If ${\;^{n - 1}}{C_r} = \left( {{k^2} - 3} \right).{\;^n}{C_{r + 1}}$, then $k$ is belongs to

$\left( { - \infty , - 2} \right]$

$\left[ { - \sqrt 3 ,\sqrt 3 } \right]$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Application +4 | -1

{\;^{n - 1}}{C_r} = \left( {{k^2} - 3} \right).{\;^n}{C_{r + 1}}

, then

k

is belongs to

View answer and explanation

Correct option: D

\left( {\sqrt 3 ,2} \right]

Option D (

\left( {\sqrt 3 ,2} \right]

) is correct because it represents the valid range for k based on the given condition and binomial coefficient properties.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n