English
Hindi
6.Permutation and Combination
hard

જો $\left( {\begin{array}{*{20}{c}}
  {n\, - \,1} \\ 
  r 
\end{array}} \right)\,\, = \,\,\left( {\,{k^2}\, - \,3\,} \right)\,\,\left( {\begin{array}{*{20}{c}}
  n \\ 
  {r\, + \,1} 
\end{array}} \right)\,$  તો $k\, \in \,\,..........$

A

$[ - 2, - \sqrt 3 ]\,\, \cup \,\,[\sqrt 3 ,2]$

B

$( - 2, - \sqrt 3 )\,\, \cup \,\,(\sqrt 3 ,2)$

C

$( - 2, - \sqrt 3 ]\,\, \cup \,\,[\sqrt 3 ,2)$

D

$[ - 2,\sqrt 3 )\,\, \cup \,\,(\sqrt 3 ,2]$

Solution

$\left( {\begin{array}{*{20}{c}}
  {n\, – \,1} \\ 
  r 
\end{array}} \right)\, = \,\left( {\,{k^2}\, – \,3\,} \right)\,\left( {\begin{array}{*{20}{c}}
  n \\ 
  {r\, + \,1} 
\end{array}} \right)$

$\frac{{\left( {\begin{array}{*{20}{c}}
  {n\, – \,1} \\ 
  r 
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  n \\ 
  {r\, + \,1} 
\end{array}} \right)}}\, = \,{k^2}\, – \,3\,$

$\frac{{(n\, – \,1)!}}{{(n\, – \,1\, – \,r)\,!\,r\,!}}\, \times \,\frac{{(n\, – \,r\, – \,1)\,!\,(r\, + \,1)!}}{{n!}}\, = \,{k^2}\, – \,3\,\,$

હવે $0 \leq  r \leq  n – 1  $   $ 1 \leq r + 1 \leq  n  $

$\frac{1}{n}\, \leqslant \,\frac{{r\, + \,1}}{n}\, \leqslant \,1\,\,\,$

$\,0\, < \,\frac{{r\, + \,1}}{n}\, \leqslant \,1$

$3 < {k^2} \leqslant 4\,$

$k \in [ – 2, – \sqrt 3 ) \cup (\sqrt 3 ,2]$

Standard 11
Mathematics

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