1
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying

$$\left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right)$$

lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :
A
$\frac{11}{5}$
B
11
C
$-1$
D
$\frac{5}{4}$
2
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Let $\mathrm{H}$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is :
A
$\frac{5}{2}$
B
3
C
2
D
$\frac{3}{2}$
3
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
$$ \lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3 $$
A
is equal to 9
B
is equal to $\frac{27}{2}$
C
does not exist
D
is equal to 27
4
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Let $y=y(x)$ be the solution of the differential equation

$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$

such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to :
A
64
B
$16 \sqrt{2}$
C
32
D
$32 \sqrt{2}$
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