Calculus

Application of Derivatives

Medium

common

The absolute maximum value of the function $f(x) = 4x - \frac{1}{2}x^2$ in the interval $\left[-2, \frac{9}{2}\right]$ is

✅ Correct Option: 3

To find the absolute maximum value of $f(x) = 4x - \frac{1}{2}x^2$ on $\left[-2, \frac{9}{2}\right]$ , we need to check critical points (where the slope is zero) and endpoints of the interval.

Find the derivative:

$f'(x) = 4 - x$

Set $f'(x) = 0$ :

$4 - x = 0$

$x = 4$

The critical point $x = 4$ is inside the interval $\left[-2, \frac{9}{2}\right] = [-2, 4.5]$ , so we must evaluate at this point and both endpoints.

Evaluate at $x = -2$ :

$f(-2) = 4(-2) - \frac{1}{2}(-2)^2$

$= -8 - \frac{1}{2}(4)$

$= -8 - 2$

$= -10$

Evaluate at $x = 4$ :

$f(4) = 4(4) - \frac{1}{2}(4)^2$

$= 16 - \frac{1}{2}(16)$

$= 16 - 8$

$= 8$

Evaluate at $x = \frac{9}{2}$ :

$f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2$

$= 18 - \frac{1}{2} \times \frac{81}{4}$

$= 18 - \frac{81}{8}$

Convert $18$ to eighths: $18 = \frac{144}{8}$

$= \frac{144}{8} - \frac{81}{8}$

$= \frac{63}{8}$

$= 7.875$

Comparing the values:

At $x = -2$ : $f(-2) = -10$

At $x = 4$ : $f(4) = 8$

At $x = \frac{9}{2}$ : $f\left(\frac{9}{2}\right) = \frac{63}{8} = 7.875$

Therefore, the absolute maximum value is $8$ .

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The demand for a certain product is represented by the function $p = 300 + 25x - x^2$ (in rupees), where x is the number of units demanded and p is the price per unit, then the marginal revenue when 15 units are sold, is

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Which of the following are correct?

(A) The function $f(x) = 3x+12$ is increasing on R.

(B) The function $f(x) = e^{2x}$ is decreasing on R.

(D) The function $f(x) = x^3-3x^2+4x$ is increasing on R.

Choose the correct answer from the options given below:

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The function $f(x) = x^2 - x + 1$ is

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Consider the function $f(x) = -2x^3 + 3x^2 + 36x - 15$ , then which of the following are TRUE?

A. $f(x)$ increases on $(-2, 3)$

B. $f(x)$ decreases on $(-\infty, -2) \cup (3, \infty)$

C. $f(x)$ increases on $(-\infty, -2) \cup (3, \infty)$

D. $f(x)$ neither increases nor decreases on $(-2, 3)$

Choose the correct answer from the options given below:

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The function $f(x) = 4x + \frac{1}{x}$ is

(A). Increasing in the interval $\left(-\infty, -\frac{1}{2}\right)$

(B). Increasing in the interval $\left(-\frac{1}{2}, 0\right)$

(C). Decreasing in the interval $\left(0, \frac{1}{2}\right)$

(D). Decreasing in the interval $\left(\frac{1}{2}, \infty\right)$

Choose the correct answer from the options given below:

The absolute maximum value of the function f(x)=4x−12x2f(x) = 4x - \frac{1}{2}x^2f(x)=4x−21​x2 in the interval [−2,92]\left[-2, \frac{9}{2}\right][−2,29​] is

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CUET Mathematics Past Year Questions (Topic-Wise)

The absolute maximum value of the function $f(x) = 4x - \frac{1}{2}x^2$ in the interval $\left[-2, \frac{9}{2}\right]$ is