Exercise 5.4-Additional Questions - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 5.4-Additional Questions - Chapter 5 - Binomial Theorem, Sequences and Series - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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 }
$