Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Partial Fractions Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2A Partial Fractions Important Questions

Question 1.

\(\frac{x+4}{\left(x^{2}-4\right)(x+1)}\) (Mar. 14)

Solution:

\(\frac{x+4}{\left(x^{2}-4\right)(x+1)}\) = \(\frac{A}{x+1}\) + \(\frac{B}{x+2}\) + \(\frac{c}{x-2}\)

Multiplying with (x^{2} – 4) (x + 1)

x + 4 = A(x^{2} – 4) + B(x + 1) (x – 2) + C (x + 1) (x + 2)

x = -1 ⇒ 3 = A(1 – 4) = -3A ⇒ A = -1

x = -2 ⇒ 2 = B(-2 + 1) (-2 – 2)

= 4B ⇒ B = + \(\frac{2}{4}\) = \(\frac{1}{2}\)

x = 2 ⇒ 6 = C(2 + 1)(2 + 2)

= 12C ⇒ C = \(\frac{1}{2}\)

Question 2.

\(\frac{x^{2}-x+1}{(x+1)(x-1)^{2}}\) (TS Mar. 15)

Solution:

Let \(\frac{x^{2}-x+1}{(x+1)(x-1)^{2}}\) = \(\frac{A}{x+1}\) + \(\frac{B}{x-1}\) + \(\frac{C}{(x-1)^{2}}\)

Multiplying with (x + 1) (x – 1)^{2}

x^{2} – x + 1 = A(x – 1)^{2} + B(x + 1) (x – 1) + C(x + 1)

Put x = -1, 1 + 1 + 1 = A(4)

⇒ A = \(\frac{3}{4}\)

Put x = 1, 1 – 1 + 1 = C(2)

⇒ C = + \(\frac{1}{2}\)

Equating the coefficients of x^{2},

A + B = 1

⇒ B = 1 – A = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)

Question 3.

Resolve the \(\frac{2 x^{2}+3 x+4}{(x-1)\left(x^{2}+2\right)}\) into partial fractions. (AP Mar. ’15, ’11; May ’11) (TS Mar. ’17)

Solution:

Let \(\frac{2 x^{2}+3 x+4}{(x-1)\left(x^{2}+2\right)}\) = \(\frac{A}{x-1}\) + \(\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+2}\)

Multiplying with (x – 1) (x^{2} + 2)

2x^{2} + 3x + 4 = A(x^{2} + 2) + (Bx + C)(x – 1)

x = 1 ⇒ 2 + 3 + 4 = A(1 + 2)

9 = 3A ⇒ A = 3

Equating the coefficients of x^{2}

2 = A + B ⇒ B = 2 – A = 2 – 3 = -1

Equating constants

4 = 2A – C ⇒ C = 2A – 4 = 6 – 4 = 2

\(\frac{2 x^{2}+3 x+4}{(x-1)\left(x^{2}+2\right)}\) = \(\frac{3}{x-1}\) + \(\frac{-x+2}{x^{2}+2}\)

Question 4.

Resolve \(\frac{x^{4}}{(x-1)(x-2)}\) into partial fractions.

Solution:

Equating the coefficients of (x – 1) (x – 2),

15x – 14 = A(x – 2) + B(x – 1)

Put x = 1, 15 – 14 ⇒ A(-1) = A = -1

Put x = 2, 30 – 14 = B(1) ⇒ B = 16

∴ \(\frac{x^{4}}{(x-1)(x-2)}\) = x^{2} + 3x + 7 – \(\frac{1}{x-1}\) + \(\frac{16}{x-2}\)

Question 5.

Resolve \(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\) into partial fractions. (AP Mar. ’17, ’16)

Solution:

Let \(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\) = \(\frac{A}{x+2}\) + \(\frac{B x+C}{x^{2}+1}\)

Multiplying with (x + 2) (x^{2} + 1)

x^{2} – 3 = A(x^{2} + 1) + (Bx + C)(x + 2)

x = -2 ⇒ 4 – 3 = A(4 + 1)

1 = 5A ⇒ A = \(\frac{1}{5}\)

Equating the coefficients of x^{2}

1 = A + B ⇒ B = 1 – A

= 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)

Equating the constants – 3 = A + 2C

2C = -3 – A = -3 – \(\frac{1}{5}\) = – \(\frac{16}{5}\)

C = –\(\frac{8}{5}\)

\(\frac{x^{2}-3}{(x+2)\left(x^{2}+1\right)}\) = \(\frac{1}{5(x+2)}\) + \(\frac{4 x-8}{5\left(x^{2}+1\right)}\)

Question 6.

Resolve \(\frac{1}{(x-1)^{2}(x-2)}\) into partial fractions. (May ’13)

Solution:

Put x = 1 in (1)

1 = A(0) + B(1 – 2) + C(0)

⇒ -B = 1 ⇒ B = -1

Put x = 2 in (1)

⇒ 1 = A(0) + B(0) + C (2 – 1)^{2}

⇒ C = 1

Equating the coefficients of x^{2} in (1)

0 = A + C ⇒ A = -C = -1

A = -1

∴ \(\frac{1}{(x-1)^{2}(x-2)}\) = \(\frac{-1}{x-1}\) – \(\frac{1}{(x-1)^{2}}\) + \(\frac{1}{x-2}\)

Question 7.

Resolve \(\frac{5 x+1}{(x+2)(x-1)}\) into Partial fractions.

Solution:

⇒ 5x + 1 = A(x – 1) + B(x + 2) —– (1)

Put x = 1 in(1)

5(1) + 1,= A(0) + B(1 + 2)

⇒ 3B = 6 ⇒ B = 2

Put x = -2 in (1)

5(-2) + 1 = A (-2 – 1) + B(0)

⇒ -9 = -3A ⇒ A = 3

∴ \(\frac{5 x+1}{(x+2)(x-1)}\) = \(\frac{3}{x+2}\) + \(\frac{2}{x-1}\)

Question 8.

Resolve \(\frac{2 x+3}{5(x+2)(2 x+1)}\) into Partial fractions.

Solution:

Question 9.

Resolve \(\frac{13 x+43}{2 x^{2}+17 x+30}\) into partial fractions.

Solution:

2x^{2} + 17x + 30 = 2x^{2} + 12x + 5x + 30

= 2x(x + 6) + 5(x + 6)

= (x + 6) (2x + 5)

Let \(\frac{13 x+43}{(x+6)(2 x+5)}\) = \(\frac{A}{x+6}\) + \(\frac{B}{2x+5}\)

∴ 13x + 43 = A(2x + 5) + B(x + 6)

putting x = -6

-78 + 43 = A(-12 + 5)

-35 = -7A

⇒ A = 5

Putting x = \(\frac{-5}{2}\)

13\(\left(\frac{-5}{2}\right)\) + 43 = \(B\left(\frac{-5}{2}+6\right)\)

⇒ -65 + 86 = B(—5 + 12)

⇒ 21 = 7B

⇒ B = 3

∴ \(\frac{13 x+43}{2 x^{2}+17 x+30}\) = \(\frac{5}{x+6}\) + \(\frac{3}{2 x+5}\)

Question 10.

Resolve \(\frac{x^{2}+5 x+7}{(x-3)^{3}}\) into partial fractions.

Solution:

Let x – 3 = y ⇒ x = y + 3

\(\frac{x^{2}+5 x+7}{(x-3)^{3}}\) = \(\frac{(y+3)^{2}+5(y+3)+7}{y^{3}}\)

Question 11.

Resolve \(\frac{x^{2}+13 x+15}{(2 x+3)(x+3)^{2}}\) into partial fractions.

Solution:

Question 12.

Resolve \(\frac{1}{(x-1)^{2}(x-2)}\) into Partial fractions.

Solution:

Put x = 1 in (1)

1 = A (0) + B(1 – 2) + C(0)

⇒ -B = 1 ⇒ B = -1

Put x = 2 in (1)

1 = A(0) + B(0) + C(2 – 1)^{2}

⇒ C = 1

Equating the coefficients of x^{2} in (1)

0 = A + C ⇒ A = -C = -1

A = -1

∴ \(\frac{1}{(x-1)^{2}(x-2)}\) = \(\frac{-1}{x-1}\) – \(\frac{1}{(x-1)^{2}}\) + \(\frac{1}{x-2}\)

Question 13.

Resolve \(\frac{3 x-18}{x^{3}(x+3)}\) into Partial fractions.

Solution:

Put x = -3 in(1)

3(-3) – 18 = A(0) + B(0) + C(O) + D (-3)^{3}

⇒ -27D = -27 ⇒ D = 1

Put x = 0 in (1)

3(0) – 18 = A (0) + B(0) + C(0 + 3) + D(0)

⇒ 3C = -18 ⇒ C = -6

Equating the coefficients of x^{3} in (1)

0 = A + D

⇒ A = -D = -1 ⇒ A = -1

Equating the coefficients of x^{2} in (1)

0 = 3A + B

⇒ B = -3A = -3(-1) = 3 ⇒ B = 31

∴ \(\frac{3 x-18}{x^{3}(x+3)}\) = \(\frac{-1}{x}\) + \(\frac{3}{x^{2}}\) – \(\frac{6}{x^{3}}\) + \(\frac{1}{x+3}\)

Question 14.

Resolve \(\frac{x-1}{(x+1)(x-2)^{2}}\) into Partial fractions.

Solution:

Question 15.

Resolve \(\frac{2 x^{2}+1}{x^{3}-1}\) into partial fractions.

Solution:

Put x = 1 in (1)

2(1) + 1 = A(1 + 1 + 1) + (B + C)(0)

⇒ 3A = 3 ⇒ A = 1

Put x = 0 in (1)

0 + 1 = A(1) + (0 + C)(0 – 1)

⇒ 1 = A – C

⇒ 1 = 1 – C ⇒ C = 0

Equating the coefficients of x^{2} in (1)

2 = A + B

⇒ 2 = 1 + B ⇒ B = 1

Question 16.

Resolve \(\frac{x^{3}+x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+3\right)}\) into partial fractions.

Solution:

Comparing the coefficients of x^{3}, x^{2}, x and constant terms

A + C = 1, B + D = 1, 3A + 2C = 0, 3B + 2D = 1

Solve

Question 17.

Resolve \(\frac{3 x^{3}-2 x^{2}-1}{x^{4}+x^{2}+1}\) into Partial fractions.

Solution:

A – B + C + D = 0 —— (3)

B + D = -1 ——- (4) D = -1 – B

Substitute C, D in (2)

-A + B + 3 – A – 1 – B = -2

⇒ -2A = -4 ⇒ A = 2

Substitute C, D in (3)

A – B + 3 – A – 1 – B = 0 ⇒ 2 = 2B ⇒ B = 1

∴ C = 3 – 2 = 1, D = -1 – 1 = -2

Ax + B = 2x + 1, Cx + D = x – 2

Question 18.

Resolve \(\frac{x^{4}+24 x^{2}+28}{\left(x^{2}+1\right)^{3}}\) into partial fractions.

Solution:

Question 19.

Resolve \(\frac{x+3}{(1-x)^{2}\left(1+x^{2}\right)}\) into partial fractions.

Solution:

Let \(\frac{x+3}{(1-x)^{2}\left(1+x^{2}\right)}\)

= \(\frac{\mathrm{A}}{1-x}\) + \(\frac{B}{(1-x)^{2}}\) + \(\frac{C x+D}{1+x^{2}}\)

= x + 3 = A(1 – x) (1 – x^{2}) + B(1 + x^{2}) + (Cx + D)(1 – x^{2})

Comparing the coefficients of like powers of x, we get

A + B + D = 3 → (1)

-A + C – 2D = 1 → (2)

A + B – 2C + D = 0 → (3)

-A + C = 0 → (4)

Solving these equations, we get

Question 20.

Resolve \(\frac{x^{3}}{(2 x-1)(x+2)(x-3)}\) into partial fractions.

Solution:

\(\frac{x^{3}}{(2 x-1)(x+2)(x-3)}\)

= \(\frac{1}{2}\) + \(\frac{A}{2 x-1}\) + \(\frac{B}{x+2}\) + \(\frac{c}{x-3}\)

Multiplying with 2(2x – 1) (x + 2) (x – 3)

2x^{3} = (2x – 1) (x + 2) (x – 3) + 2A(x + 2)

(x – 3) + 2B (2x – 1) (x – 3) + 2C (2x -1) (x + 2)

Question 21.

Resolve \(\frac{x^{4}}{(x-1)(x-2)}\) into partial fractions. (T.S. Mar. ’16, March – 2013)

Solution:

Equating the coefficients of (x – 1) (x – 2),

15x – 14 = A(x – 2) + B(x – 1)

Put x = 1, 15 – 14 = A(-1) ⇒ A = -1

Put x = 2, 30 – 14 = B(1) ⇒ B = 16

∴ \(\frac{x^{4}}{(x-1)(x-2)}\) = x^{2} + 3x + 7 – \(\frac{1}{x-1}\) + \(\frac{16}{x-2}\)

Question 22.

Find the coefficient of x^{4} in the expansion of \(\frac{3 x}{(x-2)(x+1)}\) in powers of x specifying the interval in which the expansion is valid.

Solution:

\(\frac{3 x}{(x-2)(x+1)}\) = \(\frac{A}{x-2}\) + \(\frac{B}{x+1}\)

Multiplying with (x – 2) (x + 1)

3x = A(x + 1) + B(x – 2)

Put x = -1, -3 = B(-3) ⇒ B = 1

Put x = 2, 6 = A(3) ⇒ A = 2

Question 23.

Find the coefficient of x^{n} in the power series expansion of \(\frac{x}{(x-1)^{2}(x-2)}\) specifying the region in which the expansion is valid.

Solution:

\(\frac{x}{(x-1)^{2}(x-2)}\) = \(\frac{A}{x-1}\) + \(\frac{B}{(x-1)^{2}}\) + \(\frac{C}{x-2}\)

Multiplying with (x – 1)^{2} (x – 2)

x = A(x – 1) (x – 2) + B(x – 2) + C(x – 1)^{2}

Put x = 1, 1 = B(-1) ⇒ B = -1

Put x = 2, 2 = C(1) ⇒ C = 2

Put x = 0, 0 = 2A – 2B + C ⇒ 2A = 2B – C

= -2 – 2 = -4 ⇒ A = -2