TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Resolve the following fractions into partial fractions.

Question 1.
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\)
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\) = \(\frac{A}{(x-1)}+\frac{B x+C}{\left(x^2+2\right)}\)
⇒ A (x² + 2) + (Bx + C) (x – 1)
= 2x² + 3x + 4 …………(1)
Substituting x = 1 in (1),
we get 3A = 9
⇒ A = 3
Substituting x = 0 in (1),
we get 2A – C = 4
⇒ C = 2A – 4
= 2(3) – 4
⇒ C = 2
Equating coefficient of x² on both sides in (1)
∴ A + B = 2
= 3 + B = 2
⇒ B = – 1
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\).

Question 2.
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\)
Solution:
Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\) = \(\frac{A}{x+2}+\frac{B x+C}{\left(1-x+x^2\right)}\)
⇒ A (1 – x + x²) + (Bx + C) (x + 2) = 3x – 1 ………..(1)
Substituting x = – 2 in (1), we get
7A = – 7
⇒ A = – 1
Substituting x = 0 in (1), we get
A + 2C = – 1
⇒ C = 0
Equating coefficients of x² on both sides in (1)
A + B = 0
⇒ – 1 + B = 0
⇒ B = 1
∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\) = \(\frac{-1}{x+2}+\frac{x}{1-x+x^2}\).

Question 3.
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)
Solution:
Let \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\) = \(\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)
⇒ A(x² + 1) + (Bx + C) (x + 2) = x² – 3
Substituting x = – 2 in (1), we get
5A = 1
⇒ A = \(\frac{1}{54}\)
Substituting x = 0 in (1), we get
A + 2C = – 3
⇒ 2C = – 3 – \(\frac{1}{5}\)
C = \(\frac{-8}{5}\)
Equating coefficients of x² on both sides,
we get A + B = 1
⇒ B = 1 – \(\frac{1}{5}\)
⇒ B = \(\frac{4}{5}\)
∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\) = \(\frac{1}{5(x+2)}+\frac{4}{5} \frac{(x-2)}{\left(x^2+1\right)}\).

Question 4.
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}\)
Solution:
Let \(\frac{x^2+1}{\left(x^2+x+1\right)^2}\) = \(\frac{A x+B}{\left(x^2+x+1\right)}+\frac{C x+D}{\left(x^2+x+1\right)^2}\)
⇒ (Ax + B) (x² + x + 1) + Cx + D = x² + 1 …………(1)
Equating coefficient of x³ on both sides in (1) we get, A = 0.
Equating coefficient of x² on both sides in (1) we get A + B = 1
⇒ B = 1.
Equating coefficient of x on both sides in (1)
we get A + B + C = 0.
⇒ 0 + 1 + C = 0
⇒ C = – 1
Substituting x = 0 in (1),
we get B + D = 1
⇒ D = 0
∴ \(\frac{x^2+1}{\left(x^2+x+1\right)^2}\) = \(\frac{1}{x^2+x+1}-\frac{x}{\left(x^2+x+1\right)^2}\).

Question 5.
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\)
Solution:
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\) = \(\frac{x^3+x^2+1}{(x-1)^2\left(x^2+x+1\right)}\)
Let \(\frac{x^3+x^2+1}{(x-1)^2\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+x+1}\)
∴ A (x² + x + 1) (x – 1) + B(x² + x + 1) + (Cx + D) (x – 1)² = x³ + x² + 1
⇒ A (x³ – 1) + B (x² + x + 1) + (Cx + D) (x – 1)² = x³ + x² + 1 …………..(1)
Substituting x = 1 in (1),
we get 3B = 3
⇒ B = 1
Equating coefficients of x³ on both sides in (1),
We get A + C = 1
Equating coefficients of x² on both sides in (1),
weget B – 2C + D = 1
⇒ 2C = D ………….(3)
Equating coefficients of x on both sides in (1)
we get B + C – 2D = 0
⇒ B + C – 4C = 0
⇒ B = 3C
⇒ 1 = 3C
⇒ C = \(\frac{1}{3}\)
Substituting in (2) we get
A = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
also D = 2C
⇒ D = \(\frac{2}{3}\)
∴ \(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\) = \(\frac{2}{3(x-1)}+\frac{1}{(x-1)^2}+\frac{x+2}{3\left(x^2+x+1\right)}\).