JEE Main 2023 (Online) 1st February Morning Shift | Differential Equations Question 44 | Mathematics

If $$y=y(x)$$ is the solution curve of the differential equation

$$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to

$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)$$

$$\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)$$

$$\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$

$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$

JEE Main 2023 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

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

$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$

such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to :

$16 \sqrt{2}$

$32 \sqrt{2}$

JEE Main 2023 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

Let a differentiable function $$f$$ satisfy $$f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$$. Then $$12 f(8)$$ is equal to :