CUET UG 2025 — Mathematics Calculus

Given $f(x) = (x - 2)^5(x + 2)^2$f(x)=(x−2)5(x+2)2

To find critical points, find the first derivative using the product rule:

Let $u = (x - 2)^5$u=(x−2)5 and $v = (x + 2)^2$v=(x+2)2

$$u' = 5(x - 2)^4$$u′=5(x−2)4

$$v' = 2(x + 2)$$v′=2(x+2)

$$f'(x) = 5(x - 2)^4(x + 2)^2 + (x - 2)^5 \cdot 2(x + 2)$$f′(x)=5(x−2)4(x+2)2+(x−2)5⋅2(x+2)

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Factor out the common terms $(x - 2)^4$(x−2)4 and $(x + 2)$(x+2):

$$f'(x) = (x - 2)^4(x + 2)[5(x + 2) + 2(x - 2)]$$f′(x)=(x−2)4(x+2)[5(x+2)+2(x−2)]

$$f'(x) = (x - 2)^4(x + 2)[5x + 10 + 2x - 4]$$f′(x)=(x−2)4(x+2)[5x+10+2x−4]

$$f'(x) = (x - 2)^4(x + 2)(7x + 6)$$f′(x)=(x−2)4(x+2)(7x+6)

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Set $f'(x) = 0$f′(x)=0:

$$(x - 2)^4(x + 2)(7x + 6) = 0$$(x−2)4(x+2)(7x+6)=0

Critical points are:

$$x = 2$$x=2

$$x = -2$$x=−2

$$x = -\dfrac{6}{7}$$x=−76​

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Apply the first derivative test. Note that $(x - 2)^4$(x−2)4 is always positive.

For $x < -2$x<−2, test $x = -3$x=−3:

$(x + 2) < 0$(x+2)<0 and $(7x + 6) < 0$(7x+6)<0, so $f'(x) > 0$f′(x)>0 (increasing)

For $-2 < x < -\dfrac{6}{7}$−2<x<−76​, test $x = -1$x=−1:

$(x + 2) > 0$(x+2)>0 and $(7x + 6) < 0$(7x+6)<0, so $f'(x) < 0$f′(x)<0 (decreasing)

At $x = -2$x=−2, the function changes from increasing to decreasing.

Therefore, $x = -2$x=−2 is a local maximum.