Q.7. The number of elements in the set is :

Q: What is the correct answer for JEE Main Mathematics Q7?

The correct answer is B. Sol. We are given the equation $\frac{36C_{r+1}}{35C_r} = \frac{6}{(k^2-3)}$. Using the identity $\frac{nC_r}{nC_{r-1}} = \frac{n-r+1}{r}$, we can write $\frac{36C_{r+1}}{35C_r} = \frac{36}{r+1} \cdot \frac{35C_r}{35C_r} = \frac{36}{r+1}$. So, $\frac{36}{r+1} = \frac{6}{(k^2-3)}$. This implies $6(k^2-3) = r+1$. $k^2-3 = \frac{r+1}{6}$. Since $k \in Z$, $k^2$ must be a non-negative integer. Also, $

Q: Q.7. The number of elements in the set is :

The correct answer is B. Sol. We are given the equation $\frac{36C_{r+1}}{35C_r} = \frac{6}{(k^2-3)}$. Using the identity $\frac{nC_r}{nC_{r-1}} = \frac{n-r+1}{r}$, we can write $\frac{36C_{r+1}}{35C_r} = \frac{36}{r+1} \cdot \frac{35C_r}{35C_r} = \frac{36}{r+1}$. So, $\frac{36}{r+1} = \frac{6}{(k^2-3)}$. This implies $6(k^2-3) = r+1$. $k^2-3 = \frac{r+1}{6}$. Since $k \in Z$, $k^2$ must be a non-negative integer. Also, $

Question

Q.7. The number of elements in the set  is :

FirstInTest · Accepted Answer

Sol. We are given the equation $\frac{36C_{r+1}}{35C_r} = \frac{6}{(k^2-3)}$.
Using the identity $\frac{nC_r}{nC_{r-1}} = \frac{n-r+1}{r}$, we can write $\frac{36C_{r+1}}{35C_r} = \frac{36}{r+1} \cdot \frac{35C_r}{35C_r} = \frac{36}{r+1}$.
So, $\frac{36}{r+1} = \frac{6}{(k^2-3)}$.
This implies $6(k^2-3) = r+1$.
$k^2-3 = \frac{r+1}{6}$.
Since $k \in Z$, $k^2$ must be a non-negative integer. Also, $k^2-3$ must be an integer.
From the definition of combinations, $r$ must be an integer such that $0 \le r \le 35$.
So, $1 \le r+1 \le 36$.
Therefore, $\frac{1}{6} \le \frac{r+1}{6} \le \frac{36}{6}$.
$0.166... \le k^2-3 \le 6$.
Adding 3 to all parts:
$3.166... \le k^2 \le 9$.
Since $k \in Z$, possible values for $k^2$ are $4$ and $9$.
If $k^2 = 4$, then $k = \pm 2$.
If $k^2 = 9$, then $k = \pm 3$.

Now we find the corresponding values of $r$ for each $k^2$:
If $k^2 = 4$: $k^2-3 = 4-3 = 1$. So, $\frac{r+1}{6} = 1 \Rightarrow r+1 = 6 \Rightarrow r = 5$.
This value $r=5$ is valid ($0 \le 5 \le 35$).
So, the pairs $(r, k)$ are $(5, 2)$ and $(5, -2)$.

If $k^2 = 9$: $k^2-3 = 9-3 = 6$. So, $\frac{r+1}{6} = 6 \Rightarrow r+1 = 36 \Rightarrow r = 35$.
This value $r=35$ is valid ($0 \le 35 \le 35$).
So, the pairs $(r, k)$ are $(35, 3)$ and $(35, -3)$.

The set $S$ contains the elements $\{(5, 2), (5, -2), (35, 3), (35, -3)\}$.
The number of elements in $S$ is 4.

Answer

2

Answer

4

Answer

8

Answer

16

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