JEE Main 2024 — 5 April, Shift 1 — Question 22

$$(1+2x-3x^{3})\left(\frac{3}{2}x^{2}-\frac{1}{3x}\right)^{9}$$

Step 1: General term of the binomial

The general term of $\left(\frac{3}{2}x^{2}-\frac{1}{3x}\right)^{9}$ is $T_{r+1}=\binom{9}{r}\left(\frac{3}{2}x^{2}\right)^{9-r} \left(-\frac{1}{3x}\right)^{r}.$

Simplifying,

$$T_{r+1} =\binom{9}{r}(-1)^r \frac{3^{9-r}}{2^{9-r}}\frac{1}{3^r} x^{18-3r} =\binom{9}{r}(-1)^r \frac{3^{9-2r}}{2^{9-r}}x^{18-3r}.$$

Step 2: Find constant term contributions

(i) From $1$

$$18-3r=0 \Rightarrow r=6.$$

Contribution:

$$\binom{9}{6}(-1)^6\frac{3^{-3}}{2^3} = \frac{84}{216} = \frac{7}{18}.$$

(ii) From $2x$

$$18-3r+1=0 \Rightarrow r=\frac{19}{3} \quad (\text{not an integer, ignored})$$

(iii) From $-3x^3$

$$18-3r+3=0 \Rightarrow r=7.$$

Contribution:

$$-3\binom{9}{7}(-1)^7\frac{3^{-5}}{2^2} = \frac{1}{9}.$$

Step 3: Constant term

$$p=\frac{7}{18}+\frac{1}{9} =\frac{7}{18}+\frac{2}{18} =\frac{1}{2}.$$

Step 4: Final value

$$108p=108\times\frac{1}{2} =\boxed{54}.$$