dxd​[4x]=4

$$\frac{d}{dx}[1] = 0$$dxd​[1]=0

Therefore:

$$f'(x) = \frac{2}{x-2} - 2x + 4$$f′(x)=x−22​−2x+4

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For the function to be increasing:

$$f'(x) > 0$$f′(x)>0

$$\frac{2}{x-2} - 2x + 4 > 0$$x−22​−2x+4>0

$$\frac{2}{x-2} > 2x - 4$$x−22​>2x−4

$$\frac{2}{x-2} > 2(x - 2)$$x−22​>2(x−2)

$$\frac{1}{x-2} > (x - 2)$$x−21​>(x−2)

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Since $x > 2$x>2, we have $(x-2) > 0$(x−2)>0.

Multiplying both sides by $(x-2)$(x−2):

$$1 > (x-2)^2$$1>(x−2)2

$$(x-2)^2 < 1$$(x−2)2<1

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Taking the square root:

$$|x-2| < 1$$∣x−2∣<1

$$-1 < x-2 < 1$$−1<x−2<1

$$1 < x < 3$$1<x<3

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The original function has domain $x > 2$x>2.

Combining $x > 2$x>2 and $1 < x < 3$1<x<3:

$$2 < x < 3$$2<x<3

Therefore, the function is increasing on the interval $(2, 3)$(2,3).