CUET Mathematics - Calculus - Application Of Integrals | Free Questions - Page 6

Q1:

2022: 17 Aug Shift 2

Calculus

>

Application of Integrals

Medium

Core

The area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$ is :

$\frac{2}{3}(\pi - 2)$

$\frac{2}{3}(\pi + 2)$

$\frac{3}{2}(\pi - 2)$

$\frac{3}{2}(\pi + 2)$

Correct Answer

Option 3

Correct Answer

Explanation

Q2:

2022: 6 Aug Shift 2

Calculus

>

Application of Integrals

Hard

Core

A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by $y = y(x) = x + \sqrt{2x - 3x^2}$ where 'x' (in meters) is the horizontal distance moved and 'y' (in meters) is the corresponding height. Assuming that his initial point is at origin.

Based on the above information answer the Questions

The value of $x$ for which $\frac{dy}{dx} = 0$ , is

$\frac{1}{2}$

$\frac{1}{7}$

$\frac{2}{3}$

$\frac{1}{3}$

Correct Answer

Option 1

Correct Answer

Explanation

Q3:

2022: 6 Aug Shift 2

Calculus

>

Application of Integrals

Hard

Core

A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by $y = y(x) = x + \sqrt{2x - 3x^2}$ where 'x' (in meters) is the horizontal distance moved and 'y' (in meters) is the corresponding height. Assuming that his initial point is at origin.

Based on the above information answer the Questions

$y(x)$ is strictly increasing in the interval

$(0, \frac{1}{6})$

$(\frac{1}{6}, \frac{2}{3})$

$(0, \frac{1}{2})$

$(\frac{1}{2}, \frac{2}{3})$

Correct Answer

Option 3

Correct Answer

Explanation

Q4:

2022: 6 Aug Shift 2

Calculus

>

Application of Integrals

Hard

Core

A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by $y = y(x) = x + \sqrt{2x - 3x^2}$ where 'x' (in meters) is the horizontal distance moved and 'y' (in meters) is the corresponding height. Assuming that his initial point is at origin.

Based on the above information answer the Questions

The maximum height (in meters) achieved in the first jump is

$\frac{2}{3}$

$1$

$\frac{1}{2}$

$\frac{1}{6}$

Correct Answer

Option 2

Correct Answer

Explanation

Q5:

2022: 6 Aug Shift 2

Calculus

>

Application of Integrals

Hard

Core

A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by $y = y(x) = x + \sqrt{2x - 3x^2}$ where 'x' (in meters) is the horizontal distance moved and 'y' (in meters) is the corresponding height. Assuming that his initial point is at origin.

Based on the above information answer the Questions

Evaluate $\int_0^{\frac{1}{3}} y \, dx$

$\frac{1}{18} + \frac{2\sqrt{3}\pi}{27}$

$\frac{1}{18} + \sqrt{3}\frac{\pi}{36}$

$\frac{1}{18} + \frac{\sqrt{6}\pi}{9}$

$\frac{1}{18} + \frac{\pi}{36}$

Correct Answer

Option 2

Correct Answer

Explanation

Q6:

2022: 6 Aug Shift 2

Calculus

>

Application of Integrals

Hard

Core

A man plans to jump and climb the stairs. The path (as shown in the figure) he moves for the first jump is given by $y = y(x) = x + \sqrt{2x - 3x^2}$ where 'x' (in meters) is the horizontal distance moved and 'y' (in meters) is the corresponding height. Assuming that his initial point is at origin.

Based on the above information answer the Questions

The value of $\frac{dy}{dx}$ when $x = \frac{1}{6}$ , is

$2$

$0$

$3$

$1$

Correct Answer

Option 1

Correct Answer

Explanation

Q7:

2022: 4 Aug Shift 1

Calculus

>

Application of Integrals

Medium

Core

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

$\frac{3}{8}$ sq. unit

$\frac{5}{8}$ sq. unit

$\frac{7}{8}$ sq. unit

$\frac{9}{8}$ sq. unit

Correct Answer

Option 4

Correct Answer

Explanation

Q8:

2022: 10 Aug Shift 1

Calculus

>

Application of Integrals

Medium

Core

The line $y = x$ , partition the area of the circle $(x-1)^2 + y^2 = 1$ , into two segments. The area of the major segment is

$\frac{\pi}{2} + \frac{1}{4}$

$\frac{\pi}{4} + \frac{1}{2}$

$\frac{3\pi}{4} + \frac{1}{6}$

$\frac{3\pi}{4} + \frac{1}{2}$

Correct Answer

Option 4

Correct Answer

Explanation

Q9:

2022: 10 Aug Shift 1

Calculus

>

Application of Integrals

Easy

Core

The portion of the area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , that lies in the first quadrant is

$\pi ab$

$\frac{\pi ab}{2}$

$\frac{\pi ab}{4}$

$2\pi ab$

Correct Answer

Option 3

Correct Answer

Explanation

Q10:

2022: 10 Aug Shift 1

Calculus

>

Application of Integrals

Hard

Core

The area (in square units) of minor segment of the circle $x^2 + y^2 = 25$ cut off by the line $x = \frac{5}{2}$ is

$25\left(\frac{\pi}{4} - \frac{\sqrt{3}}{2}\right)$

$\frac{25}{12}(4\pi - 3\sqrt{3})$

$\frac{25}{12}(3\pi - 4\sqrt{3})$

$25\left(\frac{\sqrt{3}}{2} + \frac{\pi}{4}\right)$

Correct Answer

Option 2

Correct Answer

Explanation

Q11:

2022: 30 Aug Shift 1

Calculus

>

Application of Integrals

Medium

Core

The area enclosed by the curve $y^2 = 4ax$ and its latus-rectum is

$\frac{8}{3}a^2$

$\frac{4}{3}a^2$

$\frac{1}{3}a^2$

$\frac{1}{12}a^2$

Correct Answer

Option 1

Correct Answer

Explanation

Q12:

2022: 30 Aug Shift 1

Calculus

>

Application of Integrals

Medium

Core

The smaller of the areas enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ is

$2(\pi - 2)$

$\pi - 2$

$2\pi - 1$

$2\pi + 2$

Correct Answer

Option 2

Correct Answer

Explanation

Q13:

2022: 16 July Shift 2

Calculus

>

Application of Integrals

Hard

Core

The smaller area enclosed by the curve $y = |x|$ and the circle $(x-a)^2 + y^2 = a^2$ is:

$\frac{a^2}{4}(\pi - 1)$

$\frac{a^2}{4}(\pi - 2)$

$\frac{a^2}{2}(\pi - 2)$

$\frac{a^2}{2}(\pi - 1)$

Correct Answer

Option 2

Correct Answer

Explanation

Q14:

2022: 16 July Shift 2

Calculus

>

Application of Integrals

Easy

Core

The line $y = mx$ ( $m > 0$ ) partitions the area of the circle $x^2 + y^2 = a^2$ ( $a > 0$ ) in the ratio:

$m : a$

$a : m$

$m : \sqrt{a^2 + m^2}$

$1 : 1$

Correct Answer

Option 4

Correct Answer

Explanation

Q15:

2022: 23 Aug Shift 1

Calculus

>

Application of Integrals

Medium

Core

The area bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$ is :

$\frac{9}{8}$

$\frac{9}{4}$

$\frac{9}{2}$

$\frac{9}{7}$

Correct Answer

Option 1

Correct Answer

Explanation

Q16:

2022: 25 May Shift 1

Calculus

>

Application of Integrals

Easy

Common

The area enclosed by the ellipse $\frac{x^2}{9^2} + \frac{y^2}{6^2} = 1$ is:

$15\pi$

$54\pi$

$18\pi$

$\frac{3}{2}\pi$

Correct Answer

Option 2

Correct Answer

Explanation

Q17:

2022: 25 May Shift 1

Calculus

>

Application of Integrals

Medium

Core

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is:

$\frac{4a^2}{3}$ sq. units

$\frac{8a^2}{3}$ sq. units

$\frac{2a^2}{3}$ sq. units

$\frac{9a^2}{5}$ sq. units

Correct Answer

Option 2

Correct Answer

Explanation

Q18:

2022: 25 May Shift 1

Calculus

>

Application of Integrals

Easy

Core

The area of the region bounded by the lines $x = 2y + 3, x = 0, y = 1$ and $y = -1$ is:

4 sq. units

6 sq. units

8 sq. units

$\frac{3}{2}$ sq. units

Correct Answer

Option 2

Correct Answer

Explanation

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