Exercise 2.11 - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 2.11
Question 1.
(i) $(125)^{\frac{2}{3}}$
(ii) $16^{\frac{-3}{4}}$
(iii) $(-1000)^{\frac{-2}{3}}$
(iv) $\left(3^{-6}\right)^{\frac{1}{3}}$
(v) $\frac{27^{\frac{-2}{3}}}{27^{\frac{-1}{3}}}$
Solution:
(i)
$
\left(5^3\right)^{2 / 3}=5^{3 \times \frac{2}{3}}=5^2=25
$
(ii)
$
16^{\frac{-3}{4}}=\frac{1}{16^{\frac{3}{4}}} \frac{1}{\left(2^4\right)^{3 / 4}}=\frac{1}{2^{4 \times \frac{3}{4}}}=\frac{1}{2^3}=\frac{1}{8}
$
(iii)
$
\left[(-10)^3\right]^{-\frac{2}{3}}=(-10)^{3 \times-\frac{2}{3}}=(-10)^{-2}=\frac{1}{(-10)^2}=\frac{1}{100}
$
(iv)
$
3^{-6 \times \frac{1}{3}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}
$
(v)
$
(27)^{-\frac{2}{3}+\frac{1}{3}}=(27)^{-\frac{1}{3}}=\left(3^3\right)^{\frac{-1}{3}}=3^{-1}=\frac{1}{3^1}=\frac{1}{3}
$

Question 2.
Evaluate $\left[\left((256)^{\frac{-1}{2}}\right)^{\frac{-1}{4}}\right]^3$
Solution:
$
\begin{aligned}
& 256=16^2=\left(2^4\right)^2=2^8 \\
& \therefore 256^{-1 / 2}=\left(2^8\right)^{-1 / 2}=2^{-4} \\
& {\left[(256)^{-1 / 2}\right]^{-1 / 4}=2^{-4 \times \frac{-1}{4}}=2^1} \\
& \left\{\left[(256)^{-1 / 2}\right]^{-1 / 4}\right\}^3=\left(2^1\right)^3=2^3=8
\end{aligned}
$
Question 3.
If $\left(x^{1 / 2}+x^{-1 / 2}\right)^2=\frac{9}{2}$, then find the value of $\left(x^{1 / 2}-x^{-1 / 2}\right)$ for $x>1$.
Solution:
$
\begin{aligned}
& \text { Given }\left(x^{1 / 2}+x^{-1 / 2}\right)^2=\frac{9}{2} \Rightarrow\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=\frac{9}{2} \\
& \text { (i.e.,) } \quad x+\frac{1}{x}+2 \sqrt{x} \frac{1}{\sqrt{x}}=\frac{9}{2} \Rightarrow x+\frac{1}{x}+2=\frac{9}{2} \Rightarrow x+\frac{1}{x}=\frac{9}{2}-2=\frac{9-4}{2}=\frac{5}{2} \\
& \text { Now }\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}-2 \sqrt{x} \frac{1}{\sqrt{x}}=x+\frac{1}{x}-2=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2} \\
& \Rightarrow \sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}
\end{aligned}
$

Question 4.
Simplify and hence find the value of $n: 3^{2 n} 9^2 \mathrm{~m}^{-n} / 3^{3 n}=27$
Solution:
$
\begin{aligned}
& \frac{3^{2 n} 9^2 3^{-n}}{3^{3 n}}=\frac{3^{2 n}\left(3^2\right)^2 3^{-n}}{3^{3 n}}=3^{2 n} 3^4 3^{-n} 3^{-3 n} \\
&=3^{2 n+4-n-3 n}=3^{4-2 n}=27 \Rightarrow \text { given } 27=3^3 \\
& \Rightarrow \quad 3^{4-2 n}=3^3 \Rightarrow 4-2 n=3 \\
& 4-3=2 n \Rightarrow 2 n=1 \Rightarrow n=\frac{1}{2}
\end{aligned}
$
Question 5.
Find the radius of the spherical tank whose volume is $32 \pi / 3$ units.
Solution:
$
\begin{aligned}
& \text { Volume of the sphere }=\frac{4}{3} \pi r^3=\frac{32 \pi}{3} \\
& \Rightarrow r^3=\frac{32 \pi}{3} \times \frac{3}{4 \pi}=8=2^3 \\
& \Rightarrow r=2 \text { units }
\end{aligned}
$
Question 6.

Simplify by rationalising the denominator $\frac{7+\sqrt{6}}{3-\sqrt{2}}$
Solution:
$
\begin{aligned}
\frac{7+\sqrt{6}}{3-\sqrt{2}} & =\frac{7+\sqrt{6}}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} \\
& =\frac{21+7 \sqrt{2}+3 \sqrt{6}+\sqrt{12}}{3^2-\sqrt{2}^2} \\
& =\frac{21+7 \sqrt{2}+3 \sqrt{6}+2 \sqrt{3}}{9-2}=\frac{21+7 \sqrt{2}+3 \sqrt{6}+2 \sqrt{3}}{7}
\end{aligned}
$
Question 7.
Simplify $\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}$

Solution:
$
\begin{aligned}
& \frac{1}{3-\sqrt{8}}=\frac{1}{3-\sqrt{8}} \times \frac{3+\sqrt{8}}{3+\sqrt{8}}=\frac{3+\sqrt{8}}{3^2-8}=\frac{3+\sqrt{8}}{1}=3+\sqrt{8} \\
& \frac{1}{\sqrt{8}-\sqrt{7}}=\frac{1}{\sqrt{8}-\sqrt{7}} \times \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}=\frac{\sqrt{8}+\sqrt{7}}{8-7}=\sqrt{8}+\sqrt{7} \\
& \frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}=\sqrt{7}+\sqrt{6} \\
& \frac{1}{\sqrt{6}-\sqrt{5}}=\frac{1}{\sqrt{6}-\sqrt{5}} \times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}=\frac{\sqrt{6}+\sqrt{5}}{6-5}=\sqrt{6}+\sqrt{5} \\
& \frac{1}{\sqrt{5}-2}=\frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\frac{\sqrt{5}+2}{1}=\sqrt{5}+2 \\
& \therefore \quad \frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2} \\
& =3+\sqrt{8}-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+2) \\
& =3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2=5 \\
&
\end{aligned}
$
Question 8.
If $x=\sqrt{2}+\sqrt{3}$ find $x^2+1 / x^2-2$
Solution:

$\begin{aligned}
& x=\sqrt{3}+\sqrt{2} \Rightarrow \frac{1}{x}=\sqrt{3}-\sqrt{2} \\
& \text { Now } x^2+1=\frac{x+\frac{1}{x}}{x} \\
& =\frac{\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{x}=\frac{2 \sqrt{3}}{x} \\
& =\frac{x-2 / x}{x}=\frac{(\sqrt{3}+\sqrt{2})-2(\sqrt{3}-\sqrt{2})}{x} \\
& =\quad \sqrt{3}+\sqrt{2}-2 \sqrt{3}+2 \sqrt{2}=\frac{3 \sqrt{2}-\sqrt{3}}{x} \\
& \text { So } \frac{x^2+1}{x^2-2}=\frac{x+\frac{1}{x}}{x} / \frac{x-\frac{2}{x}}{x} \\
& =\quad \frac{2 \sqrt{3}}{x} / \frac{3 \sqrt{2}-\sqrt{3}}{x}=\frac{2 \sqrt{3}}{3 \sqrt{2}-\sqrt{3}} \\
& =\frac{2 \sqrt{3}}{3 \sqrt{2}-\sqrt{3}} \times \frac{3 \sqrt{2}+\sqrt{3}}{3 \sqrt{2}+\sqrt{3}}=\frac{6 \sqrt{6}+2 \times 3}{18-3} \\
& =\frac{6 \sqrt{6}+6}{15}=\frac{6(\sqrt{6}+1)}{15}=\frac{2(\sqrt{6}+1)}{5} \\
&
\end{aligned}$