1
JEE Main 2023 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

For all $$z \in C$$ on the curve $$C_{1}:|z|=4$$, let the locus of the point $$z+\frac{1}{z}$$ be the curve $$\mathrm{C}_{2}$$. Then :

A
the curves $$C_{1}$$ and $$C_{2}$$ intersect at 4 points
B
the curve $$C_{2}$$ lies inside $$C_{1}$$
C
the curve $$C_{1}$$ lies inside $$C_{2}$$
D
the curves $$C_{1}$$ and $$C_{2}$$ intersect at 2 points
2
JEE Main 2023 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The value of $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$ is equal to :

A
$$\frac{10}{3}-\sqrt{3}+\log _{e} \sqrt{3}$$
B
$$\frac{7}{2}-\sqrt{3}-\log _{e} \sqrt{3}$$
C
$$\frac{10}{3}-\sqrt{3}-\log _{e} \sqrt{3}$$
D
$$-2+3\sqrt{3}+\log _{e} \sqrt{3}$$
3
JEE Main 2023 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language

If the maximum distance of normal to the ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1, b < 2$$, from the origin is 1, then the eccentricity of the ellipse is :

A
$$\frac{\sqrt{3}}{4}$$
B
$$\frac{1}{2}$$
C
$$\frac{1}{\sqrt{2}}$$
D
$$\frac{\sqrt{3}}{2}$$
4
JEE Main 2023 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$$. Then, at x = 1,

A
$$2 y^{\prime}+\sqrt{3} \pi^{2} y=0$$
B
$$y^{\prime}+3 \pi^{2} y=0$$
C
$$\sqrt{2} y^{\prime}-3 \pi^{2} y=0$$
D
$$2 y^{\prime}+3 \pi^{2} y=0$$
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