Question

The degree of polynomial ${(x^2+\sqrt{x^5-1})}^{2021}+{(x^2-\sqrt{x^5-1})}^{2021}$ is

• A. $4052$
• B. $5052$
• C. $6052$
• D. $5250$

Answer 1

The correct option is B $5052$

Given polynomial is ${(x^2+\sqrt{x^5-1})}^{2021}+{(x^2-\sqrt{x^5-1})}^{2021}$
We know that,
$(a+y{)}^n+(a-y{)}^n=2({}^n{C}_0{a}^n+{}^n{C}_2{a}^{n-2}{y}^2+{}^n{C}_4{a}^{n-4}{y}^4+\cdots )$
General term is
$T_{r+1}=2{}^n{C}_{2r}\cdot a^{n-2r}\cdot y^{2r}$

Now, $a=x^2,y=\sqrt{x^5-1},n=2021$
So, $T_{r+1}=2{}^{2021}{C}_{2r}\cdot {\left(x^2\right)}^{2021-2r}\cdot {\left(\sqrt{x^5-1}\right)}^{2r}$
$\Rightarrow T_{r+1}=2{}^{2021}{C}_{2r}\cdot (x{)}^{4042-4r}\cdot {(x^5-1)}^r$
where $2r\le 2021\Rightarrow r\le 1010$

For ${(x^5-1)}^r,$ the maximum degree of $x$ is $5r.$
So, the degree of polynomial
$=4042-4r+5r=4042+r$
The maximum degree of the polynomial $=4042+1010=5052$