Question

The expression $(\sqrt{2x^2+1}+\sqrt{2x^2-1}{)}^6+{\left(\frac{2}{\sqrt{2x^2+1}+\sqrt{2x^2-1}}\right)}^6$ is a polynomial of degree

• A. $6$
• B. $8$
• C. $10$
• D. $12$

Answer 1

The correct option is A $6$

We have,
$\left(\frac{2}{\sqrt{2x^2+1}+\sqrt{2x^2-1}}\right)\Rightarrow \left(\frac{2(\sqrt{2x^2+1}-\sqrt{2x^2-1})}{(2x^2+1)-(2x^2-1)}\right)\Rightarrow (\sqrt{2x^2+1}-\sqrt{2x^2-1})$

So, the given expression can be written as
$(\sqrt{2x^2+1}+\sqrt{2x^2-1}{)}^6+(\sqrt{2x^2+1}-\sqrt{2x^2-1}{)}^6$

We know that
$(a+b{)}^6+(a-b{)}^6=2[a^6+{}^6{C}_2{a}^4{b}^2+{}^6{C}_4{a}^2{b}^4+b^6]=2[a^6+15a^4{b}^2+15a^2{b}^4+b^6]$

Putting $a=\sqrt{2x^2+1},b=\sqrt{2x^2-1}$
$(\sqrt{2x^2+1}+\sqrt{2x^2-1}{)}^6+(\sqrt{2x^2+1}-\sqrt{2x^2-1}{)}^6$
$=2\left[\right(2x^2+1{)}^3+15(2x^2+1{)}^2(2x^2-1)+15(2x^2+1\left)\right(2x^2-1{)}^2+(2x^2-1{)}^3]$

So, the degree of the polynomial is $6$.