If $\alpha \,\,{\rm{and}}\,\,\beta $ be the coefficients of ${x^4}\,\,{\rm{and}}\,\,{x^2}$ respectively in the expansion of ${(x + \sqrt {{x^2} - 1} )^6} + {(x - \sqrt {{x^2} - 1} )^6},$ then

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

\alpha \,\,\mathrm{and}\,\,\beta

be the coefficients of

{x^4}\,\,\mathrm{and}\,\,{x^2}

respectively in the expansion of

{(x + \sqrt {{x^2} - 1} )^6} + {(x - \sqrt {{x^2} - 1} )^6},

then

View answer and explanation

Correct option: B

\alpha - \beta = - 132

Option B (

\alpha - \beta = - 132

) is correct because it is the result of calculating the coefficients of x⁴ and x² in the given expansion and then subtracting them.

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