If a revenue function is given by R(x) = 2027x - 1013x^2 - 675x^3, then the marginal revenue function (MR) is: Option 1: MR = 2027 - 1013x - 675x^2 Option 2: MR = 2026 - 4050x Option 3: MR = 2027 - 2026x - 2025x^2 Option 4: MR = 1013x - 2025x^2

Answer: MR = 2027 - 2026x - 2025x^2. Solution: Marginal Revenue (MR) is the derivative of the Revenue function. Given: R(x) = 2027x - 1013x^2 - 675x^3 Find: MR = dR/dx Using the power rule, where the derivative of ax^n is n x ax^(n-1): For the term 2027x: Power of x is 1 Derivative = 1 x 2027x^(1-1) = 2027x^0 = 2027 For the term -1013x^2: Power of x is 2 Derivative = 2 x (-1013)x^(2-1) = -2026x For the term -675x^3: Power of x is 3 Derivative = 3 x (-675)x^(3-1) = -2025x^2 Combining all terms: MR = 2027 - 2026x - 2025x^2

Calculus

Application of Derivatives

Medium

applied

If a revenue function is given by $R(x) = 2027x - 1013x^2 - 675x^3$ , then the marginal revenue function (MR) is:

$MR = 2027 - 1013x - 675x^2$

$MR = 2026 - 4050x$

$MR = 2027 - 2026x - 2025x^2$

$MR = 1013x - 2025x^2$

✅ Correct Option: 3

Marginal Revenue (MR) is the derivative of the Revenue function.

Given: $R(x) = 2027x - 1013x^2 - 675x^3$

Find: $MR = \frac{dR}{dx}$

Using the power rule, where the derivative of $ax^n$ is $n \times ax^{n-1}$ :

For the term $2027x$ :

Power of $x$ is $1$

Derivative $= 1 \times 2027x^{1-1}$

$= 2027x^0$

$= 2027$

For the term $-1013x^2$ :

Power of $x$ is $2$

Derivative $= 2 \times (-1013)x^{2-1}$

$= -2026x$

For the term $-675x^3$ :

Power of $x$ is $3$

Derivative $= 3 \times (-675)x^{3-1}$

$= -2025x^2$

Combining all terms:

$MR = 2027 - 2026x - 2025x^2$

2025: 21 May Shift 2

If the maximum value of the function $f(x) = \frac{2\log_e x}{x}$ , $x > 0$ occurs at $x = e$ , then $e^3 f''(e)$ is equal to

2026: 18 May Shift 2

The function $f(x)=\sin^4(x)+\cos^4(x)$ , on interval $\left(0,\dfrac{\pi}{2}\right)$ , has

2026: 15 May Shift 1

If $f(x) = x^4 - \dfrac{2x^3}{3}$ , then $f(x)$ is

A. increasing in $\left(\dfrac{1}{2}, \infty\right)$

B. decreasing in $\left(\dfrac{1}{2}, \infty\right)$

C. increasing in $(-\infty, 0)$

D. decreasing in $(-\infty, 0) \cup \left(0, \dfrac{1}{2}\right)$

Choose the correct answer from the options given below:

2026: 29 May Shift 2

For the function $f(x) = (x-1)^3(x-2)^2$

Match the LIST-I with LIST-II

LIST-I		LIST-II
A.	$f(x) \geq 0$	I. $x \in (-\infty,1) \cup \left(1,\frac{8}{5}\right) \cup (2,\infty)$
B.	$f(x) < 0$	II. $x \geq 1$
C.	$f(x)$ is increasing if	III. $x < 1$
D.	$f(x)$ is decreasing if	IV. $x \in \left(\frac{8}{5},2\right)$

Choose the correct answer from the options given below:

2025: 2 June Shift 1

If $x = -1$ and $x = -2$ are the extreme points of $f(x) = \alpha\log|x| + \beta x^2 + x$ then

2026: 21 May Shift 2

Which of the following functions are strictly decreasing on $\left(0, \dfrac{\pi}{2}\right)$ ?

A. $\cos x$

B. $-\cot x$

C. $\cos 2x$

D. $\sin x$

Choose the correct answer from the options given below:

If a revenue function is given by R(x)=2027x−1013x2−675x3R(x) = 2027x - 1013x^2 - 675x^3R(x)=2027x−1013x2−675x3, then the marginal revenue function (MR) is:

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

If a revenue function is given by $R(x) = 2027x - 1013x^2 - 675x^3$ , then the marginal revenue function (MR) is: