JEE Main 2024 (Online) 5th April Morning ShiftQuestion 1
If the constant term in the expansion of $\left(1+2x-3x^{3}\right)\left(\dfrac{3}{2}x^{2}-\dfrac{1}{3x}\right)^{9}$ is $p$, then $108p$ is equal to $\_\_\_\_\_\_\_\_$ .
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$54$
Solution
Concept. Each term of $1+2x-3x^3$ needs a matching power from the binomial. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - General term of the binomial. $$T_{r+1}={}^{9}C_r\left(\frac32\right)^{9-r}\left(-\frac13\right)^r x^{18-3r}.$$ The powers $18-3r$ are multiples of $3$. Step 2 - From $1$: need $x^0$. $r=6$: $${}^{9}C_6\left(\frac32\right)^3\frac{1}{3^6}=84\cdot\frac{27}{8}\cdot\frac{1}{729}=\frac{7}{18}.$$ Step 3 - From $2x$: need $x^{-1}$. $18-3r=-1$ has no integer solution, so this gives $0$. Step 4 - From $-3x^3$: need $x^{-3}$. $r=7$: $$-3\cdot{}^{9}C_7\left(\frac32\right)^2\left(-\frac{1}{3^7}\right)=3\cdot36\cdot\frac94\cdot\frac{1}{2187}=\frac19.$$ Step 5 - Value. $p=\frac7{18}+\frac2{18}=\frac12$, so $108p=54$. Answer: 54