Verified Solution • 2025–26 Session moderate QJEEM24Ma426c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma426c Solution: If the constant term in the expansion of is , then is equal to… If the constant term in the expansion of (1+2x−3x3)(32x2−13x)9\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9(1+2x−3x3)(23x2−3x1)9 is p\mathrm{p}p , then 108p108 \mathrm{p}108p is equal to ________. Step-by-Step Verified Solution General term of (32x2−13x)9\text { General term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 General term of (23x2−3x1)9 Tr+1=9Cr(32x2)9−r(−13x)r=9Cr(−1)r39−2r2r−9x18−35T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}Tr+1=9Cr(23x2)9−r(−3x1)r=9Cr(−1)r39−2r2r−9x18−35 Constant term in expansion of (1+2x−3x3)\left(1+2 x-3 x^3\right)(1+2x−3x3) amp;(32x2−13x)9amp;=T7−3T8=9C63−3⋅2−3+39C7⋅3−5⋅2−2amp;=3×4×733⋅23+3×9×435×22=amp;p=42+12108=54108amp;108p=54\begin{aligned}& \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\& =T_7-3 T_8={ }^9 C_6 3^{-3} \cdot 2^{-3}+3{ }^9 C_7 \cdot 3^{-5} \cdot 2^{-2} \\& =\frac{3 \times 4 \times 7}{3^3 \cdot 2^3}+\frac{3 \times 9 \times 4}{3^5 \times 2^2}= \\& p=\frac{42+12}{108}=\frac{54}{108} \\& 108 p=54\end{aligned}amp;(23x2−3x1)9amp;=T7−3T8=9C63−3⋅2−3+39C7⋅3−5⋅2−2amp;=33⋅233×4×7+35×223×9×4=amp;p=10842+12=10854amp;108p=54 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring