1+/1- d = Option 1: /2 + /2 + c, where c is an arbitrary constant Option 2: /2 + + c, where c is an arbitrary constant Option 3: (/2) + + c, where c is an arbitrary constant Option 4: -2(/2) - + c, where c is an arbitrary constant

The answer is '-2(/2) - + c, where c is an arbitrary constant'. View full solution:

-2(2) - + c, where c is an arbitrary constant

2 + 2 + c, where c is an arbitrary constant

-2(2) - + c, where c is an arbitrary constant

CUET Mathematics - 1+/1- d = | PYQs + Solutions

CUET Mathematics 2026 31st May Shift 2

Calculus

Medium

core

$\tan\dfrac{\theta}{2} + \sec\dfrac{\theta}{2} + c$ , where $c$ is an arbitrary constant

$\tan\dfrac{\theta}{2} + \theta + c$ , where $c$ is an arbitrary constant

$\cot\left(\dfrac{\theta}{2}\right) + \theta + c$ , where $c$ is an arbitrary constant

$-2\cot\left(\dfrac{\theta}{2}\right) - \theta + c$ , where $c$ is an arbitrary constant

✅ Correct Option: 4

Related questions:

21 May Shift 1

Match List-I with List-II

List-I	List-II
Definite integral	Value
---	---
(A) $\int_{1}^{e} \frac{\log x}{x} dx$	(I) 4
(B) $\int_{-2}^{2} x^3(1 - x^2) dx$	(II) $\frac{1}{2}$
(C) $\int_{1}^{2} x \, dx$	(III) 0
(D) $\int_{-2}^{2} \vert x\vert dx$	(IV) $\frac{3}{2}$

Choose the correct answer from the options given below:

22 May Shift 2

If $\int \frac{2x - 5}{(2x - 3)^3} e^{2x} dx = \frac{\lambda e^{2x}}{(2x - 3)^2} + C$ , where $C$ is an arbitrary constant then the value of $\lambda$ is

10 Aug Shift 1

If $\int_0^{\pi/2} \sqrt{\tan x} \, dx = \frac{\lambda}{\sqrt{2}}$ , then the value of $\lambda$ is

CUET Mathematics 2026 31st May Shift 2 PYQs

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