1
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
$$\sqrt{3} y=x$$
2
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
$$-\frac{4}{3}\left(\frac{x-2}{x+1}\right)^{1 / 4}+C$$
3
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
$$\frac{79}{16}$$
4
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

Let $$\mathbf{a}=2 \mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}+2 \mathbf{j}-\mathbf{k}$$ and $$a$$ unit vector $$\mathbf{c}$$ be coplanar. If $$\mathbf{c}$$ is perpendicular to $$\mathbf{a}$$, then c equals to

A
$$\frac{1}{\sqrt{5}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}})$$
B
$$\frac{1}{\sqrt{2}}(-\hat{\mathbf{j}}+\hat{\mathbf{k}})$$
C
$$\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$$
D
$$\frac{1}{\sqrt{3}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$$
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