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

Answer: (A)

$6$

$(\sqrt{2x^2+1}+\sqrt{2x^2+1}{)}^6+(\frac{2}{\sqrt{2x^2+1}+\sqrt{2x^2+1}}{)}^6$
$=(\sqrt{2x^2+1}+\sqrt{2x^2+1}{)}^6+(\frac{2(\sqrt{2x^2+1}-\sqrt{2x^2+1})}{2x^2+1-2x^2+1}{)}^6$
$=(\sqrt{2x^2+1}+\sqrt{2x^2+1}{)}^6+(\sqrt{2x^2+1}-\sqrt{2x^2+1}{)}^6$
Hence, upon expansion we get a polynomial of degree 6.