CUET UG 2025 — Mathematics Calculus

The function $y = ax^2 + bx$y=ax2+bx has a minimum at $x = 2$x=2 with minimum value $-12$−12.

When a parabola has a minimum (lowest point), it opens upward.

Therefore, $a > 0$a>0.

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At a minimum point, the derivative equals zero.

$$\frac{dy}{dx} = 2ax + b$$dxdy​=2ax+b

At $x = 2$x=2:

$$2a(2) + b = 0$$2a(2)+b=0

$$4a + b = 0$$4a+b=0

$b = -4a$b=−4a ...(Equation 1)

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At $x = 2$x=2, the value $y = -12$y=−12:

$$y = ax^2 + bx$$y=ax2+bx

$$-12 = a(2)^2 + b(2)$$−12=a(2)2+b(2)

$-12 = 4a + 2b$−12=4a+2b ...(Equation 2)

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From Equation 1: $b = -4a$b=−4a

Substituting into Equation 2:

$$-12 = 4a + 2(-4a)$$−12=4a+2(−4a)

$$-12 = 4a - 8a$$−12=4a−8a

$$-12 = -4a$$−12=−4a

$$a = 3$$a=3

Now finding $b$b:

$$b = -4a$$b=−4a

$$b = -4(3)$$b=−4(3)

$$b = -12$$b=−12

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Checking the options:

(A) $a = 3$a=3 ✓ Correct

(B) $a = -3$a=−3 ✗ Incorrect (would give maximum, not minimum)

(C) $b = 12$b=12 ✗ Incorrect

(D) $b = -12$b=−12 ✓ Correct

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The correct options are (A) and (D).

Therefore, $a = 3$a=3 and $b = -12$b=−12.