CUET UG 2025 — Mathematics Calculus

Given: $f(x) = a \log_e|x| + bx^2 + x$f(x)=aloge​∣x∣+bx2+x has critical points at $x = -2$x=−2 and $x = 1$x=1

At critical points, the derivative equals zero.

The derivative of $f(x)$f(x) is:

$$f'(x) = \frac{a}{x} + 2bx + 1$$f′(x)=xa​+2bx+1

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At $x = 1$x=1, we have $f'(1) = 0$f′(1)=0:

$$\frac{a}{1} + 2b(1) + 1 = 0$$1a​+2b(1)+1=0

$a + 2b + 1 = 0$a+2b+1=0 ... (1)

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At $x = -2$x=−2, we have $f'(-2) = 0$f′(−2)=0:

$$\frac{a}{-2} + 2b(-2) + 1 = 0$$−2a​+2b(−2)+1=0

$$-\frac{a}{2} - 4b + 1 = 0$$−2a​−4b+1=0

$$-a - 8b + 2 = 0$$−a−8b+2=0

$a + 8b = 2$a+8b=2 ... (2)

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From equation (1):

$$a = -2b - 1$$a=−2b−1

Substituting into equation (2):

$$(-2b - 1) + 8b = 2$$(−2b−1)+8b=2

$$6b - 1 = 2$$6b−1=2

$$6b = 3$$6b=3

$$b = \frac{1}{2}$$b=21​

Therefore:

$$a = -2\left(\frac{1}{2}\right) - 1$$a=−2(21​)−1

$$a = -2$$a=−2

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With $a = -2$a=−2 and $b = \frac{1}{2}$b=21​:

$$a + 4b = -2 + 4\left(\frac{1}{2}\right)$$a+4b=−2+4(21​)

$$a + 4b = -2 + 2$$a+4b=−2+2

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