CUET UG 2025 — Mathematics Calculus

2025mcqmedium

If $y = \sqrt{2024x + 2025}$ , then which of the following is correct?

Official previous-year question

Held on 13 May 2025 · Verified 13 Jul 2026.

Options

A
$y\frac{d^2y}{dx^2} - 2\frac{dy}{dx} = 0$
B
$y\frac{d^2y}{dx^2} - 2\left(\frac{dy}{dx}\right)^2 = 0$
C
$y\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 = 0$
D
$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$

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Solution

The ellipse equation is $16x^2 + 25y^2 = 400$ .

The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .

Dividing the entire equation by 400:

$\frac{16x^2}{400} + \frac{25y^2}{400} = \frac{400}{400}$

$\frac{x^2}{25} + \frac{y^2}{16} = 1$

From $\frac{x^2}{25} + \frac{y^2}{16} = 1$ :

$a^2 = 25$

$a = 5$

$b^2 = 16$

$b = 4$

The area of an ellipse is given by $\pi ab$ .

Area $= \pi \times 5 \times 4$

Area $= 20\pi$ sq. units

Therefore, the area of the region enclosed by the ellipse is $20\pi$ square units.

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Options

$y\frac{d^2y}{dx^2} - 2\frac{dy}{dx} = 0$

$y\frac{d^2y}{dx^2} - 2\left(\frac{dy}{dx}\right)^2 = 0$

$y\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 = 0$

$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$

Solution

The ellipse equation is $16x^2 + 25y^2 = 400$ .

The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .

Dividing the entire equation by 400:

$\frac{16x^2}{400} + \frac{25y^2}{400} = \frac{400}{400}$

$\frac{x^2}{25} + \frac{y^2}{16} = 1$

From $\frac{x^2}{25} + \frac{y^2}{16} = 1$ :

$a^2 = 25$

$a = 5$

$b^2 = 16$

$b = 4$

The area of an ellipse is given by $\pi ab$ .

Area $= \pi \times 5 \times 4$

Area $= 20\pi$ sq. units

Therefore, the area of the region enclosed by the ellipse is $20\pi$ square units.

CUET UG 2025 — Mathematics Calculus

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CUET UG 2025 — Mathematics Calculus

Options

Did you get this right?

Solution

Your note

CUET UG Calculus in other years

More CUET UG Calculus questions

Other Mathematics topics for CUET UG

Algebra

Applied-Mathematics

Statistics & Applications

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Calculus