CUET UG 2025 — Mathematics Calculus

If $f'(x) > 0$f′(x)>0 on an interval, the function is increasing there. If $f'(x) < 0$f′(x)<0, it is decreasing.

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$$f(x) = x|x|$$f(x)=x∣x∣

$$f(x) = \begin{cases} x^2, & \text{if } x \geq 0 \\ -x^2, & \text{if } x < 0 \end{cases}$$f(x)={x2,−x2,​if x≥0if x<0​

$$f'(x) = \begin{cases} 2x \geq 0, & \text{when } x \geq 0 \\ -2x > 0, & \text{when } x < 0 \end{cases}$$f′(x)={2x≥0,−2x>0,​when x≥0when x<0​

$f'(x) \geq 0$f′(x)≥0 everywhere, so $f(x)$f(x) is increasing on $(-\infty, \infty)$(−∞,∞)

$\Rightarrow$⇒ (A) → (IV)

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$$f(x) = x^2 + 2x - 5$$f(x)=x2+2x−5

$$f'(x) = 2x + 2 = 2(x + 1)$$f′(x)=2x+2=2(x+1)

For $x < -1$x<−1, $f'(x) < 0$f′(x)<0, so $f(x)$f(x) is decreasing on $(-\infty, -1)$(−∞,−1)

$\Rightarrow$⇒ (B) → (III)

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$$f(x) = x^2 - 6x + 9 = (x - 3)^2$$f(x)=x2−6x+9=(x−3)2

$$f'(x) = 2x - 6 = 2(x - 3)$$f′(x)=2x−6=2(x−3)

For $x > 3$x>3, $f'(x) > 0$f′(x)>0, so $f(x)$f(x) is increasing on $(3, \infty)$(3,∞)

$\Rightarrow$⇒ (C) → (II)

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$$f(x) = -x^2$$f(x)=−x2

$$f'(x) = -2x$$f′(x)=−2x

For $x > 0$x>0, $f'(x) < 0$f′(x)<0, so $f(x)$f(x) is decreasing on $(0, \infty)$(0,∞)

$\Rightarrow$⇒ (D) → (I)

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(A) → (IV), (B) → (III), (C) → (II), (D) → (I)