If f(x) = a _e, x, + bx^2 + x has critical points at x = -2 and x = 1, then Option 1: a + 2b = 0 Option 2: a - 2b = 0 Option 3: a + 4b = 0 Option 4: a - 4b = 0

Answer: a + 4b = 0. Solution: Given: f(x) = a _e, x, + bx^2 + x has critical points at x = -2 and x = 1 At critical points, the derivative equals zero. The derivative of f(x) is: f'(x) = a/x + 2bx + 1 At x = 1, we have f'(1) = 0: a/1 + 2b(1) + 1 = 0 a + 2b + 1 = 0 ... (1) At x = -2, we have f'(-2) = 0: a/(-2) + 2b(-2) + 1 = 0 -a/2 - 4b + 1 = 0 -a - 8b + 2 = 0 a + 8b = 2 ... (2) From equation (1): a = -2b - 1 Substituting into equation (2): (-2b - 1) + 8b = 2 6b - 1 = 2 6b = 3 b = 1/2 Therefore: a = -2(1/2) - 1 a = -2 With a = -2 and b = 1/2: a + 4b = -2 + 4(1/2) a + 4b = -2 + 2 a + 4b = 0

Calculus

Application of Derivatives

Medium

common

If $f(x) = a \log_e|x| + bx^2 + x$ has critical points at $x = -2$ and $x = 1$ , then

$a + 2b = 0$

$a - 2b = 0$

$a + 4b = 0$

$a - 4b = 0$

✅ Correct Option: 3

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

At critical points, the derivative equals zero.

The derivative of $f(x)$ is:

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

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

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

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

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

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

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

$-a - 8b + 2 = 0$

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

From equation (1):

$a = -2b - 1$

Substituting into equation (2):

$(-2b - 1) + 8b = 2$

$6b - 1 = 2$

$6b = 3$

$b = \frac{1}{2}$

Therefore:

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

$a = -2$

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

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

$a + 4b = -2 + 2$

$a + 4b = 0$

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(B) $f$ has no critical points

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Choose the correct answer from the options given below:

If f(x)=alog⁡e∣x∣+bx2+xf(x) = a \log_e|x| + bx^2 + xf(x)=aloge​∣x∣+bx2+x has critical points at x=−2x = -2x=−2 and x=1x = 1x=1, then

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CUET Mathematics Past Year Questions (Topic-Wise)

If $f(x) = a \log_e|x| + bx^2 + x$ has critical points at $x = -2$ and $x = 1$ , then