1
JEE Main 2024 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the system of equations

$$\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}$$

has a non-trivial solution, then $$\alpha \in\left(0, \frac{\pi}{2}\right)$$ is equal to :

A
$$\frac{5 \pi}{24}$$
B
$$\frac{11 \pi}{24}$$
C
$$\frac{7 \pi}{24}$$
D
$$\frac{3 \pi}{4}$$
2
JEE Main 2024 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\alpha$$ and $$\beta$$ be the sum and the product of all the non-zero solutions of the equation $$(\bar{z})^2+|z|=0, z \in C$$. Then $$4(\alpha^2+\beta^2)$$ is equal to :

A
4
B
2
C
6
D
8
3
JEE Main 2024 (Online) 4th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f: \mathbf{R} \rightarrow \mathbf{R}$$ be a function given by

$$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x < 0 \\ \alpha, & x=0, \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0\end{cases}$$

where $$\alpha, \beta \in \mathbf{R}$$. If $$f$$ is continuous at $$x=0$$, then $$\alpha^2+\beta^2$$ is equal to :

A
48
B
6
C
3
D
12
4
JEE Main 2024 (Online) 4th April Morning Shift
Numerical
+4
-1
Change Language

Let the solution $$y=y(x)$$ of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}-y=1+4 \sin x$$ satisfy $$y(\pi)=1$$. Then $$y\left(\frac{\pi}{2}\right)+10$$ is equal to __________.

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