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

Updated On May 15, 2024

By SaraNextGen

EX.2.9

Question 1.
$
\frac{1}{x^2-a^2}
$
Solution:
Factorizing the denominator
$
\begin{aligned}
& \mathrm{Dr}=x^2-a^2=(x-a)(x+a) \\
& \text { Let } \frac{1}{x^2-a^2}=\frac{\mathrm{A}}{x-a}+\frac{\mathrm{B}}{x+a}=\frac{\mathrm{A}(x+a)+\mathrm{B}(x-a)}{(x-a)(x+a)}
\end{aligned}
$
Equating the numerator we get
$
1=\mathrm{A}(x+a)+\mathrm{B}(x-a)
$
This equation is true for any value of $x$
To find $A$ and $B$

$\begin{aligned}
& x-a=0 \\
& \Rightarrow x=a \\
& x+a=0 \\
& \Rightarrow x=-a
\end{aligned}$
$
\begin{aligned}
& \text { Put } x=a \\
& 1=\mathrm{A}(2 a)+\mathrm{B}(0) \\
& \Rightarrow \quad \mathrm{A}=\frac{1}{2} a \\
& \text { Put } x=-a \\
& 1=\mathrm{A}(0)+\mathrm{B}(-2 a) \\
& \Rightarrow \quad \mathrm{B}=-\frac{1}{2} a \\
& \therefore \frac{1}{x^2-a^2}=\frac{1 / 2 a}{(x-a)}+\frac{-1 / 2 a}{x+a} \\
& =\frac{1}{2 a(x-a)}-\frac{1}{2 a(x+a)} \\
&
\end{aligned}
$

Question 2.
$
\frac{3 x+1}{(x-2)(x+1)}
$
Solution:
Let $\frac{3 x+1}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{\mathrm{B}}{x+1}$
(i.e., $\frac{3 x+1}{(x-2)(x+1)}=\frac{\mathrm{A}(x+1)+\mathrm{B}(x-2)}{(x-2)(x+1)}$
Equating numerator parts
$
3 x+1=\mathrm{A}(x+1)+\mathrm{B}(x-2)
$
This equation is true for any value of $x$.
To find $A$ and $B$
$
\begin{aligned}
\text { Put } x & =-1 \\
-3+1 & =\mathrm{A}(0)+\mathrm{B}(-1-2) \\
-3 \mathrm{~B} & =-2 \Rightarrow \mathrm{B}=2 / 3
\end{aligned}
$
Put $x=2$
$\begin{aligned} 3(2)+1 & =\mathrm{A}(2+1)+\mathrm{B}(0) \\ 3 \mathrm{~A} & =7 \Rightarrow \mathrm{A}=7 / 3 \\ \text { Hence } \frac{3 x+1}{(x-2)(x+1)} & =\frac{7}{3(x-2)}+\frac{2}{3(x+1)}\end{aligned}$
Question 3.
$
\frac{x}{\left(x^2+1\right)(x-1)(x+2)}
$
Solution:

Let $\frac{x}{\left(x^2+1\right)(x-1)(x+2)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C} x+\mathrm{D}}{x^2+1}$
(i.e.,) $\frac{x}{\left(x^2+1\right)(x-1)(x+2)}=\frac{\mathrm{A}(x+2)\left(x^2+1\right)+\mathrm{B}(x-1)\left(x^2+1\right)+(\mathrm{C} x+\mathrm{D})(x-1)(x+2)}{(x-1)(x+2)\left(x^2+1\right)}$
Equating numerator on both sides
$
x=\mathrm{A}(x+2)\left(x^2+1\right)+\mathrm{B}(x-1)\left(x^2+1\right)+(\mathrm{C} x+\mathrm{D})(x-1)(x+2)
$
This equations is true for any value of $x$ to find $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$.
Put $x=1$
$
\begin{aligned}
& 1=\mathrm{A}(3)(2)+\mathrm{B}(0)+(0) \\
& 6 \mathrm{~A}=1 \Rightarrow \mathrm{A}=1 / 6
\end{aligned}
$
Put $x=-2$
$
\begin{aligned}
& -2=\mathrm{A}(0)+\mathrm{C}(0)+\mathrm{B}(-3)(5) \\
& \Rightarrow \quad-15 \mathrm{~B}=-2 \Rightarrow \mathrm{B}=2 / 15 \\
& \Rightarrow \quad 2 \mathrm{~A}-\mathrm{B}-2 \mathrm{D}=0 \\
& \text { (i.e.,) } \frac{2}{6}-\frac{2}{15}-2 \mathrm{D}=0 \\
& \Rightarrow \quad 2 \mathrm{D}=\frac{2}{6}-\frac{2}{15}=\frac{10-4}{30}=\frac{6}{30}=\frac{1}{5} \\
& \Rightarrow \quad \mathrm{D}=\frac{1}{5 \times 2}=\frac{1}{10} \\
& \mathrm{D}=\frac{1}{10} \\
& \text { Equating co-efficient of } x^3 \\
& \mathrm{~A}+\mathrm{B}+\mathrm{C}=0 \\
& \frac{1}{6}+\frac{2}{15}+\mathrm{C}=0 \Rightarrow \mathrm{C}=\frac{-1}{6}-\frac{2}{15}=\frac{-5-4}{30} \\
& \therefore \frac{x^2=\frac{-9}{30}=\frac{-3}{10}}{\left(x^2+1\right)(x-1)(x+2)}=\frac{1}{6(x-1)}+\frac{2}{15(x+2)}+\frac{\frac{-3}{10} x+\frac{1}{10}}{x^2+1} \\
& =\frac{1}{6(x-1)}+\frac{2}{15(x+2)}+\frac{1-3 x}{10\left(x^2+1\right)} \\
&
\end{aligned}
$

Question 4.
$
\frac{x}{(x-1)^3}
$
Solution:
Equating numerator on both sides
$
x=\mathrm{A}(x-1)^2+\mathrm{B}(x-1)+\mathrm{C}
$
Put $x=1$
$
1=\mathrm{A}(0)+\mathrm{B}(0)+\mathrm{C} \Rightarrow \mathrm{C}=1
$
Equating co-eff-of $x^2 \Rightarrow A=0$
Put $x=0$
$
\begin{aligned}
& \mathrm{A}-\mathrm{B}+\mathrm{C}=0 \\
& 0-\mathrm{B}+1=0 \Rightarrow \mathrm{B}=1 \\
& \therefore \frac{x}{(x-1)^3}=\frac{0}{x-1}+\frac{1}{(x-1)^2}+\frac{1}{(x-1)^3}=\frac{1}{(x-1)^2}+\frac{1}{(x-1)^3}
\end{aligned}
$
Question 5.
$
\frac{1}{x^4-1}=\frac{1}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{1}{(x-1)(x+1)\left(x^2+1\right)}
$
Solution:
Let $\frac{1}{x^4-1}=\frac{1}{(x-1)(x+1)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C} x+\mathrm{D}}{x^2+1}$
So $\frac{1}{x^4-1}=\frac{\mathrm{A}(x+1)\left(x^2+1\right)+\mathrm{B}(x-1)\left(x^2+1\right)+(\mathrm{C} x+\mathrm{D})(x-1)(x+1)}{x^4-1}$

Equating nuemarator on bothsides we get
$
1=\mathrm{A}(x+1)\left(x^2+1\right)+\mathrm{B}(x-1)\left(x^2+1\right)+(\mathrm{C} x+\mathrm{D})(x-1)(x+1)
$
Put $x=1$
$
\begin{aligned}
1 & =\mathrm{A}(2)(2) \Rightarrow 4 \mathrm{~A}=1 \Rightarrow \mathrm{A}=\frac{1}{4} \\
\text { Put } x & =-1 \Rightarrow \mathrm{B}(-4)=1 \Rightarrow \mathrm{B}=\frac{-1}{4} \\
\text { Put } x & =0 \Rightarrow \mathrm{A}-\mathrm{B}-\mathrm{D}=1 \\
(\text { ie.,) } & \frac{1}{4}+\frac{1}{4}-\mathrm{D}=1 \Rightarrow \mathrm{D}=\frac{1}{4}+\frac{1}{4}-1=\frac{-1}{2}
\end{aligned}
$
Equating co-eff-of $x^3$
$
\begin{aligned}
& \mathrm{A}+\mathrm{B}+\mathrm{C}=0 \\
& \frac{1}{4}-\frac{1}{4}+\mathrm{C}=0 \Rightarrow \mathrm{C}=0 \\
& \therefore \frac{1}{x^4-1}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}+\frac{0-1 / 2}{x^2+1} \\
& \text { (ie.) } \frac{1}{x^4-1}=\frac{1}{4(x-1)}-\frac{1}{4(x+1)}-\frac{1}{2\left(x^2+1\right)}
\end{aligned}
$
Question 6.
$
\frac{(x-1)^2}{x^3+x}
$
Solution:
$
\begin{aligned}
& x^3+x=x\left(x^2+1\right) \\
& \therefore \frac{(x-1)^2}{x^3+x}=\frac{(x-1)^2}{x\left(x^2+1\right)} \\
& \text { Let } \frac{(x-1)^2}{x\left(x^2+1\right)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1} . \\
& \therefore \frac{(x-1)^2}{x\left(x^2+1\right)}=\frac{\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(x)}{x\left(x^2+1\right)}
\end{aligned}
$
Equating numerator on both sides
$
(\mathrm{x}-2)^2=\mathrm{A}\left(\mathrm{x}^2+1\right)+(\mathrm{Bx}+\mathrm{c})(\mathrm{x})
$

Put $\mathrm{x}=0$
$
1=\mathrm{A}
$
Equating co-eff of $x^2$
$
\begin{aligned}
& 1=\mathrm{A}+\mathrm{B} \\
& \text { (i.e.,) } 1+\mathrm{B}=1 \Rightarrow \mathrm{B}=0 \\
& \text { put } \mathrm{x}=1 \\
& \mathrm{~A}(2)+\mathrm{B}+\mathrm{C}=0 \text { (i.e.,) } 2 \mathrm{~A}+\mathrm{B}+\mathrm{C}=0 \\
& 2+0+\mathrm{C}=0 \Rightarrow \mathrm{C}=-2 \\
& \therefore \frac{(x-1)^2}{x\left(x^2+1\right)}=\frac{1}{x}+\frac{0 x-2}{x^2+1} \\
& =\frac{1}{x}-\frac{2}{x^2+1}
\end{aligned}
$
Question 7.
$
\frac{x^2+x+1}{x^2-5 x+6}
$
Solution
Since numerator and denominator are of same degree we have divide the numerator by the denominator
$
\begin{gathered}
\left.x^2-5 x+6\right) x^2+x+1 \\
\frac{x^2-5 x+6}{6 x-5} \\
\therefore \frac{x^2+x+1}{x^2-5 x+6}=1+\frac{6 x-5}{x^2-5 x+6}
\end{gathered}
$

$\begin{aligned}
& \text { Now } \frac{6 x-5}{x^2-5 x+6}=\frac{6 x-5}{(x-2)(x-3)} \\
& \text { Let } \frac{6 x-5}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}=\frac{\mathrm{A}(x-3)+\mathrm{B}(x-2)}{(x-2)(x-3)} \\
& 6 x-5=\mathrm{A}(x-3)+\mathrm{B}(x-2) \\
& \text { Put } x=3 \\
& 18-5=\mathrm{A}(0)+\mathrm{B}(3-2) \\
& \mathrm{B}=13 \\
& \text { Put } x=2 \\
& 12-5=\mathrm{A}(-1)+\mathrm{B}(0) \\
& \Rightarrow \quad-\mathrm{A}=7 \\
& \mathrm{~A}=-7 \\
& \therefore \frac{6 x-5}{(x-2)(x-3)}=\frac{-7}{x-2}+\frac{13}{x-3} \\
&
\end{aligned}$

Substituting the value in ....(1)
$
\begin{aligned}
\frac{x^2+x+1}{x^2-5 x+6} & =1+\frac{-7}{x-2}+\frac{13}{x-3} \\
& =1-\frac{7}{x-2}+\frac{13}{x-3}
\end{aligned}
$
Question 8.
$
\frac{x^3+2 x+1}{x^2+5 x+6}
$
Solution:
Numerator is of greater degree than the denominator So dividing Numerator by the denominator

So $\frac{x^3+2 x+1}{x^2+5 x+6}=(x-5)+\frac{21 x+31}{x^2+5 x+6}$
$\begin{aligned} & \text { Now } \frac{21 x+31}{x^2+5 x+6}=\frac{21 x+31}{(x+2)(x+3)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B}}{x+3} \\ &=\frac{\mathrm{A}(x+3)+\mathrm{B}(x+2)}{(x+2)(x+3)} \\ & \text { Equating Numerator on both sides }\end{aligned}$
$
\begin{aligned}
& \Rightarrow 21 \mathrm{x}+31=\mathrm{A}(\mathrm{x}+3)+\mathrm{B}(\mathrm{x}+2) \\
& \text { Put } \mathrm{x}=-3 \\
& -63+31=\mathrm{B}(-1) \\
& \mathrm{B}=32 \\
& \text { Put } \mathrm{x}=-2 \\
& -42+31=\mathrm{A}(1)+\mathrm{B}(0)
\end{aligned}
$

$
\begin{aligned}
& A=-11 \\
& \therefore \frac{21 x+31}{(x+2)(x+3)}=\frac{-11}{x+2}+\frac{32}{x+3}
\end{aligned}
$
Substituting (2) in (1) we get
$
\begin{aligned}
\frac{x^3+2 x+1}{x^2+5 x+6} & =(x-5)+\frac{-11}{x+2}+\frac{32}{x+3} \\
& =(x-5)+\frac{32}{x+3}-\frac{11}{x+2}
\end{aligned}
$
Question 9.
$
\frac{x+12}{(x+1)^2(x-2)}
$
Solution:

Let $\frac{x+12}{(x+1)^2(x-2)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x+1}+\frac{\mathrm{C}}{(x+1)^2}$
$
=\frac{\mathrm{A}(x+1)^2+\mathrm{B}(x-2)(x+1)+\mathrm{C}(x-2)}{(x-2)(x+1)^2}
$
equating Numerator on both sides
$
\begin{aligned}
& -3 \mathrm{C}=11 \\
& \mathrm{C}=\frac{-11}{3} \\
& \text { put } x=0 \Rightarrow \mathrm{A}-2 \mathrm{~B}-2 \mathrm{C}=12 \\
& \frac{14}{9}-2 B+\frac{22}{3}=12 \\
& -2 B=12-\frac{14}{9}-\frac{22}{3} \\
& =\frac{108-14-66}{9} \\
& =\frac{28}{9} \\
& \Rightarrow B=-\frac{28}{9 \times 2}=\frac{-14}{9} \\
& \therefore \frac{x+12}{(x+1)^2(x-2)}=\frac{14}{9(x-2)}-\frac{14}{9(x+1)}-\frac{11}{3(x+1)^2} \\
&
\end{aligned}
$
Question10.
$
\frac{6 x^2-x+1}{x^3+x^2+x+1}
$
Solution:

Factorizing $x^3+x^2+x+1$
We get $x^3+x^2+x+1=(x+1)\left(x^2+1\right)$
$
\begin{aligned}
\therefore \frac{6 x^2-x+1}{x^3+x^2+x+1} & =\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1} \\
& =\frac{\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(x+1)}{(x+1)\left(x^2+1\right)}
\end{aligned}
$
Equating Numerator on both sides we get
$
\begin{aligned}
& 6 \mathrm{x}^2-\mathrm{x}+1=\mathrm{A}\left(\mathrm{x}^2+1\right)+(\mathrm{Bx}+\mathrm{c})(\mathrm{x}+1) \\
& 6+1+1=\mathrm{A}(2)+0 \Rightarrow 2 \mathrm{~A}=8 \Rightarrow \mathrm{A}=4
\end{aligned}
$
Equating co-eff of $x^2$
$
\begin{aligned}
& 6=\mathrm{A}+\mathrm{B} \\
& \text { (i.e.,) } 4+\mathrm{B}=6 \Rightarrow \mathrm{B}=6-4=2 \\
& \text { put } \mathrm{x}=0 \\
& 1=\mathrm{A}+\mathrm{C} \\
& 4+\mathrm{C}=1 \Rightarrow \mathrm{C}=1-4=-3 \\
& \therefore \frac{6 x^2-x+1}{x^3+x^2+x+1}=\frac{4}{x+1}+\frac{2 x-3}{x^2+1}
\end{aligned}
$
Question 11.
$
\frac{2 x^2+5 x-11}{x^2+2 x-3}
$
Solution:
Since Numerator and are of same degree divide Numerator by the denominator

$
\begin{aligned}
\therefore \frac{2 x^2+5 x-11}{x^2+2 x-3} & =2+\frac{x-5}{x^2+2 x-3} \\
\text { Now } \frac{x-5}{x^2+2 x-3} & =\frac{x-5}{(x-1)(x+3)} \\
& =\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+3}=\frac{\mathrm{A}(x+3)+\mathrm{B}(x-1)}{(x-1)(x+3)}
\end{aligned}
$
equating Numerator on both sides we get
$
\mathrm{x}-5=\mathrm{A}(\mathrm{x}+3)+\mathrm{B}(\mathrm{x}-1)
$
Put $\mathrm{x}=-3$
$
-3-5=\mathrm{A}(0)+\mathrm{B}(-4)
$
$
\begin{aligned}
& -4 \mathrm{~B}=-8 \Rightarrow \mathrm{B}=2 \\
& \text { Put } \mathrm{x}=1 \\
& 1-5=\mathrm{A}(4)+\mathrm{B}(0) \\
& 4 A=-4 \Rightarrow A=-1 \\
& \therefore \frac{x-5}{x^2+2 x-3}=\frac{-1}{x-1}+\frac{2}{x+3} \\
&
\end{aligned}
$
Substituting (2) in (1) we get
$
\begin{aligned}
\frac{2 x^2+5 x-11}{x^2+2 x-3} & =2+\left(\frac{-1}{x-1}+\frac{2}{x+3}\right) \\
& =2+\frac{2}{x+3}-\frac{1}{x-1}
\end{aligned}
$

Question 12.
$
\frac{7+x}{(1+x)\left(1+x^2\right)}
$
Solution:
Let $\frac{7+x}{(1+x)\left(1+x^2\right)}=\frac{\mathrm{A}}{1+x}+\frac{(\mathrm{B} x+\mathrm{C})}{\left(1+x^2\right)}=\frac{\mathrm{A}\left(x^2+1\right)+(\mathrm{B} x+\mathrm{C})(1+x)}{(1+x)\left(1+x^2\right)}$
Equating Numerator on both sides we get
$
\begin{aligned}
7+x & =\mathrm{A}\left(1+x^2\right)+(\mathrm{B} x+C)(1+x) \\
\text { put } x & =-1 \\
7-1 & =\mathrm{A}(2)+0 \Rightarrow 2 \mathrm{~A}=6 \Rightarrow \mathrm{A}=3
\end{aligned}
$
Equating co-eff of $x^2$
$
\begin{aligned}
\mathrm{A}+\mathrm{B} & =0 \\
3+\mathrm{B} & =0 \\
\Rightarrow \mathrm{B} & =-3 \\
\text { put } x & =0 \\
\mathrm{~A}+\mathrm{C} & =7 \\
3+\mathrm{C} & =7 \\
\mathrm{C} & =7-3=4 \\
\therefore \frac{7+x}{(1+x)\left(1+x^2\right)} & =\frac{3}{1+x}+\frac{-3 x+4}{1+x^2} \\
\therefore \quad & =\frac{3}{1+x}+\frac{4-3 x}{1+x^2}
\end{aligned}
$