JEE Main 2017 (Online) 9th AprilQuestion 1
The coefficient of $x^{-5}$ in the binomial expansion of $\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)^{10}$, where $x\neq0,1$ is
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Both fractions cancel with a factorisation, leaving $x^{1/3}-x^{-1/2}$. Formulas used. (1) $t^3+1=(t+1)(t^2-t+1)$ and $x-1=(\sqrt x-1)(\sqrt x+1)$(2) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - Simplify. With $t=x^{1/3}$, $\dfrac{x+1}{x^{2/3}-x^{1/3}+1}=t+1$. Also $\dfrac{x-1}{x-\sqrt x}=\dfrac{\sqrt x+1}{\sqrt x}=1+x^{-1/2}$. The base is $$x^{1/3}-x^{-1/2}.$$ Step 2 - General term. $$T_{r+1}={}^{10}C_r(-1)^r\,x^{\frac{10-r}{3}-\frac r2}.$$ Step 3 - Power $-5$. $\frac{10-r}{3}-\frac r2=-5$ gives $20-5r=-30$, $r=10$. Step 4 - Coefficient. $(-1)^{10}\,{}^{10}C_{10}=1$. Answer: (C)