Exercise 2.8 - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 2.8
Question 1.
Find all values of $\mathrm{x}$ for which $\frac{x^3(x-1)}{(x-2)}>0$
Solution:
$
\frac{x^3(x-1)}{(x-2)}>0
$
Now we have to find the signs of $x^3, x-1$ and $x-2$ as follows
$
\mathrm{x}^3=0 ; \mathrm{x}-1=0 \Rightarrow \mathrm{x}=1 ; \mathrm{x}-2=0 \Rightarrow \mathrm{x}=2
$
Plotting the points in a number line and finding intervals

So the solution set $=(0,1) \cup(2, \infty)$
Question 2.
Find all values of $x$ that satisfies the inequality $\frac{2 x-3}{(x-2)(x-4)}<0$
Solution:
$
\begin{aligned}
& \frac{2 x-3}{(x-2)(x-4)}<0 \\
& 2 x-3=0 \Rightarrow x=\frac{3}{2} \\
& x-2=0 \Rightarrow x=2 \\
& x-4=0 \Rightarrow x=4
\end{aligned}
$
Plotting the points $3 / 2,2$ and 4 on the number line and taking the intervals.

The solution for $\frac{2 x-3}{(x-2)(x-4)}<0$ is in the interval $\left(-\infty, \frac{3}{2}\right) \cup(2,4)$
Question 3.
Solve $\frac{x^2-4}{x^2-2 x-15} \leq 0$
Solution:
$
\begin{aligned}
& \frac{x^2-4}{x^2-2 x-15} \leq 0 \Rightarrow \frac{(x-2)(x+2)}{(x+3)(x-5)} \leq 0 \\
& x-2=0 \Rightarrow x=2 ; x+2=0 \Rightarrow x=-2 \\
& x+3=0 \Rightarrow x=-3 ; x-5=0 \Rightarrow x=5
\end{aligned}
$

$\text { The intervals are }(-\infty,-3),(-3,-2),[-2,2],(2,5),(5, \infty)$

So the solution for the inequality $\frac{x^2-4}{x^2-2 x-15} \leq 0$ in the interval $(-3,-2) \cup(2,5)$