jee-main 2025 Q70

jee-main · India · session2_04apr_shift1 Probability Definitions Combinatorial Counting (Non-Probability)
Q70. Let the relations $R _ { 1 }$ and $R _ { 2 }$ on the set $X = \{ 1,2,3 , \ldots , 20 \}$ be given by $R _ { 1 } = \{ ( x , y ) : 2 x - 3 y = 2 \}$ and $R _ { 2 } = \{ ( x , y ) : - 5 x + 4 y = 0 \}$. If $M$ and $N$ be the minimum number of elements required to be added in $R _ { 1 }$ and $R _ { 2 }$, respectively, in order to make the relations symmetric, then $M + N$ equals
(1) 12
(2) 16
(3) 8
(4) 10
Q71. For $\alpha , \beta \in \mathbb { R }$ and a natural number $n$, let $A _ { r } = \left| \begin{array} { c c c } r & 1 & \frac { n ^ { 2 } } { 2 } + \alpha \\ 2 r & 2 & n ^ { 2 } - \beta \\ 3 r - 2 & 3 & \frac { n ( 3 n - 1 ) } { 2 } \end{array} \right|$. Then
(1) 0
(2) $4 \alpha + 2 \beta$
(3) $2 \alpha + 4 \beta$
(4) $2 n$
Q70. Let the relations $R _ { 1 }$ and $R _ { 2 }$ on the set $X = \{ 1,2,3 , \ldots , 20 \}$ be given by $R _ { 1 } = \{ ( x , y ) : 2 x - 3 y = 2 \}$ and $R _ { 2 } = \{ ( x , y ) : - 5 x + 4 y = 0 \}$. If $M$ and $N$ be the minimum number of elements required to be added in $R _ { 1 }$ and $R _ { 2 }$, respectively, in order to make the relations symmetric, then $M + N$ equals\\
(1) 12\\
(2) 16\\
(3) 8\\
(4) 10

Q71.\\
For $\alpha , \beta \in \mathbb { R }$ and a natural number $n$, let $A _ { r } = \left| \begin{array} { c c c } r & 1 & \frac { n ^ { 2 } } { 2 } + \alpha \\ 2 r & 2 & n ^ { 2 } - \beta \\ 3 r - 2 & 3 & \frac { n ( 3 n - 1 ) } { 2 } \end{array} \right|$. Then\\
(1) 0\\
(2) $4 \alpha + 2 \beta$\\
(3) $2 \alpha + 4 \beta$\\
(4) $2 n$