Exercise 2.2-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 $\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x>0$
Solution:
Given that $\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}, x>0$
$
\begin{aligned}
& \Rightarrow \quad 4 \leq 3(x+1) \leq 6 \quad \Rightarrow \quad 4 \quad 4 \leq 3 x+3 \leq 6 \\
& \Rightarrow \quad 4-3 \leq 3 x \leq 6-3 \quad \Rightarrow 1 \leq 3 x \leq 3 . \\
& \Rightarrow \quad \frac{1}{3} \leq x \leq 1
\end{aligned}
$
Hence, the solution is $\frac{1}{3} \leq x \leq 1$.

Question 2.
Solve $\frac{|x-2|-1}{|x-2|-2} \leq 0$
Solution:
Given that $\frac{|x-2|-1}{|x-2|-2} \leq 0$
Question 3.
Solve: $\frac{1}{|x|-3} \leq \frac{1}{2}$
Solution:
Given that $\frac{1}{|x|-3} \leq \frac{1}{2}$
$
\Rightarrow \quad|x|-3 \geq 2 \quad\left[\because \frac{1}{x}<\frac{1}{y} \Rightarrow x>y\right]
$

$
\begin{array}{ll}
\Rightarrow & |x| \geq 5 \\
\Rightarrow & x \leq-5 \text { (or) } x \geq 5 \\
\text { So, } & x \in(-\infty,-5)] \cup[5, \infty)
\end{array}
$
Here $\quad|x|-3 \neq 0$
$
\begin{array}{lll}
\Rightarrow & |x|-3<0 & \text { (or) }|x|-3>0 \\
\Rightarrow & |x|<3 & \text { (or) } \quad|x|>3 \\
\Rightarrow & -3<x<3 & \text { (or) } x<-3 \text { or } x>3
\end{array}
$
From (i) and (ii) we get
$
x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty)
$

Question 4.
Solve: $|x-1| \leq 5,|x| \geq 2$
Solution:
$
\begin{aligned}
& |\mathrm{x}-1| \leq 5 \text { and }|\mathrm{x}| \geq 2 \\
& \Rightarrow-5 \leq \mathrm{x}-1 \leq 5 \text { and } \mathrm{x} \leq-2 \text { (or) } \mathrm{x}>2 \\
& \Rightarrow-5+1 \leq \mathrm{x} \leq 5+1 \\
& \Rightarrow-4 \leq \mathrm{x} \leq 6 \text { and } \mathrm{x} \leq-2 \text { (or) } \mathrm{x} \geq 2 \\
& \text { Hence } \mathrm{x}<[-4,-2] \cup[2,6]
\end{aligned}
$
Question 5.
Solve: $\left|x-\frac{1}{4}\right|<\left|\frac{1}{2} x-\frac{3}{4}\right|$
Solution:
$
\begin{aligned}
& x-\frac{1}{4}>\frac{1}{2} x-\frac{3}{4} \text { (or) } x-\frac{1}{4}<\frac{3}{4}-\frac{1}{2} x \\
& x-\frac{1}{2} x>-\frac{3}{4}+\frac{1}{4}\left(=-\frac{2}{4}=\frac{-1}{2}\right) \quad x+\frac{1}{2} x<\frac{3}{4}+\frac{1}{4}=1 \\
& \frac{1}{2} x>\frac{-1}{2} \\
& \frac{3}{2} x<1 \\
& \Rightarrow x>-1 \\
& \text { From (1) and (2) } \\
& x<\frac{2}{3} \\
& -1<x<\frac{2}{3} \\
&
\end{aligned}
$