Match List-I with List-II | List-I | List-II | |---|---| | (A) A fire in a factory causing production delay for some time is | (I) Secular trend | | (B) Technological progress is | (II) Seasonal trend | | (C) The rise in prices before big festival is example of | (III) Irregular trend | | (D) Rise and fall of share market is | (IV) Cyclic trend | Choose the correct answer from the options given below:

CUET UG 2025 — Mathematics Statistics & Applications

2025mcqmedium

Match List-I with List-II

List-I	List-II
(A) A fire in a factory causing production delay for some time is	(I) Secular trend
(B) Technological progress is	(II) Seasonal trend
(C) The rise in prices before big festival is example of	(III) Irregular trend
(D) Rise and fall of share market is	(IV) Cyclic trend

Choose the correct answer from the options given below:

Official previous-year question

Held on 2 Jun 2025 · Verified 13 Jul 2026.

Options

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

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Solution

The area between the curve $y = \cos x$ and the x-axis from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$ needs to be found.

The curve $y = \cos x$ has the following values:

At $x = -\frac{\pi}{2}$ : $\cos(-\frac{\pi}{2}) = 0$

At $x = 0$ : $\cos(0) = 1$

At $x = \frac{\pi}{2}$ : $\cos(\frac{\pi}{2}) = 0$

Between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ , the cosine curve stays above the x-axis (all values are positive).

The area bounded by the curve and the x-axis is:

Area $\displaystyle = \int_{-\pi/2}^{\pi/2} \cos x \, dx$

The antiderivative of $\cos x$ is $\sin x$ :

$\int \cos x \, dx = \sin x + C$

Applying the limits:

Area $= [\sin x]_{-\pi/2}^{\pi/2}$

Area $= \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})$

Area $= 1 - (-1)$

Area $= 2$

Therefore, the area of the region bounded by the curve $y = \cos x$ between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ and the x-axis is $2$ square units.

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List-I

List-II

(A) A fire in a factory causing production delay for some time is

(I) Secular trend

(B) Technological progress is

(II) Seasonal trend

(III) Irregular trend

(D) Rise and fall of share market is

(IV) Cyclic trend

Solution

The area between the curve $y = \cos x$ and the x-axis from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$ needs to be found.

The curve $y = \cos x$ has the following values:

At $x = -\frac{\pi}{2}$ : $\cos(-\frac{\pi}{2}) = 0$

At $x = 0$ : $\cos(0) = 1$

At $x = \frac{\pi}{2}$ : $\cos(\frac{\pi}{2}) = 0$

Between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ , the cosine curve stays above the x-axis (all values are positive).

The area bounded by the curve and the x-axis is:

Area $\displaystyle = \int_{-\pi/2}^{\pi/2} \cos x \, dx$

The antiderivative of $\cos x$ is $\sin x$ :

$\int \cos x \, dx = \sin x + C$

Applying the limits:

Area $= [\sin x]_{-\pi/2}^{\pi/2}$

Area $= \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})$

Area $= 1 - (-1)$

Area $= 2$

Therefore, the area of the region bounded by the curve $y = \cos x$ between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ and the x-axis is $2$ square units.

CUET UG 2025 — Mathematics Statistics & Applications

Options

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Solution

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CUET UG Statistics & Applications in other years

More CUET UG Statistics & Applications questions

Other Mathematics topics for CUET UG

Algebra

Calculus

Applied-Mathematics

Geometry

Arithmetic

Coordinate Geometry

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CUET UG 2025 — Mathematics Statistics & Applications

Options

Did you get this right?

Solution

Your note

CUET UG Statistics & Applications in other years

More CUET UG Statistics & Applications questions

Other Mathematics topics for CUET UG

Algebra

Calculus

Applied-Mathematics

Geometry

Arithmetic

Coordinate Geometry

Keep practicing Statistics & Applications