Q. The expression $\{ x + (x^{2} - 1)^{1/2}\}^{5} + \{ x - (x^{2} - {1)}^{1/2}\}^{5}$ is a polynomial of degree

A) 5
B) 6
C) 7
D) 8
Explanation

(c)
We know that,
${(x + a)}^{n} + {(x - a)}^{n} = 2\lbrack\ ^{n}C_{0}x^{n} + \ ^{n}C_{2}x^{n - 2}a^{2} +$….]
Here, $n = 5,\ x = x\ and\ a = (x^{3} - 1)^{1/2}$
$\therefore\lbrack x + (x^{3} - {1)}^{1/2}\rbrack^{5}\lbrack x - (x^{3} - 1)^{1/2}\rbrack^{5}$
$= 2\lbrack\ ^{5}C_{0}x^{5} + \ ^{5}C_{2}x^{3}\left( x^{3} - 1 \right) + \ ^{5}C_{4}x(x^{3} - 1)^{2}\rbrack$
$= 2\lbrack x^{5} + 10x^{3}\left( x^{3} - 1 \right) + 5\ x\left( x^{3} - 1)^{2} \right\rbrack$
$\therefore$ Given expression is a polynomial of degree 7.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.