1
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

The general solution of the linear differential equation $$\frac{d y}{d x}+\sec x \cdot y=\tan x\left(0 \leq x \leq \frac{\pi}{2}\right)$$ is

A
$$y=-x(\sec x+\tan x)^{-1}+\frac{c}{\sec x+\tan x}+1$$
B
$$y=x+\frac{C}{\sec x+\tan x}+\frac{1}{\tan x}$$
C
$$y=\frac{x+1}{\sec x+\tan x}+C$$
D
$$y=x+\sec x+\tan x+C$$
2
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

On solving the differential equation $$x^2 y d x-\left(x^3+y^3\right) d y=0$$, the value of $$\log y$$ is

A
$$\frac{x^3}{3 y^3}+C$$
B
$$\frac{x^2}{y^2}+C$$
C
$$\frac{x^2}{3 y^3}+C$$
D
$$\frac{x^3}{x^3+y^3}+C$$
3
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

The particular solution of the differential equation $$\frac{d y}{d x}+y \cot x=2 x+x^2 \cot x$$, such that $$y(\pi / 2)=0$$ is

A
$$y=\frac{\pi^2}{4 \cos x},(x \neq 0)$$
B
$$y=x^2-\frac{\pi}{2} \tan x$$
C
$$y=\frac{2 x}{\sin x}+\frac{1}{x^2},(x \neq 0)$$
D
$$y=x^2-\frac{\pi^2}{4 \sin x}(\sin x \neq 0)$$
4
VITEEE 2021
MCQ (Single Correct Answer)
+1
-0

The area bounded by the circle $$x^2+y^2=16$$ and the line $$y=x$$ in the first quadrant is

A
$$4 \pi$$ sq units
B
$$8 \pi$$ sq units
C
$$2 \pi$$ sq units
D
$$\pi$$ sq unit
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