Exercise 5.4-Additional Questions - Chapter 5 Binomial Theorem Sequences and Series 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Find the $\sqrt[3]{126}$ approximately to two decimal places.
Solution:
$
\begin{gathered}
\sqrt{126}=(126)^{1 / 3}=(125+1)^{1 / 3}=\left\{125\left(1+\frac{1}{125}\right)\right\}^{\prime}=(125)^{1 / 3}\left[1+\frac{1}{125}\right] \\
=5\left[1+\frac{1}{3} \times \frac{1}{125}+\ldots\right]\left(\because \frac{1}{125}<1\right) \\
=5\left[1+\frac{1}{3}(0.008)\right]=5(1+0.002666)=5.01
\end{gathered}
$

Question 2.
Write the four terms in the expansions of the following:
(i) $\frac{1}{(2+x)^4}$ where $|x|>2$
(ii) $\frac{1}{\sqrt[3]{6-3 x}}$ where $|x|<2$
Solution:

(i)
$
\begin{aligned}
\frac{1}{(2+x)^4} & =\frac{1}{[x(2 / x+1)]^4}|x|>2 \Rightarrow 1>\left|\frac{2}{x}\right| \\
& =\frac{1}{x^4(1+2 / x)^4}=\frac{1}{x^4}(1+2 / x)^{-4}=\frac{1}{x^4}\left[1-4\left(\frac{2}{x}\right)+\frac{4.5}{12}\left(\frac{2}{x}\right)^2-\frac{4.5 .6}{1.2 .3}\left(\frac{2}{x}\right)^3\right] \\
& =\frac{1}{x^4}\left[1-\frac{8}{x}+\frac{40}{x^2}-\frac{160}{x^3}+\ldots \ldots . .\right]=\frac{1}{x^5}\left[x-8+\frac{40}{x}-\frac{160}{x^2}+\ldots \ldots \ldots\right]
\end{aligned}
$
(ii)
$
\begin{aligned}
\frac{1}{\sqrt[3]{6-3 x}} & =\frac{1}{(6-7 x)^{1 / 3}}=\frac{1}{6^{1 / 3}\left(1-\frac{x}{2}\right)^{1 / 3}}=\frac{1}{6^{1 / 3}}\left[1-\frac{x}{2}\right]^{-\frac{1}{3}} \\
& =\frac{1}{6^{1 / 3}}\left[1+\frac{1}{3}\left(\frac{x}{2}\right)+\frac{(1 / 3)(4 / 3)}{1.2}\left(\frac{x}{2}\right)^2+\frac{(1 / 3)(4 / 3)(7 / 3)}{1.2 .3}\left(\frac{x}{2}\right)^3\right] \\
& =\frac{1}{6^{1 / 3}}\left[1+\frac{x}{6}+\frac{x^2}{18}+\frac{7}{324} x^3+\ldots\right]
\end{aligned}
$

Question 3.
Evaluate the following:
(i) $\sqrt[3]{1003}$ correct to 4 places of decimals (ii) $\frac{1}{\sqrt[3]{128}}$ correct to 4 places of decimals.
Solution:
(i) $\sqrt[3]{1003}=(1003)^{\frac{1}{3}}=(1000+3)^{1 / 3}=(1000)^{1 / 3}\left[1+\frac{3}{1000}\right]^{1 / 3}=10[1+0.003]^{1 / 3}$
$
\begin{aligned}
& =10\left[1+\frac{1}{3}(0.003)+\frac{\left(\frac{1}{3}\right)\left(\frac{-2}{3}\right)}{1.2}(0.003)^2+\ldots\right] \\
& =10[1+0.001-0.0000001+\ldots]=10.00999=10.0100
\end{aligned}
$
(ii) $\frac{1}{\sqrt[3]{128}}=\frac{1}{(128)^{\frac{1}{3}}}=\frac{1}{(125+3)^{\frac{1}{3}}}=\frac{1}{5\left(1+\frac{3}{125}\right)^{\frac{1}{3}}}=\frac{1}{5}\left(1+\frac{3}{125}\right)^{-\frac{1}{3}}$
$
\begin{aligned}
& =\frac{1}{5}\left[1-\frac{1}{3}(0.024)+\frac{\frac{1}{3}\left(\frac{4}{3}\right)}{1.2}(0.024)^2 \ldots\right] \\
& =\frac{1}{5}[1-0.008+0.000128]=0.1984256
\end{aligned}
$
Question 4.
If $x$ so large prove that $\sqrt{x^2+25}-\sqrt{x^2+9}=\frac{8}{x}$ nearly.

Solution:
$
\begin{aligned}
\sqrt{x^2+25} & -\sqrt{x^2+9}=x\left(1+\frac{25}{x^2}\right)^{1 / 2}-x\left(1+\frac{9}{x^2}\right)^{1 / 2} \\
= & x\left[1+\frac{1}{2}\left(\frac{25}{x^2}\right)+\frac{\frac{1}{2}\left(\frac{-1}{2}\right)}{1.2}\left(\frac{25}{x^2}\right)^2+\ldots .-x\left[1+\frac{1}{2}\left(\frac{9}{x^2}\right)+\frac{2\left(\frac{-1}{2}\right)}{1.2}\left(\frac{9}{x^2}\right)^2+\ldots\right]\right. \\
= & x+\frac{25}{2 x}-\frac{625}{8 x^3}+\ldots-x-\frac{9}{2 x}+\frac{81}{8 x^3}+\ldots=\frac{16}{2 x}=\frac{8}{x} \text { approximately }
\end{aligned}
$
Question 5.
Show that $x^n=1+n\left(1-\frac{1}{x}\right)+\frac{n(n+1)}{1.2}\left(1-\frac{1}{x}\right)^2+\ldots$
Solution:
$
\text { RHS }=1+n\left(1-\frac{1}{x}\right)+\frac{n(n+1)}{1.2}\left(1-\frac{1}{x}\right)+\ldots
$
Put $y=1-\frac{1}{x}=1+n y+\frac{n(n+1)}{1.2} y^2+\ldots \ldots=(1-y)^{-n}$
$
\left[1-\left(1-\frac{1}{x}\right)\right]^{-n}=\left(\frac{1}{x}\right)^{-n}=x^n=\text { LHS }
$