1
JEE Advanced 2024 Paper 1 Online
Numerical
+4
-0
Change Language

Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in \mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is _____________.

Your input ____
2
JEE Advanced 2024 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Change Language
Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=\left(a_{i j}\right)_{3 \times 3}$, define $R_i=a_{i 1}+a_{i 2}+a_{i 3}$ and $C_j=a_{1 j}+a_{2 j}+a_{3 j}$ for $i=1,2,3$ and $j=1,2,3$.

Match each entry in List-I to the correct entry in List-II.

List-I List-II
(P) The number of matrices $ M = (a_{ij})_{3x3} $ with all entries in $ T $ such that $ R_i = C_j = 0 $ for all $ i, j $, is (1) 1
(Q) The number of symmetric matrices $ M = (a_{ij})_{3x3} $ with all entries in $ T $ such that $ C_j = 0 $ for all $ j $, is (2) 12
(R) Let $ M = (a_{ij})_{3x3} $ be a skew symmetric matrix such that $ a_{ij} \in T $ for $ i > j $.

Then the number of elements in the set

$ \left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in \mathbb{R}, M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ a_{13} \end{pmatrix} \right\} $ is
(3) infinite
(S) Let $ M = (a_{ij})_{3x3} $ be a matrix with all entries in $ T $ such that $ R_i = 0 $ for all $ i $. Then the absolute value of the determinant of $ M $ is (4) 6

The correct option is
A
(P) $\rightarrow$ (4) $\quad$ (Q) $\rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad$ (S) $\rightarrow$ (1)
B
$(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)$
C
$(\mathrm{P}) \rightarrow(2) \quad$ (Q) $\rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad$ (S) $\rightarrow$ (5)
D
(P) $\rightarrow$ (1) $\quad$ (Q) $\rightarrow$ (5) $\quad$ (R) $\rightarrow$ (3) $\quad$ (S) $\rightarrow$ (4)
3
JEE Advanced 2024 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

Let the straight line $y=2 x$ touch a circle with center $(0, \alpha), \alpha>0$, and radius $r$ at a point $A_1$. Let $B_1$ be the point on the circle such that the line segment $A_1 B_1$ is a diameter of the circle. Let $\alpha+r=5+\sqrt{5}$.

Match each entry in List-I to the correct entry in List-II.

List-I List-II
(P) $\alpha$ equals (1) $(-2, 4)$
(Q) $r$ equals (2) $\sqrt{5}$
(R) $A_1$ equals (3) $(-2, 6)$
(S) $B_1$ equals (4) $5$
(5) $(2, 4)$

The correct option is
A
$(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$
B
$(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(3)$
C
$(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(3)$
D
$(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)$
4
JEE Advanced 2024 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

Let $\gamma \in \mathbb{R}$ be such that the lines $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O=(0,0,0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$.

Match each entry in List-I to the correct entry in List-II.

List-I List-II
(P) $\gamma$ equals (1) $-\hat{i} - \hat{j} + \hat{k}$
(Q) A possible choice for $\hat{n}$ is (2) $\sqrt{\frac{3}{2}}$
(R) $\overrightarrow{OR_1}$ equals (3) $1$
(S) A possible value of $\overrightarrow{OR_1} \cdot \hat{n}$ is (4) $\frac{1}{\sqrt{6}} \hat{i} - \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}$
(5) $\sqrt{\frac{2}{3}}$

The correct option is :
A
$(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
B
$(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)$
C
$(\mathrm{P}) \rightarrow(3) \quad$ (Q) $\rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad$ (S) $\rightarrow(5)$
D
$(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(1) \quad(\mathrm{R}) \rightarrow(4) \quad$ (S) $\rightarrow(5)$
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