1
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If for the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+(\tan x) y=\frac{2+\sec x}{(1+2 \sec x)^2}$, $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$, then $f\left(\frac{\pi}{4}\right)$ is equal to:

A
$\frac{5-\sqrt{3}}{2 \sqrt{2}}$
B

$\frac{4 - \sqrt{2}}{14}$

C

$\frac{9\sqrt{3} + 3}{10(4 + \sqrt{3})}$

D

$\frac{\sqrt{3} + 1}{10(4 + \sqrt{3})}$

2
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let y = y(x) be the solution of the differential equation :

$\cos x\left(\log _e(\cos x)\right)^2 d y+\left(\sin x-3 y \sin x \log _e(\cos x)\right) d x=0$, x ∈ (0, $\frac{\pi}{2}$ ). If $ y(\frac{\pi}{4}) $ = $-\frac{1}{\log_{e}2}$, then $ y(\frac{\pi}{6}) $ is equal to :

A

$\frac{2}{\log_{e}(3)−\log_{e}(4)}$

B

$-\frac{1}{\log_{e}(4)}$

C

$\frac{1}{\log_{e}(4)−\log_{e}(3)}$

D

$\frac{1}{\log_{e}(3)−\log_{e}(4)}$

3
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let for some function $\mathrm{y}=f(x), \int_0^x t f(t) d t=x^2 f(x), x>0$ and $f(2)=3$. Then $f(6)$ is equal to

A
1
B
6
C
2
D
3
4
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $\left(x y-5 x^2 \sqrt{1+x^2}\right) d x+\left(1+x^2\right) d y=0, y(0)=0$. Then $y(\sqrt{3})$ is equal to

A
$\frac{5 \sqrt{3}}{2}$
B
$\sqrt{\frac{15}{2}}$
C
$\sqrt{\frac{14}{3}}$
D
$2 \sqrt{2}$
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