Assuming x to be so small that x3 and higher powers of x can be neglected, then value ofE=1−32×5(2+3x)6, is

# Assuming x to be so small that ${x}^{3}$ and higher powers of x can be neglected, then value of$E={\left(1-\frac{3}{2}x\right)}^{5}\left(2+3x{\right)}^{6}$, is

1. A

$64+96x+720{x}^{2}$

2. B

$65+97x+721{x}^{2}$

3. C

$64-96x+720{x}^{2}$

4. D

$64+96x-720{x}^{2}$

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### Solution:

We have

$\begin{array}{l}{\left(1-\frac{3}{2}x\right)}^{5}\left(2+3x{\right)}^{6}={2}^{6}{\left(1-\frac{3}{2}x\right)}^{5}{\left(1+\frac{3}{2}x\right)}^{6}\\ ={2}^{6}\left(1+\frac{3}{2}x\right){\left[\left(1-\frac{3}{2}x\right)\left(1+\frac{3}{2}x\right)\right]}^{5}\\ ={2}^{6}\left(1+\frac{3}{2}x\right){\left(1-\frac{9}{4}{x}^{2}\right)}^{5}\\ ={2}^{6}\left(1+\frac{3}{2}x\right)\left(1-\frac{45}{4}{x}^{2}\right)\end{array}$

$={2}^{6}\left(1+\frac{3}{2}x-\frac{45}{4}{x}^{2}\right)=64+96x-720{x}^{2}$

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