1
JEE Main 2025 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $ A = \left\{ \theta \in [0, 2\pi] : 1 + 10\operatorname{Re}\left( \frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta} \right) = 0 \right\} $. Then $ \sum\limits_{\theta \in A} \theta^2 $ is equal to

A

$ \frac{21}{4} \pi^2 $

B

$ 6\pi^2 $

C

$ \frac{27}{4} \pi^2 $

D

$ 8\pi^2 $

2
JEE Main 2025 (Online) 7th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

If the locus of z ∈ ℂ, such that Re$ \left( \frac{z - 1}{2z + i} \right) + \text{Re} \left( \frac{\overline{z} - 1}{2\overline{z} - i} \right) = 2 $, is a circle of radius r and center $(a, b)$, then $ \frac{15ab}{r^2} $ is equal to :

A

16

B

24

C

12

D

18

3
JEE Main 2025 (Online) 7th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Among the statements

(S1) : The set $\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right.$ and $\frac{z-i}{z+i}$ is purely real $\}$ contains exactly two elements, and

(S2) : The set $\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right.$ and $\frac{z-1}{z+1}$ is purely imaginary $\}$ contains infinitely many elements.

A
both are incorrect
B
both are correct
C
only (S2) is correct
D
only (S1) is correct
4
JEE Main 2025 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let the product of $\omega_1=(8+i) \sin \theta+(7+4 i) \cos \theta$ and $\omega_2=(1+8 i) \sin \theta+(4+7 i) \cos \theta$ be $\alpha+i \beta$, $i=\sqrt{-1}$. Let p and q be the maximum and the minimum values of $\alpha+\beta$ respectively. Then $\mathrm{p}+\mathrm{q}$ is equal to :

A
130
B
150
C
160
D
140
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