1
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

If $$\left(1+x^2\right) d y+2 x y d x=\cot x d x$$, then the general solution be

A
$$y=\frac{\log |\sin x|}{1+x^2}+\frac{C}{1+x^2}$$
B
$$y=\frac{\log |\sin x|}{1-x^2}+\frac{C}{1-x^2}$$
C
$$y=\frac{\log |\cos x|}{1+x^2}+\frac{C}{1+x^2}$$
D
$$y=\frac{\log |\cos x|}{1-x^2}+\frac{C}{1-x^2}$$
2
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

If the straight line $$y=m x+c$$, touches the circle $$x^2+y^2=a^2$$ at a point, then $$c^2$$ is

A
$$m^2\left(1-a^2\right)$$
B
$$m^2\left(1+a^2\right)$$
C
$$a^2\left(1-m^2\right)$$
D
$$a^2\left(1+m^2\right)$$
3
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

A normal is drawn at the point $$P$$ to the circle $$x^2+y^2=25$$, which is inclined at $$45^{\circ}$$ with the straight line $$y=6$$. Then, the point lies on the straight line

A
$$y=x$$
B
$$y=-x$$
C
$$y=\sqrt{3} x$$
D
$$\sqrt{3} y=x$$
4
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

The value of integral $$\int \frac{d x}{(1+x)^{3 / 4}(x-2)^{5 / 4}}$$ is is equal to

A
$$-\frac{3}{4}\left(\frac{x+1}{x-2}\right)^{1 / 4}+C$$
B
$$-\frac{3}{4}\left(\frac{x-2}{x+1}\right)^{1 / 4}+C$$
C
$$-\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{1 / 4}+C$$
D
$$-\frac{4}{3}\left(\frac{x-2}{x+1}\right)^{1 / 4}+C$$
EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12