Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.1

Resolve Into Partial Fractions for the Following:

Question 1. Resolve \( \frac{3x+7}{x^2 - 3x + 2} \) into partial fractions.
Answer: First, we need to factor the denominator. The quadratic expression \( x^2 - 3x + 2 \) can be factored as \( (x-1)(x-2) \). So, the fraction becomes \( \frac{3x+7}{(x-1)(x-2)} \).
We set up the partial fraction decomposition as follows:
\( \frac{3x+7}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \) (Equation 1)
To find A and B, we multiply both sides by \( (x-1)(x-2) \):
\( 3x+7 = A(x-2) + B(x-1) \) (Equation 2)
Now, we substitute values for x that make the terms zero:
Put \( x = 2 \) into Equation 2:
\( 3(2)+7 = A(2-2) + B(2-1) \)
\( 6+7 = A(0) + B(1) \)
\( 13 = B \)
So, \( B = 13 \).
Put \( x = 1 \) into Equation 2:
\( 3(1)+7 = A(1-2) + B(1-1) \)
\( 3+7 = A(-1) + B(0) \)
\( 10 = -A \)
So, \( A = -10 \).
Substitute the values of A and B back into Equation 1:
\( \frac{3x+7}{(x-1)(x-2)} = \frac{-10}{x-1} + \frac{13}{x-2} \)
This can also be written as:
\( \frac{3x+7}{(x-1)(x-2)} = \frac{13}{x-2} - \frac{10}{x-1} \). This process simplifies complex fractions into easier-to-handle terms, which is especially useful in calculus for integration.
In simple words: First, break the bottom part of the fraction into simpler factors. Then, write the fraction as a sum of two new fractions, each with one of the simpler factors at the bottom. Find the missing numbers (A and B) by plugging in values of x that make parts of the equation zero.

🎯 Exam Tip: Always factor the denominator completely before setting up the partial fractions. When substituting values for x, pick the roots of the linear factors to quickly solve for the unknown coefficients.

Question 2. Resolve \( \frac{4 x+1}{(x-2)(x+1)} \) into partial fractions.
Answer: The denominator is already factored into linear factors, \( (x-2) \) and \( (x+1) \).
We set up the partial fraction decomposition as:
\( \frac{4 x+1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} \) (Equation 1)
Multiply both sides by \( (x-2)(x+1) \) to clear the denominators:
\( 4x+1 = A(x+1) + B(x-2) \) (Equation 2)
Now, we substitute values for x to find A and B:
Put \( x = -1 \) into Equation 2:
\( 4(-1)+1 = A(-1+1) + B(-1-2) \)
\( -4+1 = A(0) + B(-3) \)
\( -3 = -3B \)
So, \( B = 1 \).
Put \( x = 2 \) into Equation 2:
\( 4(2)+1 = A(2+1) + B(2-2) \)
\( 8+1 = A(3) + B(0) \)
\( 9 = 3A \)
So, \( A = 3 \).
Substitute the values of A and B back into Equation 1:
\( \frac{4 x+1}{(x-2)(x+1)} = \frac{3}{x-2} + \frac{1}{x+1} \). Choosing values for x that make a factor zero simplifies the equation significantly, allowing for direct calculation of one of the coefficients.
In simple words: The bottom part of the fraction is already split up. Write the fraction as two simpler ones. Then, find the unknown numbers by plugging in x-values that make the bottom parts zero.

🎯 Exam Tip: When the denominator has distinct linear factors, the "cover-up method" can also be used for a quicker mental check, but showing the full substitution steps is important for exam marks.

Question 3. Resolve \( \frac{1}{(x-1)(x+2)^{2}} \) into partial fractions.
Answer: The denominator has a linear factor \( (x-1) \) and a repeated linear factor \( (x+2)^2 \).
The partial fraction decomposition is set up as:
\( \frac{1}{(x-1)(x+2)^{2}} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \) (Equation 1)
Multiply both sides by \( (x-1)(x+2)^2 \):
\( 1 = A(x+2)^2 + B(x-1)(x+2) + C(x-1) \) (Equation 2)
Now, we substitute values for x to find A, B, and C:
Put \( x = 1 \) into Equation 2:
\( 1 = A(1+2)^2 + B(1-1)(1+2) + C(1-1) \)
\( 1 = A(3^2) + B(0) + C(0) \)
\( 1 = 9A \)
So, \( A = \frac{1}{9} \).
Put \( x = -2 \) into Equation 2:
\( 1 = A(-2+2)^2 + B(-2-1)(-2+2) + C(-2-1) \)
\( 1 = A(0) + B(0) + C(-3) \)
\( 1 = -3C \)
So, \( C = -\frac{1}{3} \).
To find B, we can compare coefficients. Expand Equation 2:
\( 1 = A(x^2+4x+4) + B(x^2+x-2) + C(x-1) \)
\( 1 = Ax^2 + 4Ax + 4A + Bx^2 + Bx - 2B + Cx - C \)
Rearrange by powers of x:
\( 1 = (A+B)x^2 + (4A+B+C)x + (4A-2B-C) \)
Compare the coefficient of \( x^2 \) on both sides. On the left side, the coefficient of \( x^2 \) is 0.
\( 0 = A+B \)
Since we know \( A = \frac{1}{9} \):
\( 0 = \frac{1}{9} + B \)
So, \( B = -\frac{1}{9} \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{1}{(x-1)(x+2)^{2}} = \frac{1/9}{x-1} + \frac{-1/9}{x+2} + \frac{-1/3}{(x+2)^2} \)
\( \frac{1}{(x-1)(x+2)^{2}} = \frac{1}{9(x-1)} - \frac{1}{9(x+2)} - \frac{1}{3(x+2)^2} \). When a linear factor is repeated, it needs a separate term for each power up to its highest power in the denominator.
In simple words: When a factor in the bottom repeats (like \( (x+2)^2 \)), you need one fraction for each power of that factor (like \( \frac{B}{x+2} \) and \( \frac{C}{(x+2)^2} \)). Solve for A, B, and C by using specific x-values and by matching up the numbers next to \( x^2 \), x, and the constant terms on both sides of the equation.

🎯 Exam Tip: For repeated linear factors, make sure to include terms for each power up to the highest power. Using a combination of substitution and comparing coefficients is often the most efficient way to find all unknowns.

Question 4. Resolve \( \frac{1}{x^{2}-1} \) into partial fractions.
Answer: First, factor the denominator. The expression \( x^2-1 \) is a difference of squares and can be factored as \( (x+1)(x-1) \).
So the fraction is \( \frac{1}{(x+1)(x-1)} \).
Set up the partial fraction decomposition:
\( \frac{1}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1} \) (Equation 1)
Multiply both sides by \( (x+1)(x-1) \):
\( 1 = A(x-1) + B(x+1) \) (Equation 2)
Substitute values for x to find A and B:
Put \( x = 1 \) into Equation 2:
\( 1 = A(1-1) + B(1+1) \)
\( 1 = A(0) + B(2) \)
\( 1 = 2B \)
So, \( B = \frac{1}{2} \).
Put \( x = -1 \) into Equation 2:
\( 1 = A(-1-1) + B(-1+1) \)
\( 1 = A(-2) + B(0) \)
\( 1 = -2A \)
So, \( A = -\frac{1}{2} \).
Substitute the values of A and B back into Equation 1:
\( \frac{1}{(x+1)(x-1)} = \frac{-1/2}{x+1} + \frac{1/2}{x-1} \)
\( \frac{1}{(x+1)(x-1)} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)} \). The ability to factor expressions like difference of squares is a fundamental skill for solving such problems.
In simple words: First, break down the bottom of the fraction using the "difference of squares" rule. Then, split the fraction into two simpler ones. Find the missing numbers (A and B) by picking x-values that make the denominators zero.

🎯 Exam Tip: Always remember that \( x^2 - a^2 \) factors into \( (x-a)(x+a) \). This is a common pattern that can simplify many partial fraction problems quickly.

Question 5. Resolve \( \frac{x-2}{(x+2)(x-1)^{2}} \) into partial fractions.
Answer: The denominator has a linear factor \( (x+2) \) and a repeated linear factor \( (x-1)^2 \).
Set up the partial fraction decomposition:
\( \frac{x-2}{(x+2)(x-1)^{2}} = \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \) (Equation 1)
Multiply both sides by \( (x+2)(x-1)^2 \):
\( x-2 = A(x-1)^2 + B(x+2)(x-1) + C(x+2) \) (Equation 2)
Substitute values for x to find A, B, and C:
Put \( x = 1 \) into Equation 2:
\( 1-2 = A(1-1)^2 + B(1+2)(1-1) + C(1+2) \)
\( -1 = A(0) + B(0) + C(3) \)
\( -1 = 3C \)
So, \( C = -\frac{1}{3} \).
Put \( x = -2 \) into Equation 2:
\( -2-2 = A(-2-1)^2 + B(-2+2)(-2-1) + C(-2+2) \)
\( -4 = A(-3)^2 + B(0) + C(0) \)
\( -4 = 9A \)
So, \( A = -\frac{4}{9} \).
To find B, compare coefficients. Expand Equation 2:
\( x-2 = A(x^2-2x+1) + B(x^2+x-2) + C(x+2) \)
\( x-2 = Ax^2 - 2Ax + A + Bx^2 + Bx - 2B + Cx + 2C \)
Group terms by powers of x:
\( x-2 = (A+B)x^2 + (-2A+B+C)x + (A-2B+2C) \)
Compare the coefficient of \( x^2 \) on both sides. On the left, the coefficient of \( x^2 \) is 0.
\( 0 = A+B \)
Since \( A = -\frac{4}{9} \):
\( 0 = -\frac{4}{9} + B \)
So, \( B = \frac{4}{9} \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{x-2}{(x+2)(x-1)^{2}} = \frac{-4/9}{x+2} + \frac{4/9}{x-1} + \frac{-1/3}{(x-1)^2} \)
\( \frac{x-2}{(x+2)(x-1)^{2}} = -\frac{4}{9(x+2)} + \frac{4}{9(x-1)} - \frac{1}{3(x-1)^2} \). Combining substitution with coefficient comparison is often the most efficient approach for problems with repeated factors.
In simple words: Break the fraction into three simpler ones because there's one normal factor and one factor that repeats twice on the bottom. Find the three missing numbers by plugging in specific x-values and matching the numbers that sit next to \( x^2 \), x, and the plain numbers.

🎯 Exam Tip: Remember to set up terms for each power of a repeated linear factor. For example, \( (x-1)^2 \) requires both \( \frac{B}{x-1} \) and \( \frac{C}{(x-1)^2} \).

Question 6. Resolve \( \frac{2 x^{2}-5 x-7}{(x-2)^{3}} \) into partial fractions.
Answer: The denominator is a repeated linear factor \( (x-2)^3 \).
The partial fraction decomposition is:
\( \frac{2 x^{2}-5 x-7}{(x-2)^{3}} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3} \) (Equation 1)
Multiply both sides by \( (x-2)^3 \):
\( 2x^2-5x-7 = A(x-2)^2 + B(x-2) + C \) (Equation 2)
Substitute values for x to find C:
Put \( x = 2 \) into Equation 2:
\( 2(2^2) - 5(2) - 7 = A(2-2)^2 + B(2-2) + C \)
\( 2(4) - 10 - 7 = A(0) + B(0) + C \)
\( 8 - 10 - 7 = C \)
\( -9 = C \).
To find A and B, compare coefficients. Expand Equation 2:
\( 2x^2-5x-7 = A(x^2-4x+4) + Bx - 2B + C \)
\( 2x^2-5x-7 = Ax^2 - 4Ax + 4A + Bx - 2B + C \)
Group terms by powers of x:
\( 2x^2-5x-7 = Ax^2 + (-4A+B)x + (4A-2B+C) \)
Compare the coefficient of \( x^2 \):
\( 2 = A \). So, \( A = 2 \).
Compare the coefficient of x:
\( -5 = -4A+B \)
Substitute \( A = 2 \):
\( -5 = -4(2)+B \)
\( -5 = -8+B \)
\( B = 8-5 \)
So, \( B = 3 \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{2 x^{2}-5 x-7}{(x-2)^{3}} = \frac{2}{x-2} + \frac{3}{(x-2)^2} + \frac{-9}{(x-2)^3} \)
\( \frac{2 x^{2}-5 x-7}{(x-2)^{3}} = \frac{2}{x-2} + \frac{3}{(x-2)^2} - \frac{9}{(x-2)^3} \). This method is direct and efficient for highly repeated factors.
In simple words: When the bottom factor is repeated three times (like \( (x-2)^3 \)), you need three fractions, one for each power (like \( (x-2) \), \( (x-2)^2 \), and \( (x-2)^3 \)). Find the missing numbers (A, B, C) by using an x-value that makes the factor zero, then match up the numbers next to \( x^2 \), x, and the plain numbers.

🎯 Exam Tip: For denominators with a single linear factor raised to a power, like \( (x-a)^n \), the numerator will always be constants (A, B, C, etc.) over each increasing power of the factor.

Question 7. Resolve \( \frac{x^{2}-6x+2}{x^{2}(x+2)} \) into partial fractions.
Answer: The denominator has a repeated linear factor \( x^2 \) and a linear factor \( (x+2) \).
The partial fraction decomposition is set up as:
\( \frac{x^{2}-6x+2}{x^{2}(x+2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+2} \) (Equation 1)
Multiply both sides by \( x^2(x+2) \):
\( x^2-6x+2 = Ax(x+2) + B(x+2) + C(x^2) \) (Equation 2)
Substitute values for x to find B and C:
Put \( x = 0 \) into Equation 2:
\( 0^2-6(0)+2 = A(0)(0+2) + B(0+2) + C(0^2) \)
\( 2 = A(0) + B(2) + C(0) \)
\( 2 = 2B \)
So, \( B = 1 \).
Put \( x = -2 \) into Equation 2:
\( (-2)^2-6(-2)+2 = A(-2)(-2+2) + B(-2+2) + C(-2)^2 \)
\( 4+12+2 = A(0) + B(0) + C(4) \)
\( 18 = 4C \)
So, \( C = \frac{18}{4} = \frac{9}{2} \).
To find A, compare coefficients. Expand Equation 2:
\( x^2-6x+2 = Ax^2+2Ax + Bx+2B + Cx^2 \)
Group terms by powers of x:
\( x^2-6x+2 = (A+C)x^2 + (2A+B)x + 2B \)
Compare the coefficient of \( x^2 \):
\( 1 = A+C \)
Substitute \( C = \frac{9}{2} \):
\( 1 = A + \frac{9}{2} \)
\( A = 1 - \frac{9}{2} = \frac{2-9}{2} = -\frac{7}{2} \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{x^{2}-6x+2}{x^{2}(x+2)} = \frac{-7/2}{x} + \frac{1}{x^2} + \frac{9/2}{x+2} \)
\( \frac{x^{2}-6x+2}{x^{2}(x+2)} = -\frac{7}{2x} + \frac{1}{x^2} + \frac{9}{2(x+2)} \). This breakdown makes complex rational functions more manageable for further mathematical operations.
In simple words: This fraction has \( x^2 \) at the bottom, which means you need two fractions for x (one for x, one for \( x^2 \)), plus another for the \( (x+2) \) part. Find the numbers (A, B, C) by using x-values that make parts zero and then by matching the numbers for \( x^2 \) and x.

🎯 Exam Tip: When the denominator includes \( x^n \), remember to include terms for \( x, x^2, \dots, x^n \) in your partial fraction setup, not just \( x^n \).

Question 8. Resolve \( \frac{x^{2}-3}{(x+2)(x^{2}+1)} \) into partial fractions.
Answer: The denominator has a linear factor \( (x+2) \) and an irreducible quadratic factor \( (x^2+1) \). An irreducible quadratic factor cannot be factored further into real linear terms.
The partial fraction decomposition is set up as:
\( \frac{x^{2}-3}{(x+2)(x^{2}+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^{2}+1} \) (Equation 1)
Multiply both sides by \( (x+2)(x^2+1) \):
\( x^2-3 = A(x^2+1) + (Bx+C)(x+2) \) (Equation 2)
Substitute values for x to find A:
Put \( x = -2 \) into Equation 2:
\( (-2)^2-3 = A((-2)^2+1) + (B(-2)+C)(-2+2) \)
\( 4-3 = A(4+1) + (0) \)
\( 1 = 5A \)
So, \( A = \frac{1}{5} \).
To find B and C, compare coefficients. Expand Equation 2:
\( x^2-3 = Ax^2+A + Bx^2+2Bx+Cx+2C \)
Group terms by powers of x:
\( x^2-3 = (A+B)x^2 + (2B+C)x + (A+2C) \)
Compare the coefficient of \( x^2 \):
\( 1 = A+B \)
Substitute \( A = \frac{1}{5} \):
\( 1 = \frac{1}{5} + B \)
\( B = 1 - \frac{1}{5} = \frac{4}{5} \).
Compare the coefficient of x:
\( 0 = 2B+C \)
Substitute \( B = \frac{4}{5} \):
\( 0 = 2(\frac{4}{5}) + C \)
\( 0 = \frac{8}{5} + C \)
So, \( C = -\frac{8}{5} \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{x^{2}-3}{(x+2)(x^{2}+1)} = \frac{1/5}{x+2} + \frac{(4/5)x + (-8/5)}{x^{2}+1} \)
\( \frac{x^{2}-3}{(x+2)(x^{2}+1)} = \frac{1}{5(x+2)} + \frac{4x-8}{5(x^{2}+1)} \). The presence of an irreducible quadratic factor changes the form of its numerator in the partial fraction.
In simple words: This fraction has a regular factor and one that cannot be easily broken down (like \( x^2+1 \)). For the regular factor, use a normal fraction with a number on top. For the special factor, use a fraction with an x-term and a number on top (like \( Bx+C \)). Find all the missing numbers by plugging in special x-values and matching numbers for \( x^2 \), x, and constants.

🎯 Exam Tip: For irreducible quadratic factors like \( ax^2+bx+c \), the numerator in the partial fraction must be of the form \( Px+Q \), not just a constant.

Question 9. Resolve \( \frac{x+2}{(x-1)(x+3)^{2}} \) into partial fractions.
Answer: The denominator has a linear factor \( (x-1) \) and a repeated linear factor \( (x+3)^2 \).
The partial fraction decomposition is set up as:
\( \frac{x+2}{(x-1)(x+3)^{2}} = \frac{A}{x-1} + \frac{B}{x+3} + \frac{C}{(x+3)^2} \) (Equation 1)
Multiply both sides by \( (x-1)(x+3)^2 \):
\( x+2 = A(x+3)^2 + B(x-1)(x+3) + C(x-1) \) (Equation 2)
Substitute values for x to find A and C:
Put \( x = 1 \) into Equation 2:
\( 1+2 = A(1+3)^2 + B(1-1)(1+3) + C(1-1) \)
\( 3 = A(4^2) + B(0) + C(0) \)
\( 3 = 16A \)
So, \( A = \frac{3}{16} \).
Put \( x = -3 \) into Equation 2:
\( -3+2 = A(-3+3)^2 + B(-3-1)(-3+3) + C(-3-1) \)
\( -1 = A(0) + B(0) + C(-4) \)
\( -1 = -4C \)
So, \( C = \frac{1}{4} \).
To find B, compare coefficients. Expand Equation 2:
\( x+2 = A(x^2+6x+9) + B(x^2+2x-3) + C(x-1) \)
\( x+2 = Ax^2+6Ax+9A + Bx^2+2Bx-3B + Cx-C \)
Group terms by powers of x:
\( x+2 = (A+B)x^2 + (6A+2B+C)x + (9A-3B-C) \)
Compare the coefficient of \( x^2 \):
\( 0 = A+B \)
Substitute \( A = \frac{3}{16} \):
\( 0 = \frac{3}{16} + B \)
So, \( B = -\frac{3}{16} \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{x+2}{(x-1)(x+3)^{2}} = \frac{3/16}{x-1} + \frac{-3/16}{x+3} + \frac{1/4}{(x+3)^2} \)
\( \frac{x+2}{(x-1)(x+3)^{2}} = \frac{3}{16(x-1)} - \frac{3}{16(x+3)} + \frac{1}{4(x+3)^2} \). Each term is handled separately based on its factor type.
In simple words: We have one simple factor and one factor that repeats twice. Set up three fractions, one for each part. Find the numbers by using specific x-values and by matching the number of \( x^2 \)'s, x's, and plain numbers on both sides.

🎯 Exam Tip: Always verify your final answer by combining the partial fractions back into a single fraction to ensure it matches the original expression.

Question 10. Resolve \( \frac{1}{\left(x^{2}+4\right)(x+1)} \) into partial fractions.
Answer: The denominator has a linear factor \( (x+1) \) and an irreducible quadratic factor \( (x^2+4) \).
The partial fraction decomposition is set up as:
\( \frac{1}{(x+1)(x^{2}+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+4} \) (Equation 1)
Multiply both sides by \( (x+1)(x^2+4) \):
\( 1 = A(x^2+4) + (Bx+C)(x+1) \) (Equation 2)
Substitute values for x to find A:
Put \( x = -1 \) into Equation 2:
\( 1 = A((-1)^2+4) + (B(-1)+C)(-1+1) \)
\( 1 = A(1+4) + (0) \)
\( 1 = 5A \)
So, \( A = \frac{1}{5} \).
To find B and C, compare coefficients. Expand Equation 2:
\( 1 = Ax^2+4A + Bx^2+Bx+Cx+C \)
Group terms by powers of x:
\( 1 = (A+B)x^2 + (B+C)x + (4A+C) \)
Compare the coefficient of \( x^2 \):
\( 0 = A+B \)
Substitute \( A = \frac{1}{5} \):
\( 0 = \frac{1}{5} + B \)
So, \( B = -\frac{1}{5} \).
Compare the coefficient of x:
\( 0 = B+C \)
Substitute \( B = -\frac{1}{5} \):
\( 0 = -\frac{1}{5} + C \)
So, \( C = \frac{1}{5} \).
Substitute the values of A, B, and C back into Equation 1:
\( \frac{1}{(x+1)(x^{2}+4)} = \frac{1/5}{x+1} + \frac{(-1/5)x + (1/5)}{x^{2}+4} \)
\( \frac{1}{(x+1)(x^{2}+4)} = \frac{1}{5(x+1)} + \frac{1-x}{5(x^{2}+4)} \). Recognizing irreducible quadratic factors is key to setting up the correct partial fraction form.
In simple words: This fraction has a simple factor and a special factor that cannot be broken down (like \( x^2+4 \)). Make one fraction for the simple factor with a number on top. For the special factor, put \( Bx+C \) on top. Find these numbers using a special x-value and by matching up the numbers next to \( x^2 \), x, and the plain numbers.

🎯 Exam Tip: An irreducible quadratic factor means you cannot find two real numbers that multiply to the constant term and add to the coefficient of x. In such cases, its numerator in the partial fraction must be a linear expression.