Exercise 2.5-Additional Questions - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
Solve for $\mathrm{x}$.
$
\frac{x-2}{3 x+1}>\frac{x-3}{3 x-2}
$
Solution:
$
\begin{aligned}
& \frac{x-2}{3 x+1}-\frac{x-3}{3 x-2}>0 \\
& \frac{(3 x-2)(x-2)-(x-3)(3 x+1)}{(3 x+1)(3 x-2)}>0 \\
& \frac{3 x^2-6 x-2 x+4-\left[3 x^2+x-9 x-3\right]}{(3 x+1)(3 x-2)}>0 \\
& \frac{7}{(3 x+1)(3 x-2)}>0 \\
& (\because 7>0,(3 x+1)(3 x-2)>0)
\end{aligned}
$
Select the intervals in which $(3 x+1)(3 x-2)$ is positive $(3 \mathrm{x}+1)>0$ and $(3 \mathrm{x}-2)>0$ or $3 \mathrm{x}+1<0$ and $3 \mathrm{x}-2<0$
$\Rightarrow \quad x>\frac{-1}{3}$ and $x>\frac{2}{3}$ (or) $x<\frac{-1}{3}$ and $x<\frac{2}{3}$ (or) $x<\frac{-1}{3}$ and $x>\frac{2}{3}$
(i.e.) $\quad x \in\left(-\infty, \frac{-1}{3}\right) \cup\left(\frac{2}{3}, \infty\right)$

Question 2.
Solve $\frac{2 x-1}{x}>-1$
Solution:
$
\begin{aligned}
& \frac{2 x-1}{x}>-1 \\
& \Rightarrow \quad \frac{2 x-1}{x}+1>-1+1 \\
& \text { (i.e.,) } \frac{2 x-1+x}{x}>0 \\
& \frac{(3 x-1)}{x}>C \\
& \Rightarrow \frac{(3 x-1)(x)}{x}>0 \times x \\
& \text { (i.e.) } x(3 x-1)>0 \\
& x>0 \text { and } x<\frac{1}{3} \\
&
\end{aligned}
$
Question 3.
Solve $\frac{x+1}{x-1}>0$

Solution:
$
\begin{aligned}
& \frac{x+1}{x-1}>0 \Rightarrow \frac{(x+1)(x-1)}{(x-1)^2}>0 \\
\Rightarrow \quad & (x+1)(x-1)>0 \quad\left(\because(x-1)^2>0 \text { for all } x \neq 1\right) \\
\Rightarrow & x \in(-\infty,-1) \cup(1, \infty)
\end{aligned}
$