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Model Paper 1 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 13 Marks
Question
Using binomial theorem, expand: $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^6$
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Answer

To find: Expansion of $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^6$ by means of binomial theorem 
Formula used: ${ }^n C_r=\frac{n!}{(n-r)(r)!}$
$(a+b)^{n}={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots \ldots+{ }^n C_{n-1} a b^{n-1}+n C_n b^n$
Now here We have, $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^6$
$\begin{array}{l}=\left[6 c_0\left(\frac{2 x}{3}\right)^{6-0}\right]+\left[6 c_1\left(\frac{2 x}{3}\right)^{6-1}\left(-\frac{3}{2 x}\right)^1\right]+\left[6 c_2\left(\frac{2 x}{3}\right)^{6-2}\left(-\frac{3}{2 x}\right)^2\right] \\ +\left[6 c_3\left(\frac{2 x}{3}\right)^{6-3}\left(-\frac{3}{2 x}\right)^3\right]+\left[6 C _4\left(\frac{2 x}{3}\right)^{6-4}\left(-\frac{3}{2 x}\right)^4\right] \\ +\left[6 c_5\left(\frac{2 x}{3}\right)^{6-5}\left(-\frac{3}{2 x}\right)^5\right]+\left[6 c_6\left(-\frac{3}{2 x}\right)^6\right] \\ =\left[\frac{6!}{01(6-0)!}\left(\frac{2 x}{3}\right)^6\right]-\left[\frac{6!}{11(6-1)!}\left(\frac{2 x}{3}\right)^5\left(\frac{3}{2 x}\right)\right]+\left[\frac{6!}{2!(6-2)!}\left(\frac{2 x}{3}\right)^4\left(\frac{9}{4 x^2}\right)\right]-\left[\frac{6!}{3!(6-3)!}\left(\frac{2 x}{3}\right)^3\left(\frac{27}{8 x^3}\right)\right] \\ +\left[\frac{6!}{4!(6-4)!}\left(\frac{2 x}{3}\right)^2\left(\frac{81}{16 x^4}\right)\right]-\left[\frac{6!}{5!(6-5)!}\left(\frac{2 x}{3}\right)^1\left(\frac{243}{32 x^5}\right)\right]+\left[\frac{6!}{6!(6-6)!}\left(\frac{729}{64 x^6}\right)\right] \\ =\left[1\left(\frac{64 x^5}{729}\right)\right]-\left[6\left(\frac{32 x^5}{243}\right)\left(\frac{3}{2 x}\right)\right]+\left[15\left(\frac{16 x^4}{81}\right)\left(\frac{9}{4 x^2}\right)\right]-\left[20\left(\frac{8 x^3}{27}\right)\right] \\ \left.\left(\frac{27}{8 x^3}\right)\right]+\left[15\left(\frac{4 x^2}{9}\right)\left(\frac{81}{16 x^4}\right)\right]-\left[6\left(\frac{2 x}{3}\right)\left(\frac{243}{32 x^5}\right)\right]+\left[1\left(\frac{729}{64 x^6}\right)\right] \\ =\frac{64}{729} x^6-\frac{32}{27} x^4+\frac{20}{3} x^2-20+\frac{135}{4} \frac{1}{x^2}-\frac{243}{8} \frac{1}{x^4}+\frac{729}{64} \frac{1}{x^6}\end{array}$

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