Exercise 2.9-Additional Questions - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

AdditionalQuestions
Question 1.
$
\frac{3 x+7}{x^2-3 x+2}
$
Solution:
$
\begin{aligned}
x^2-3 x+2 & =(x-2)(x-1) \\
\therefore \frac{3 x+7}{(x-2)(x-1)} & =\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-1}=\frac{\mathrm{A}(x-1)+\mathrm{B}(x-2)}{(x-2)(x-1)} \\
3 x+7 & =\mathrm{A}(x-1)+\mathrm{B}(x-2) \\
\text { put } x & =2 \\
(3)(2)+7 & =\mathrm{A}(2-1)+\mathrm{B}(0) \\
\mathrm{A} & =13 \\
\text { put } x & =1 \\
3+7 & =\mathrm{A}(0)+\mathrm{B}(-1) \Rightarrow \mathrm{B}=-10 \\
\Rightarrow \quad \therefore \quad \frac{3 x+7}{(x-2)(x-1)} & =\frac{13}{x-2}-\frac{10}{x-1}
\end{aligned}
$
Question 2.
$
\frac{x+4}{\left(x^2-4\right)(x+1)}
$

Solution:
$
\begin{aligned}
\frac{x+4}{\left(x^2-4\right)(x+1)} & =\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{x+1} \\
& =\frac{\mathrm{A}(x+2)(x+1)+\mathrm{B}(x-2)(x+1)+\mathrm{C}(x-2)(x+2)}{(x-2)(x+2)(x+1)}
\end{aligned}
$
Equating numerator on both sides
$
(x+4)=\mathrm{A}(x+2)(x+1)+\mathrm{B}(x-2)(x+1)+\mathrm{C}(x-2)(x+2)
$
Put $x=-2$
$
\begin{aligned}
-2+4 & =\mathrm{A}(0)+\mathrm{B}(-2-2)(-2+1)+\mathrm{C}(0) \\
2=4 \mathrm{~B} & \Rightarrow \mathrm{B}=2 / 4 \Rightarrow \mathrm{B}=1 / 2 \\
\text { Put } x & =2 \\
6 & =\mathrm{A}(4)(3)+\mathrm{B}(0)+\mathrm{C}(0) \\
12 \mathrm{~A} & =6 \\
\mathrm{~A} & =1 / 2
\end{aligned}
$
Put $x=-1$
$
\begin{aligned}
-1+4 & =\mathrm{A}(0)+\mathrm{B}(0)+\mathrm{C}(-3)(1) \Rightarrow-3 \mathrm{C}=3 \Rightarrow \mathrm{C}=-1 \\
\therefore \quad \frac{x+4}{\left(x^2-4\right)(x+1)} & =\frac{1}{2(x-2)}+\frac{1}{2(x+2)}-\frac{1}{x+1}
\end{aligned}
$
Question 3.
$
\frac{9}{(x-1)(x+2)^2}
$
Solution:
$
\begin{aligned}
\frac{9}{(x-1)(x+2)^2} & =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}}{(x+2)^2} \\
& =\frac{\mathrm{A}(x+2)^2+\mathrm{B}(x-1)(x+2)+\mathrm{C}(x-1)}{(x-1)(x+2)(x+2)^2}
\end{aligned}
$
Equating nemerator on $\mathrm{b} / \mathrm{s}$
$
9=A(x+2)^2+B(x-1)(x+2)+C(x-1)
$
Put $\mathrm{x}=-2$
$
\begin{aligned}
& 9=\mathrm{A}(0)+\mathrm{B}(0)+\mathrm{C}(-3) \\
& -3 \mathrm{C}=9 \Rightarrow \mathrm{C}=-3
\end{aligned}
$
Put $x=1$
$
9=\mathrm{A}(1+2)^2+\mathrm{B}(0)+\mathrm{C}(0)
$

$
9 \mathrm{~A}=9
$
$
A=1
$
Put $\mathrm{x}=0$
$
\begin{aligned}
& 9=4 \mathrm{~A}-2 \mathrm{~B}-\mathrm{C} \\
& 9=4(1)-2 \mathrm{~B}+3 \\
& 9-7=-2 \mathrm{~B} \\
& 2=-2 \mathrm{~B} \\
& \mathrm{~B}=-1 \\
& \therefore \quad \frac{9}{(x-1)(x+2)^2}=\frac{1}{x-1}-\frac{1}{x+2}-\frac{3}{(x+2)^2}
\end{aligned}
$
Question 4.
$
\frac{x^2-2 x-9}{(x+1)\left(x^2+x+6\right)}
$
Solution:
$
\begin{aligned}
\frac{x^2-2 x-9}{(x+1)\left(x^2+x+6\right)} & =\frac{\mathrm{A}}{(x+1)}+\frac{\mathrm{B} x+\mathrm{C}}{\left(x^2+x+6\right)} \\
& =\frac{\mathrm{A}\left(x^2+x+6\right)+(\mathrm{B} x+\mathrm{C})(x+1)}{(x+1)\left(x^2+x+6\right)}
\end{aligned}
$$Equating numerator on $\mathrm{b} / \mathrm{s}$
$
x^2-2 x-9=\mathrm{A}\left(x^2+x+6\right)+(\mathrm{B} x+\mathrm{C})(x+1)
$
Put $x=-1$
$
\begin{aligned}
1+2-9 & =\mathrm{A}(1-1+6)+0 \\
-6 & =6 \mathrm{~A} \Rightarrow \mathrm{A}=-1
\end{aligned}
$
Equating co-eff of $x^2$

$
\begin{aligned}
& 1=A+B \\
& 1=-1+B \Rightarrow B=2
\end{aligned}
$
Equating co-eff of $x$
$
\begin{aligned}
& -2=\mathrm{A}+\mathrm{B}+\mathrm{C} \\
& -2=-1+2+\mathrm{C} \\
& -2=1+\mathrm{C} \Rightarrow \mathrm{C}=-2-1=-3 \\
& \frac{x^2-2 x-9}{(x+1)\left(x^2+x+6\right)}=\frac{-1}{x+1}+\frac{2 x-3}{x^2+x+6}
\end{aligned}
$
Question 5.
$
\frac{x^3-1}{x^2+x+1}=\frac{x^3-1}{(x+2)(x-1)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B}}{x-1}
$

Solution:
$
\begin{gathered}
\frac{x^3-1}{x^2+x+1}=\frac{x^3-1}{(x+2)(x-1)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B}}{x-1} \\
\frac{x-1}{x+x+}=\frac{\mathrm{A}(x-1)}{(x+2)}+\frac{\mathrm{B}(x+2)}{(x-1)}
\end{gathered}
$
Equating numerator on both sides
$
x^3-1=\mathrm{A}(x-1)+\mathrm{B}(x+2)
$
Put $x=1$
$
\begin{aligned}
& 0=0+\mathrm{B}(1+2) \\
& 3 \mathrm{~B}=0 \Rightarrow \mathrm{B}=0
\end{aligned}
$
Put $\mathrm{x}=-2$
$
\begin{aligned}
& (-2)^3-1=\mathrm{A}(-2-1)+\mathrm{B}(0) \\
& -8-1=-3 \mathrm{~A} \\
& -9=-3 \mathrm{~A} \\
& \mathrm{~A}=9 / 3 \Rightarrow \mathrm{A}=3 \\
& \therefore \frac{x^3-1}{x^2+x+1}=\frac{3}{x+2}+\frac{0}{x-1} \\
& \frac{x^3-1}{x^2+x+1}=\frac{3}{x+2}
\end{aligned}
$