Miscellaneous Exercise (Revised) - Chapter 8 - Binomial Theorem - Ncert Solutions class 11 - Maths

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Miscellaneous Exercise Question 1.

If $a$ and $b$ are distinct integers, prove that $a-b$ is a factor of $a^n-b^n$, whenever $n$ is a positive integer.

Answer:

$
\begin{aligned}
& \text {We have, } a^n=[(a-b)+b]^n \\
& \Rightarrow a^n={ }^n \mathrm{C}_0(a-b)^n+{ }^n \mathrm{C}_1(a-b)^{n-1} \cdot b+{ }^n \mathrm{C}_2(a-b)^{n-2} \cdot b^2+\ldots \ldots \ldots+{ }^n \mathrm{C}_n b^n \\
& \therefore \\
& a^n-b^n={ }^n C_0(a-b)^n+{ }^n C_1(a-b)^{n-1} b+\ldots \ldots \ldots \ldots \ldots+{ }^n C_{n-1}(a-b) b^{n-1}+b^n-b^n
\end{aligned}
$

$\Rightarrow a^n-b^n=(a-b)^n+{ }^n \mathrm{C}_1(a-b)^{n-1} \cdot b+{ }^n \mathrm{C}_2(a-b)^{n-2} \cdot b^2+\ldots \ldots \ldots .+{ }^n \mathrm{C}_{n-1}(a-b) b^{n-1}$

$a^n-b^n=(a-b)\left[(a-b)^{n-1}+{ }^n C_1(a-b)^{n-2} b+{ }^n C_2(a-b)^{n-3} b^2+\ldots \ldots .+{ }^n C_{n-1} b^{n-1}\right]$

$\Rightarrow a^n-b^n=k(a-b)$,where
$k=\left[(a-b)^{n-1}+{ }^n C_1(a-b)^{n-2} b+\ldots \ldots \ldots+{ }^n C_{n-1} b^{n-1}\right]$ is an integer
which shows that $(a-b)$ is a factor of $a^n-b^n$.

Miscellaneous Exercise 2.

Evaluate: $(\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6$

Answer.

First we will consider $(a+b)^6-(a-b)^6$

Using Binomial Theorem, we have
$
\begin{aligned}
& (a+b)^6-(a-b)^6= \\
& {\left[{ }^6 C_0 a^6+{ }^6 C_1 a^5 b+{ }^6 C_2 a^4 b^2+{ }^6 C_3 a^3 b^3+{ }^6 C_4 a^2 b^4+{ }^6 C_5 a b^5+{ }^6 C_6 b^6\right]} \\
& -\left[{ }^6 C_0 a^6-{ }^6 C_1 a^5 b+{ }^6 C_2 a^4 b^2-{ }^6 C_3 a^3 b^3+{ }^6 C_4 a^2 b^4-{ }^6 C_5 a b^5+{ }^6 C_6 b^6\right] \\
& =\left[a^6+6 a^5 b+15 a^4 b^2+20 a^3 b^3+15 a^2 b^4+6 a b^5+b^6\right] \\
& -\left[a^6-6 a^5 b+15 a^4 b^2-20 a^3 b^3+15 a^2 b^4-6 a b^5+b^6\right] \\
& =12 a^5 b+40 a^3 b^3+12 a b^5=4 a b\left(3 a^4+10 a^2 b^2+3 b^4\right) \\
& \therefore(a+b)^6-(a-b)^6=4 a b\left(3 a^4+10 a^2 b^2+3 b^4\right) \\
&
\end{aligned}
$

Putting $a=\sqrt{3}$ and $b=\sqrt{2}$ in equation(i),we get
$
\begin{aligned}
& (\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6=4 \sqrt{3} \sqrt{2}\left[3(\sqrt{3})^4+10(\sqrt{3})^2(\sqrt{2})^2+3(\sqrt{2})^4\right] \\
& =4 \sqrt{6}(27+60+12)=4 \sqrt{6} \times 99=396 \sqrt{6} \\
&
\end{aligned}
$

Miscellaneous Exercise 3

Find the value of $\left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4$.

Answer.

Putting $a^2=x$ and $\sqrt{a^2-1}=y$, we have
$
\begin{aligned}
& \left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4=(x+y)^4+(x-y)^4 \\
& =\left[{ }^4 C_0 x^4+{ }^4 C_1 x^3 y+{ }^4 C_2 x^2 y^2+{ }^4 C_3 x^1 y^3+{ }^4 C_4 y^4\right]+\left[{ }^4 C_0 x^4-{ }^4 C_1 x^3 y+{ }^4 C_2 x^2 y^2-\right. \\
& =2\left[{ }^4 C_0 x^4+{ }^4 C_2 x^2 y^2+{ }^4 C_4 y^4\right]=2\left[x^4+6 x^2 y^2+y^4\right] \\
& =2\left[\left(a^2\right)^4+6\left(a^2\right)^2\left(\sqrt{a^2-1}\right)^2+\left(\sqrt{a^2-1}\right)^4\right] \\
& =2\left[a^8+6 a^4\left(a^2-1\right)+\left(a^2-1\right)^2\right] \\
& =2\left[a^8+6 a^6-6 a^4+a^4-2 a^2+1\right] \\
& =2\left[a^8+6 a^6-5 a^4-2 a^2+1\right]
\end{aligned}
$

Miscellaneous Exercise 4

Find an approximation of $(0.99)^5$ using the first three terms of its expansion.

Answer.

We have,
$
(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+{ }^n C_n x^n
$

Here $(0.99)^5=(1-0.01)^5=\left[1+(-0.01)^5\right]={ }^5 \mathrm{C}_0-{ }^5 \mathrm{C}_1(0.01)-{ }^5 \mathrm{C}_2(0.01)^2$
$
=1-5 \times(0.01)-10 \times(0.01)^2 \ldots \ldots \ldots \ldots=1-0.05+0.001 \ldots \ldots \ldots . .=1.001-0.05=0.951
$
Miscellaneous Exercise 5

Find $n$, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n$ is $\sqrt{6}: 1$.

Answer.

We have the general term of the binomial expansion
$
\begin{aligned}
& (a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+{ }^n C_3 a^{n-3} b^3+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+{ }^n C_n b^n \text { is } \\
& T_{r+1}={ }^n C_r a^{n-r} b^r
\end{aligned}
$

Also $\mathrm{r}^{\text {th }}$ term from the end $=(\mathrm{n}-\mathrm{r}+2)^{\text {th }}$ term from the beginning
$\therefore 5^{\text {th }}$ term from the beginning $=T_5=T_{4+1}={ }^n C_4 a^{n-4} b^4$ and
$5^{\text {th }}$ term from the end $=T_{n-5+2}=T_{n-3}=T_{(n-4)+1}={ }^n C_{n-4} a^4 b^{n-4}$
Hence in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n$
$5^{\text {th }}$ term from the beginning $={ }^n C_4(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^4$ and
$
5^{\text {th }} \text { term from the end }={ }^n \mathrm{C}_{n-4}(\sqrt[4]{2})^4\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}
$

Given that

$\begin{aligned}
& \frac{{ }^n C_4(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^4}{{ }^n C_{n-4}(\sqrt[4]{2})^4\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}}=\frac{\sqrt{6}}{1} \\
& \Rightarrow(\sqrt[4]{2})^{n-4-4}\left(\frac{1}{\sqrt[4]{3}}\right)^{4-n+4}=\frac{\sqrt{6}}{1} \text {, since }{ }^n C_r={ }^n C_{n-r} \\
& (2)^{\frac{n-8}{4}} \cdot(3)^{\frac{n-8}{4}}=2^{\frac{1}{2}} \times 3^{\frac{1}{2}} \\
& \Rightarrow \frac{n-8}{4}=\frac{1}{2} \\
&
\end{aligned}$

$
\begin{aligned}
& \Rightarrow n-8=2 \\
& \Rightarrow n=10
\end{aligned}
$
Miscellaneous Exercise 6

Expand using binomial theorem $\left[1+\frac{x}{2}-\frac{2}{x}\right]^4=x \neq 0$.

Answer.

We have, $(1+x)^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 x^3+\ldots \ldots \ldots \ldots \ldots+{ }^n C_n x^n$
Now, $\left[1+\frac{x}{2}-\frac{2}{x}\right]^4=\left[1+\left(\frac{x}{2}-\frac{2}{x}\right)\right]^4$
$
\begin{aligned}
& ={ }^4 \mathrm{C}_0+{ }^4 \mathrm{C}_1\left(\frac{x}{2}-\frac{2}{x}\right)+{ }^4 \mathrm{C}_2\left(\frac{x}{2}-\frac{2}{x}\right)^2+{ }^4 \mathrm{C}_3\left(\frac{x}{2}-\frac{2}{x}\right)^3+{ }^4 \mathrm{C}_4\left(\frac{x}{2}-\frac{2}{x}\right)^4 \\
& =1+4\left(\frac{x}{2}-\frac{2}{x}\right)+6\left(\frac{x^2}{4}+\frac{4}{x^2}-2\right)+4\left(\frac{x^3}{8}-\frac{8}{x^3}-\frac{3 x}{2}+\frac{6}{x}\right) \\
& +\left[{ }^4 \mathrm{C}_0\left(\frac{x}{2}\right)^4-{ }^4 \mathrm{C}_1\left(\frac{x}{2}\right)^3\left(\frac{2}{x}\right)+{ }^4 \mathrm{C}_2\left(\frac{x}{2}\right)^2\left(\frac{2}{x}\right)^2-{ }^4 \mathrm{C}_3\left(\frac{x}{2}\right)\left(\frac{2}{x}\right)^3+{ }^4 \mathrm{C}_4\left(\frac{2}{x}\right)^4\right] \\
& =1+\left(2 x-\frac{8}{x}\right)+\left(\frac{3}{2} x^2+\frac{24}{x^2}-12\right)+\left(\frac{x^3}{2}-\frac{32}{x^3}-6 x+\frac{24}{x}\right)+\left(\frac{x^4}{16}-x^2+6-\frac{16}{x^2}+\frac{16}{x^4}\right)
\end{aligned}
$

$
\begin{aligned}
& =\left(\frac{24}{x}-\frac{8}{x}\right)+\left(\frac{24}{x^2}-\frac{16}{x^2}\right)-\frac{32}{x^3}+\frac{16}{x^4}+(2 x-6 x)+\left(\frac{3}{2} x^2-x^2\right)+\left(\frac{x^3}{2}\right)+\left(\frac{x^4}{16}\right)+( \\
& =\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4}-4 x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5
\end{aligned}
$
Miscellaneous Exercise 7

Find the expansion of $\left(3 x^2-2 \alpha x+3 a^2\right)^3$ using binomial theorem.

Answer.

Here $\left(3 x^2-2 a x+3 a^2\right)^3=\left[\left(3 x^2-2 a x\right)+3 a^2\right]^3$
$
={ }^3 \mathrm{C}_0\left(3 x^2-2 a x\right)^3+{ }^3 \mathrm{C}_1\left(3 x^2-2 a x\right)^2\left(3 a^2\right)+{ }^3 \mathrm{C}_2\left(3 x^2-2 a x\right)\left(3 a^2\right)^2+{ }^3 \mathrm{C}_3\left(3 a^2\right)^3
$

$\begin{aligned}
& =\left(3 x^2-2 a x\right)^3+3 \times 3 a^2\left(3 x^2-2 a x\right)^2+3 \times 9 a^4\left(3 x^2-2 a x\right)+27 a^6 \\
& = \\
& 27 x^6-8 a^3 x^3-54 a x^5+36 a^2 x^4+9 a^2\left(9 x^4+4 a^2 x^2-12 a x^3\right)+27 a^4\left(3 x^2-2 a x\right)+27 a^6 \\
& = \\
& 27 x^6-8 a^3 x^3-54 a x^5+36 a^2 x^4+81 a^2 x^4+36 a^4 x^2-108 a^3 x^3+81 a^4 x^2-54 a^5 x+27 a^6 \\
& =27 x^6-54 a x^5+117 a^2 x^4-116 a^3 x^3+117 a^4 x^2-54 a^5 x+27 a^6
\end{aligned}$