Question

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

• A. $6$
• B. $8$
• C. $10$
• D. $12$

Answer 1

Answer: (A)

$6$

$(a+{b)}^n+(a-{b)}^n$

$=2[{}^n{C}_0{a}^n{+}^n{C}_2{a}^{n-2}{b}^2+....]$

Take $a=\sqrt{{2x}^2+1};b=\sqrt{{2x}^2-1}$ $\&n=6$

$f\left(x\right)=2[{}^6{C}_0{({2x}^2+1)}^{\frac{6}{2}}+{}^6{C}_2{({2x}^2+1)}^{\frac{4}{2}}.{({2x}^2-1)}^{\frac{2}{2}}+{}^6{C}_4{({2x}^2+1)}^{\frac{2}{2}}.{({2x}^2-1)}^{\frac{4}{2}}+{}^6{C}_4{({2x}^2-1)}^{\frac{6}{2}}]$

Hence, we can see that $f\left(x\right)$ is a polynomial of degree $6.$

Hence, the answer is $6.$