JEE Main 2023 (Online) 30th January Morning ShiftQuestion 1
If the coefficient of $x^{15}$ in the expansion of $\left(ax^3+\frac{1}{bx^{1/3}}\right)^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $\left(ax^{1/3}-\frac{1}{bx^3}\right)^{15}$, where $a$ and $b$ are positive real numbers, then for each such ordered pair $(a,b)$:
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Find each coefficient from its general term and equate. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$ and ${}^{15}C_9={}^{15}C_6$Step 1 - First expansion. The power of $x$ is $3(15-r)-\frac r3$. Setting it to $15$: $30=\frac{10r}{3}$, so $r=9$; coefficient ${}^{15}C_9\,a^6\,b^{-9}$. Step 2 - Second expansion. The power is $\frac{15-r}{3}-3r$. Setting it to $-15$: $15-10r=-45$, so $r=6$; coefficient ${}^{15}C_6\,a^9\,(-1)^6\,b^{-6}$. Step 3 - Equate. By (1), $$\frac{a^6}{b^9}=\frac{a^9}{b^6}\ \Longrightarrow\ a^3b^3=1.$$ Since $a,b>0$, $ab=1$. Answer: (B)