Match List - I with List - II., List - I (Functions), List - II (Maximum value), (A) f(x) = -x^2, x (-∞, ∞), (I) 8, (B) f(x) = -x^2 + 1, x (-∞, ∞), (II) 7, (C) f(x) = x + 1, x [0, 6], (III) 1, (D) f(x) = x^3, x [0, 2], (IV) 0, Choose the correct answer from the options given below : Option 1: (A)-(IV), (B)-(III), (C)-(II), (D)-(I) Option 2: (A)-(III), (B)-(II), (C)-(I), (D)-(IV) Option 3: (A)-(II), (B)-(I), (C)-(III), (D)-(IV) Option 4: (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Answer: (A)-(IV), (B)-(III), (C)-(II), (D)-(I). Solution: (A) -x^2 on R: max = 0 (IV). (B) -x^2 + 1 on R: max = 1 (III). (C) x + 1 on [0, 6]: max = 7 (II). (D) x^3 on [0, 2]: max = 8 (I).

Applied Mathematics

Calculus

Easy

applied

Match List - I with List - II.

List - I (Functions) List - II (Maximum value)
(A) $f(x) = -x^2, x \in (-\infty, \infty)$ (I) 8
(B) $f(x) = -x^2 + 1, x \in (-\infty, \infty)$ (II) 7
(C) $f(x) = x + 1, x \in [0, 6]$ (III) 1
(D) $f(x) = x^3, x \in [0, 2]$ (IV) 0

Choose the correct answer from the options given below :

List - I (Functions)	List - II (Maximum value)
(A) $f(x) = -x^2, x \in (-\infty, \infty)$	(I) 8
(B) $f(x) = -x^2 + 1, x \in (-\infty, \infty)$	(II) 7
(C) $f(x) = x + 1, x \in [0, 6]$	(III) 1
(D) $f(x) = x^3, x \in [0, 2]$	(IV) 0

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(III), (B)-(II), (C)-(I), (D)-(IV)

(A)-(II), (B)-(I), (C)-(III), (D)-(IV)

(A)-(I), (B)-(IV), (C)-(II), (D)-(III)

✅ Correct Option: 1

(A) $-x^2$ on $\mathbb{R}$ : max $= 0$ (IV).

(B) $-x^2 + 1$ on $\mathbb{R}$ : max $= 1$ (III).

(D) $x^3$ on $[0, 2]$ : max $= 8$ (I).

2023: 22 May Shift 3

Match List - I with List - II.

	List - I		List - II
(A)	If $y = x^2 - 8$ and $\frac{dy}{dx} = 0$ , then $x = ?$	(I)	1
(B)	If $p(x) = 3x + 1$ , then $R(x)$ at $x = 2$	(II)	0
(C)	If $y = x^3$ , then $\frac{dy}{dx}$ at $x = -1$	(III)	14
(D)	If $C(x) = 100 + 5x$ , $R(x) = 102 + 3x$ , then break-even point	(IV)	3

Choose the correct answer from the options given below :

2023: 7 June Shift 1

If $y = \log_3(\log_3 x)$ , then $\frac{dy}{dx}$

2023: 23 May Shift 3

The total cost of producing $x$ generators is given by

TC = $x^3 - 60x^2 + 1500x + 2000$ .

The Marginal Cost (MC), when $x = 10$ units is:

2026: 14 May Shift 2

The area of the region bounded by curves $y=x^2$ and line $y=4$ is:

2024: 16 May Shift 1

A company is selling a certain commodity $x$ . The demand function for the commodity is linear. The company can sell $2000$ units when the price is ₹ $8$ per unit and it can sell $3000$ units when the price is ₹ $4$ per unit. The Marginal revenue at $x=5$ is:

2023: 7 June Shift 1

If the cost function and the profit function for a company is given by $C = 10 - 0.3x^2$ and $P = 0.3x^2 + 2x - 10$ respectively , where X represent units of output, then the revenue function is given by :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)