Question

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

• A. $\alpha +\beta =-30$
• B. $\alpha -\beta =-132$
• C. $\alpha +\beta =60$
• D. $\alpha -\beta =-30$

Answer 1

The correct option is B

$\alpha -\beta =-132$

Explanation for correct option:

Finding the correct relation:

By binomial expansion of ${\left(a+b\right)}^n+{\left(a-b\right)}^n$

$$\begin{gathered} {\left(x+\sqrt{x^2-1}\right)}^6+{\left(x-\sqrt{x^2-1}\right)}^6 \\ =2\left[{}^{6}C_0{x}^6+{}^{6}C_2{x}^4{\left(\sqrt{x^2-1}\right)}^2+{}^{6}C_4{x}^2{\left(\sqrt{x^2-1}\right)}^4+{}^{6}C_6{\left(\sqrt{x^2-1}\right)}^6\right] \\ =2\left[x^6+15x^4\left(x^2-1\right)+15x^2{\left(x^2-1\right)}^2+{\left(x^2-1\right)}^3\right] \\ =2\left[x^6+15x^6-15x^4+15x^6-30x^4+15x^2+x^6-3x^4+3x^2-1\right] \\ =2\left[32x^6-48x^4+18x^2-1\right] \\ =64x^6-96x^4+36x^2-2 \end{gathered}$$

Since $\alpha$and $\beta$are the coefficients of $x^4$and$x^2$, Thus

$\alpha =-96$ and $\beta =36$

$\Rightarrow \alpha -\beta =-132$

Hence, the correct option is B.