mathematics-class-11-unit-09-chapter-04-binomial-theorem-l-6-9-xcbppqm-yzu

### <Success>Problem 1</Success>
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<p>Find the numerically greatest term in $(2+3 x)^9, x=3 / 2$</p>
<p><strong>Solution:</strong> ${ }^nC_0 ~2^9+{ }^nC_1 ~2^8 . 3 x+{ }^nC_2 . 2^7 .(3 x)^2+….. {^nC_n}(3 x)^9$</p>
<p>$\frac{T_{r+1}}{T_r}=\frac{{ }^9 C_{r+1}}{{ }^9 C_r} \cdot \frac{3 x}{2}$</p>
<p>$=\frac{9 !}{(r+1) !(8-r) !} \cdot \frac{r !(9-r) !}{9 !} \cdot \frac{3 x}{2}$</p>
<p>$
=\left(\frac{9-r}{r+1}\right) . \frac{3x}{2}<1
$</p>
<p>[$\because x = \frac{3}{2}$]</p>
<p>$
=\left(\frac{9-r}{r+1}\right) . \frac{9}{4}<1
$</p>
<p>What is $r$ s.t. $\frac{9-r}{r+1} \cdot \frac{9}{4}<1$</p>
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<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>$ 81-9 r <4 r+4$</p>
<p>$77  <13 r $</p>
<p>$ r  > 77 / 13$</p>
<p>$r = 6, 7, 8, 9$</p>
<p>$\frac{3}{7} \cdot \frac{9}{4}=\frac{27}{28}$</p>
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<p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/96e1ef2e-c76f-4cda-b51b-25a407a06500/public"></p>
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### <Success>Problem 2</Success>
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<p>Find the term independent of $x$ in:</p>
<p>$(1+x+2 x^3) (\frac{3 x^2}{2}-\frac{1}{3 x})^9$</p>
<p><strong>Solution</strong>
$ x^0 , \frac{1}{x} , \frac{1}{x^3} $</p>
<p>$ { }^n C_r \cdot\left(\frac{3 x^2}{2}\right)^{9-r}\left(\frac{1}{3 x}\right)^r$</p>
<p>$x^{18-2 r-r}=x^{18-3 r}$</p>
<p>$ x^0 \Rightarrow r=6$         <br>
$ x^{-3} \Rightarrow r=7$</p>
<p>$\begin{array}{r}{ }^9 C_6 \cdot\left(\frac{3 x^2}{2}\right)^3\left(-\frac{1}{3 x}\right)^6 + 2 x^3{ }~^9 C_7 \cdot\left(\frac{3 x^2}{2}\right)^2\left(-\frac{1}{3 x}\right)^7\end{array}$</p>
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### <Success>Solution</Success>
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<p>$\begin{aligned} & =\frac{9 \times 8 \times 7}{3 \times 2} \cdot \frac{27 \cdot x^6}{8} \cdot \frac{1}{27 \times 27 \cdot x^6} -2 \times \frac{9 \times 8}{2} x^3 \cdot \frac{9 x^4}{4} \cdot \frac{1}{27 \times 27 \times 3 \times x^7}\end{aligned}$</p>
<p>$=\frac{7}  {18}-\frac{2}  {27}$</p>
<p>$=\frac{21-4}{54}=\frac{17} {54}$</p>
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### <Success>Solution</Success>
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<p>$ \frac{d f}{d x}  =-1+2 x-3 x^2+\cdots+16 x^{15}-17 x^{16} $</p>
<p>$ =a_1+2 a_2(1+x)+3 a_3(1+x)^2+4 a_4(1+x)^3 +\ldots+17 a_{17}(1+x)^{16}$</p>
<p>$\ f’’ =\frac{d^2 f}{d x^2}=  2-6 x+4 \times 3 x^2-5 \times 4 x^3+\cdots+16 \times 15 x^{14}-17 \times 16 x^{15}$</p>
<p>$ =  2 a_2+3 a_3 \times 2(1+x)+4 a_4 \times 3(1+x)^2+\cdots +17 a_{17} \times 16(1+x)^1$</p>
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<p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-09-chapter-06-binomial-theorem-l-6_9-xcbppqm_yzu/img/335-2743.2.jpg"></p>
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