CUET UG 2025 — Mathematics Calculus

A function is increasing when the derivative is positive, $f'(x) > 0$f′(x)>0.

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

$$f'(x) = 3x^2 + 4x$$f′(x)=3x2+4x

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To find critical points, set the derivative equal to zero:

$$3x^2 + 4x = 0$$3x2+4x=0

$$x(3x + 4) = 0$$x(3x+4)=0

$x = 0$x=0 or $x = -\dfrac{4}{3}$x=−34​

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The critical points divide the number line into three intervals: $(-\infty, -\dfrac{4}{3})$(−∞,−34​), $(-\dfrac{4}{3}, 0)$(−34​,0), and $(0, \infty)$(0,∞).

Test $x = -2$x=−2 in the interval $(-\infty, -\dfrac{4}{3})$(−∞,−34​):

$$f'(-2) = 3(4) + 4(-2) = 12 - 8 = 4 > 0$$f′(−2)=3(4)+4(−2)=12−8=4>0

The function is increasing on $(-\infty, -\dfrac{4}{3})$(−∞,−34​).

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Test $x = -1$x=−1 in the interval $(-\dfrac{4}{3}, 0)$(−34​,0):

$$f'(-1) = 3(1) + 4(-1) = 3 - 4 = -1 < 0$$f′(−1)=3(1)+4(−1)=3−4=−1<0

The function is decreasing on $(-\dfrac{4}{3}, 0)$(−34​,0).

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Test $x = 1$x=1 in the interval $(0, \infty)$(0,∞):

$$f'(1) = 3(1) + 4(1) = 3 + 4 = 7 > 0$$f′(1)=3(1)+4(1)=3+4=7>0

The function is increasing on $(0, \infty)$(0,∞).

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The function is increasing on $(-\infty, -\dfrac{4}{3}] \cup [0, \infty)$(−∞,−34​]∪[0,∞).