Question

If \[{}^{n - 1}{C_r} = \left( {{k^2} - 3} \right) \cdot {}^n{C_{r + 1}}\], then find the value of the interval where \[k\] lies.
A. \[\left[ { - \sqrt 3 ,\sqrt 3 } \right]\]
B. \[\left( { - \infty , - 2} \right)\]
C. \[\left( {2,\infty } \right)\]
D. \[\left( {\sqrt 3 ,2} \right]\]

Answer 1

Hint: Here, an equation of the combination term is given. First, simplify the equation by using the combination formula. Then, check the condition required for the \[r\] in the combination formula. After that, solve that inequality condition and find the required answer.



Formula Used:The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
The condition required for the \[r\] In \[{}^n{C_r}\]: \[0 < r \le n - 1\]



Complete step by step solution:The given equation is \[{}^{n - 1}{C_r} = \left( {{k^2} - 3} \right) \cdot {}^n{C_{r + 1}}\].
Let’s simplify the above equation.
\[\left( {{k^2} - 3} \right) = \dfrac{{{}^{n - 1}{C_r}}}{{{}^n{C_{r + 1}}}}\]
 Simplify the right-hand side by applying the formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
\[{k^2} - 3 = \dfrac{{\dfrac{{\left( {n - 1} \right)!}}{{r!\left( {n - 1 - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}}}\]
Cancel out the common terms and simplify the terms using factorial property \[n! = n\left( {n - 1} \right)!\].
\[ \Rightarrow {k^2} - 3 = \dfrac{{\dfrac{{\left( {n - 1} \right)!}}{{r!}}}}{{\dfrac{{n\left( {n - 1} \right)!}}{{\left( {r + 1} \right)r!}}}}\]
Again, cancel out the common terms.
\[ \Rightarrow {k^2} - 3 = \dfrac{1}{{\dfrac{n}{{\left( {r + 1} \right)}}}}\]
\[ \Rightarrow {k^2} - 3 = \dfrac{{r + 1}}{n}\] \[.....\left( 1 \right)\]

We know that the condition required for the \[r\] In \[{}^n{C_r}\] is \[0 < r \le n - 1\].
Solve this condition.
Add 1 on each side.
\[1 < r + 1 \le n\]
Divide each term by \[n\].
\[ \Rightarrow \dfrac{1}{n} < \dfrac{{r + 1}}{n} \le \dfrac{n}{n}\]
\[ \Rightarrow \dfrac{1}{n} < \dfrac{{r + 1}}{n} \le 1\]
Substitute the equation \[\left( 1 \right)\] in the above inequality.
\[ \Rightarrow \dfrac{1}{n} < {k^2} - 3 \le 1\]
Add 3 on each side.
\[ \Rightarrow \dfrac{1}{n} + 3 < {k^2} - 3 + 3 \le 1 + 3\]
\[ \Rightarrow \dfrac{1}{n} + 3 < {k^2} \le 4\]
Take square root of each term.
\[ \Rightarrow \sqrt {\dfrac{1}{n} + 3} < \sqrt {{k^2}} \le \sqrt 4 \]
\[ \Rightarrow \sqrt {\dfrac{1}{n} + 3} < k \le 2\]
As \[n \to \infty \], \[\dfrac{1}{n} \to 0\] .
\[ \Rightarrow \sqrt 3 < k \le 2\]
\[ \Rightarrow k \in \left( {\sqrt 3 ,2} \right]\]



Option ‘D’ is correct



Note: Students often get confused about the inequality condition of \[r\] in \[{}^n{C_r}\]. They, consider it as \[0 \le r \le n - 1\] or \[0 \le r \le n\]. Both are the incorrect equations. Because of that, they get incorrect solution.