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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Binomial Theorem

If the constant term in the expansion of (1+2 x-3 x^3) ((3)/(2)...

If the constant term in the expansion of (1+2x−3x3)\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9isis\mathrm{p},then, then108 \mathrm{p} is equal to

Correct answer

0

Step-by-step solution

(1+2x−3x3)(32x2−13x)9 General term m(32x2−13x)9=9Cr⋅39−2r29−r(−1)r⋅x18−3r\begin{aligned} & \left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\ & \text{ General term } m\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\ & ={ }^9 C_r \cdot \frac{3^{9-2 r}}{2^{9-r}}(-1)^r \cdot x^{18-3 r} \end{aligned} Put r=6r=6 to get coeff. of x0=9C6⋅163⋅x0=718x0x^0={ }^9 C_6 \cdot \frac{1}{6^3} \cdot x^0=\frac{7}{18} x^0 Put r=7\mathrm{r}=7 to get coeff. of x−3=9Cr⋅3−522(−1)7⋅x−3\mathrm{x}^{-3}={ }^9 \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{-5}}{2^2}(-1)^7 \cdot \mathrm{x}^{-3}=−9C7⋅135⋅22⋅x−3=−127x−3(1+2x−3x3)(718x0−127x−3)718+327=718+19=7+218=918=12∴108⋅12=54\begin{aligned} & =-{ }^9 C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3}=\frac{-1}{27} x^{-3} \\ & \left(1+2 x-3 x^3\right)\left(\frac{7}{18} x^0-\frac{1}{27} x^{-3}\right) \\ & \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2} \\ & \therefore 108 \cdot \frac{1}{2}=54 \end{aligned}$$
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