IIT-JEE 1992Question 1
The expansion $\left(x+\left(x^{3}-1\right)^{\frac12}\right)^{5}+\left(x-\left(x^{3}-1\right)^{\frac12}\right)^{5}$ is a polynomial of degree
Binomial Theorem · Miscellaneous
Solution
Concept. The odd powers of $s=\sqrt{x^3-1}$ cancel, so the result is a polynomial; find its highest power. Formulas used. (1) $(x+s)^5+(x-s)^5=2\left[x^5+10x^3s^2+5xs^4\right]$Step 1 - Substitute $s^2=x^3-1$. $$2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right].$$ Step 2 - Highest powers. $10x^3(x^3-1)$ gives $x^6$ at most; $5x(x^3-1)^2$ gives $5x^7$. So the polynomial starts $10x^7+\cdots$ Step 3 - Degree. $7$. Answer: (C)