JEE Main 2018 (Offline)Question 1
The sum of the co-efficient of all odd degree terms in the expansion of $\left(x+\sqrt{x^3-1}\right)^5+\left(x-\sqrt{x^3-1}\right)^5$, $(x>1)$ is
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. In $(x+s)^5+(x-s)^5$ with $s=\sqrt{x^3-1}$ the odd powers of $s$ cancel, so the result is a polynomial in $x$. Formulas used. (1) $(x+s)^5+(x-s)^5=2\left[x^5+10x^3s^2+5xs^4\right]$Step 1 - Substitute $s^2=x^3-1$. $$2\left[x^5+10x^3(x^3-1)+5x(x^3-1)^2\right]$$ $$=2\left[5x^7+10x^6+x^5-10x^4-10x^3+5x\right].$$ Step 2 - Odd-degree terms. $x^7,x^5,x^3,x$ have coefficients $10,\,2,\,-20,\,10$. Step 3 - Sum. $10+2-20+10=2$. Answer: (A)