CUET UG 2025 — Mathematics Calculus

Given: $f(x) = 2x^3 + 3x^2 - 12x + 1$f(x)=2x3+3x2−12x+1

A function is strictly increasing when its derivative is positive.

$$f'(x) = 6x^2 + 6x - 12$$f′(x)=6x2+6x−12

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For strictly increasing:

$$f'(x) > 0$$f′(x)>0

$$6x^2 + 6x - 12 > 0$$6x2+6x−12>0

$$x^2 + x - 2 > 0$$x2+x−2>0

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Factoring the quadratic:

$$x^2 + x - 2 = (x + 2)(x - 1)$$x2+x−2=(x+2)(x−1)

$$(x + 2)(x - 1) > 0$$(x+2)(x−1)>0

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Critical points where each factor equals zero:

$$x + 2 = 0 \Rightarrow x = -2$$x+2=0⇒x=−2

$$x - 1 = 0 \Rightarrow x = 1$$x−1=0⇒x=1

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Testing the sign of $(x + 2)(x - 1)$(x+2)(x−1) in each region:

For $x < -2$x<−2 (test $x = -3$x=−3): $(-)(-) = (+)$(−)(−)=(+) ✓

For $-2 < x < 1$−2<x<1 (test $x = 0$x=0): $(+)(-) = (-)$(+)(−)=(−) ✗

For $x > 1$x>1 (test $x = 2$x=2): $(+)(+) = (+)$(+)(+)=(+) ✓

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The function is strictly increasing on $(-\infty, -2) \cup (1, \infty)$(−∞,−2)∪(1,∞)