TS EAMCET PYQs for Integration by Partial Fractions with Solutions: Practice TS EAMCET Previous Year Questions

Updated 3+ months ago

Integration by Partial Fractions is an important topic in the Mathematics section in TS EAMCET exam. Practising this topic will increase your score overall and make your conceptual grip on TS EAMCET exam stronger.

This article gives you a full set of TS EAMCET PYQs for Integration by Partial Fractions with explanations for effective preparation. Practice of TS EAMCET Mathematics PYQs including Integration by Partial Fractions questions regularly will improve accuracy, speed, and confidence in the TS EAMCET 2026 exam.

TS EAMCET PYQs for Integration by Partial Fractions with Solutions

1.
If $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, then $A+B+C+D=$
- 0
- 1
- -1
- 6
View Solution
2.
If \[ \frac{3x^4 - 2x^2 +1}{(x-2)^4} = A + \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}, \] then $2A + 3B - C - D + E =\;?$
- $0$
- $1$
- $-11$
- $-39$
View Solution
3.
If $\frac{x^2}{2x^2 + 7x + 6} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{ax^2 + 3}$, then $A + B + C - 2D =$
- $2a$
- $-2a$
- $-4a$
- $4a$
View Solution
4.
If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$ then $a-b+c =$
- 0
- 1
- 3
- 2
View Solution
5.
If \[ \frac{x^4}{(x^2+1)(x-2)} = f(x) + \frac{Ax+B}{x^2+1} + \frac{C}{x-2} \] then $ f(14) + 2A - B = $ ?
- $5C$
- $4C$
- $6C$
- $7C$
View Solution
6.
If $\frac{x^{2} + 5}{\left(x^{2} + 1 \right)\left(x-2\right)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2} + 1 } , $ then $A + B + C = $
- -1
- $\frac{2}{5}$
- $\frac{-3}{5}$
- 0
View Solution
7.
If $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$
- 0
- 1
- 2
- -1
View Solution

View more TS EAMCET Questions

SHARE ON FACEBOOK SHARE ON TWITTER

Comments

No Comments To Show

SUBSCRIBE TO OUR NEWS LETTER

COLLEGE NOTIFICATIONS

EXAM NOTIFICATIONS

NEWS UPDATES

Download the Collegedunia app on

TS EAMCET PYQs for Integration by Partial Fractions with Solutions: Practice TS EAMCET Previous Year Questions

TS EAMCET PYQs for Integration by Partial Fractions with Solutions

1.
If $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, then $A+B+C+D=$

2.
If \[ \frac{3x^4 - 2x^2 +1}{(x-2)^4} = A + \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}, \] then \(2A + 3B - C - D + E =\;?\)

3.
If \(\frac{x^2}{2x^2 + 7x + 6} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{ax^2 + 3}\), then \(A + B + C - 2D =\)

4.
If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$ then $a-b+c =$

5.
If \[ \frac{x^4}{(x^2+1)(x-2)} = f(x) + \frac{Ax+B}{x^2+1} + \frac{C}{x-2} \] then \( f(14) + 2A - B = \) ?

6.
If $\frac{x^{2} + 5}{\left(x^{2} + 1 \right)\left(x-2\right)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2} + 1 } , $ then $A + B + C = $

7.
If $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$

Comments

SUBSCRIBE TO OUR NEWS LETTER

TS EAMCET PYQs for Integration by Partial Fractions with Solutions

1.If $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, then $A+B+C+D=$

2.If \[ \frac{3x^4 - 2x^2 +1}{(x-2)^4} = A + \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}, \] then \(2A + 3B - C - D + E =\;?\)

3. If \(\frac{x^2}{2x^2 + 7x + 6} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{ax^2 + 3}\), then \(A + B + C - 2D =\)

4.If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$ then $a-b+c =$

5.If \[ \frac{x^4}{(x^2+1)(x-2)} = f(x) + \frac{Ax+B}{x^2+1} + \frac{C}{x-2} \] then \( f(14) + 2A - B = \) ?

6.If $\frac{x^{2} + 5}{\left(x^{2} + 1 \right)\left(x-2\right)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2} + 1 } , $ then $A + B + C = $

7.If $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, then $A+B+C+D=$

2.
If \[ \frac{3x^4 - 2x^2 +1}{(x-2)^4} = A + \frac{B}{x-2} + \frac{C}{(x-2)^2} + \frac{D}{(x-2)^3} + \frac{E}{(x-2)^4}, \] then \(2A + 3B - C - D + E =\;?\)

3.
If \(\frac{x^2}{2x^2 + 7x + 6} = \frac{Ax + B}{x^2 + a} + \frac{Cx + D}{ax^2 + 3}\), then \(A + B + C - 2D =\)

4.
If $\frac{x+3}{(x+1)(x^2+2)} = \frac{a}{x+1} + \frac{bx+c}{x^2+2}$ then $a-b+c =$

5.
If \[ \frac{x^4}{(x^2+1)(x-2)} = f(x) + \frac{Ax+B}{x^2+1} + \frac{C}{x-2} \] then \( f(14) + 2A - B = \) ?

6.
If $\frac{x^{2} + 5}{\left(x^{2} + 1 \right)\left(x-2\right)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2} + 1 } , $ then $A + B + C = $

7.
If $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$