JEE Main 2021 (Online) 25th July Morning ShiftQuestion 1
The term independent of '$x$' in the 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 equal to $\_\_\_\_\_\_$ .
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$210$
Solution
Concept. Both fractions cancel with a factorisation, leaving a simple binomial. Formulas used. (1) $t^3+1=(t+1)(t^2-t+1)$ and $x-1=(\sqrt x-1)(\sqrt x+1)$(2) $T_{k+1}={}^{n}C_k\,a^{n-k}b^k$Step 1 - First fraction. With $t=x^{1/3}$, by (1): $\dfrac{x+1}{x^{2/3}-x^{1/3}+1}=t+1=x^{1/3}+1$. Step 2 - Second fraction. Since $x\ne1$: $$\frac{x-1}{x-\sqrt x}=\frac{(\sqrt x-1)(\sqrt x+1)}{\sqrt x(\sqrt x-1)}=1+x^{-1/2}.$$ Step 3 - The binomial. The base is $x^{1/3}-x^{-1/2}$, and $$T_{k+1}={}^{10}C_k\,(-1)^k\,x^{\frac{10-k}{3}-\frac k2}.$$ Step 4 - Power zero. $\frac{10-k}{3}=\frac k2$ gives $k=4$. The term is ${}^{10}C_4=210$. Answer: 210