1
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
$$a^2\left(1+m^2\right)$$
2
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
$$\sqrt{3} y=x$$
3
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
$$-\frac{4}{3}\left(\frac{x-2}{x+1}\right)^{1 / 4}+C$$
4
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

The area of the region bounded by the parabola $$y=x^2+1$$ and lines $$y=x+1, y=0, x=\frac{1}{2}$$ and $$x=2$$ is

A
$$\frac{23}{6}$$
B
$$\frac{23}{16}$$
C
$$\frac{79}{24}$$
D
$$\frac{79}{16}$$
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