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ECAT Mathematics
Chapter 5: Partial Fractions
100 Questions
90 Minutes
Pass: 60%+
ECAT Prep
0
Partial fractions are used to decompose:
A
Products of polynomials
B
Rational fractions into simpler fractions
C
Sums of polynomials
D
Irrational expressions
1
The fraction (x+1)/((x-2)(x+3)) can be written as:
A
A/(x-2) + B/(x+3)
B
(Ax+B)/(x-2)(x+3)
C
A/(x-2)(x+3)
D
Ax/(x-2)+Bx/(x+3)
2
For a repeated linear factor (x-a)^2, partial fractions include:
A
A/(x-a)
B
A/(x-a)+B/(x-a)^2
C
Ax+B/(x-a)^2
D
A/(x-a)^2 only
3
For an irreducible quadratic factor (x^2+1), the partial fraction numerator is:
A
A constant
B
Ax+B
C
A/x+B
D
Ax^2+Bx+C
4
A proper fraction has degree of numerator:
A
Equal to denominator
B
Greater than denominator
C
Less than denominator
D
Equal to 1
5
An improper fraction must first be converted by:
A
Adding denominators
B
Performing polynomial long division
C
Cross multiplication
D
Expanding numerator
6
For (3x+5)/((x-1)(x+2)), if we write A/(x-1)+B/(x+2), then A =
A
8/3
B
3
C
8
D
3/8
7
For (3x+5)/((x-1)(x+2)), B = ?
A
-2/3
B
-1
C
2/3
D
1
8
The partial fraction of 1/(x(x+1)) is:
A
1/x - 1/(x+1)
B
1/x + 1/(x+1)
C
-1/x + 1/(x+1)
D
1/(x(x+1))
9
Which is NOT a valid case for partial fractions?
A
Linear distinct factors
B
Repeated linear factors
C
Irreducible quadratic factors
D
All choices are valid
10
For 2x/((x+1)(x-1)), partial fractions give:
A
1/(x-1)+1/(x+1)
B
1/(x+1)-1/(x-1)
C
2/(x-1)-2/(x+1)
D
1/(x^2-1)
11
The sum method for partial fractions involves:
A
Setting x to zero only
B
Multiplying both sides by factors and substituting roots
C
Using integration only
D
Differentiating both sides
12
For 1/(x^2-1) = 1/((x-1)(x+1)), partial fractions:
A
1/(2(x-1)) - 1/(2(x+1))
B
1/(x-1)+1/(x+1)
C
1/(2(x-1))+1/(2(x+1))
D
1/(x^2-1)
13
The degree of the denominator for partial fractions must be:
A
Less than degree of numerator
B
Greater than degree of numerator (proper fraction)
C
Equal to numerator
D
Zero
14
For (x+2)/((x-1)^2), partial fractions are:
A
A/(x-1)+B/(x-1)^2
B
A/(x-1)^2 only
C
A(x-1)+B
D
(Ax+B)/(x-1)^2
15
For (x+2)/((x-1)^2): multiply by (x-1)^2 and set x=1 to find B:
A
B=3
B
B=1
C
B=-1
D
B=2
16
The partial fraction form for 1/(x^2+x+1) (irreducible quadratic) numerator is:
A
A only
B
Ax
C
Ax+B
D
Ax^2+Bx+C
17
For decomposing x/((x^2+1)(x-1)), which form is correct?
A
(Ax+B)/(x^2+1) + C/(x-1)
B
A/(x^2+1)+B/(x-1)
C
(Ax+B)/(x^2+1)(x-1)
D
A/x+B/(x+1)+C/(x-1)
18
What is the number of constants in partial fraction of 2x/((x+1)(x+2)(x+3))?
A
2
B
3
C
4
D
6
19
If f(x)/g(x) is improper (deg f ≥ deg g), we write it as:
A
Quotient only
B
Quotient + proper fraction
C
Proper fraction only
D
Sum of partial fractions directly
20
The partial fraction of 3x/((x-1)(x+2)) gives A/(x-1)+B/(x+2). Find A:
A
1
B
3
C
2
D
4
21
The partial fraction of 3x/((x-1)(x+2)) gives B:
A
2
B
3
C
-1
D
1
22
(5x-3)/((x-1)(x+2)) = A/(x-1)+B/(x+2). A=?
A
2/3
B
2
C
3/2
D
1
23
For partial fractions, identity method means:
A
Setting coefficients of like powers equal after combining
B
Evaluating at specific points
C
Using calculus
D
Using matrices
24
The expression 1/(x(x^2+1)) = A/x + (Bx+C)/(x^2+1). Finding A:
A
A=1
B
A=-1
C
A=0
D
A=2
25
For 1/(x(x^2+1)), B and C are found by:
A
Setting x=i
B
Equating coefficients
C
Both of the above
D
Using calculus
26
A partial fraction decomposition is valid when:
A
Denominator is fully factored
B
Numerator degree ≥ denominator degree
C
Fraction is irrational
D
Denominator is a constant
27
For 2/(x^2-4) = 2/((x-2)(x+2)), partial fractions:
A
1/(2(x-2))-1/(2(x+2))
B
1/(x-2)+1/(x+2)
C
1/(2(x-2))+1/(2(x+2))
D
2/(x^2-4)
28
The fraction (x^2+1)/(x(x-1)) is:
A
Proper
B
Improper
C
Cannot be decomposed
D
Irrational
29
After long division, (x^2+1)/(x(x-1)) = x^2/(x^2-x) → quotient 1 remainder x+1. So:
A
1 + (x+1)/(x(x-1))
B
x + 1/(x(x-1))
C
1 + 1/(x(x-1))
D
x^2+1
30
For 5/((x-1)(x-2)), partial fractions: A/(x-1)+B/(x-2). A=?
A
-5
B
5
C
-5
D
3
31
For 5/((x-1)(x-2)), B=?
A
5
B
6
C
-5
D
1
32
For the partial fraction of 1/(x^3-x) = 1/(x(x-1)(x+1)):
A
1/x+1/(x-1)+1/(x+1)
B
−1/x+1/(2(x-1))−1/(2(x+1))
C
−1/x+1/(2(x-1))+1/(2(x+1))
D
Impossible to decompose
33
The condition for irreducible quadratic factor is:
A
b^2-4ac > 0
B
b^2-4ac = 0
C
b^2-4ac < 0
D
b=0
34
How many partial fraction constants does 1/((x-1)^3) have?
A
1
B
2
C
3
D
4
35
Partial fractions are useful in:
A
Calculus (integration), series expansion, signal processing
B
Only integration
C
Only series expansion
D
Only algebra
36
The method of partial fractions requires the denominator to be factored into:
A
Complex factors only
B
Real factors (linear and irreducible quadratic)
C
Only linear factors
D
Only quadratic factors
37
For (3x+2)/((x+1)(x^2+1)), the partial fraction form is:
A
A/(x+1)+(Bx+C)/(x^2+1)
B
A/(x+1)+B/(x^2+1)
C
(Ax+B)/(x+1)(x^2+1)
D
A/(x+1)+B/x+C/(x+1)
38
For A/(x+1)+(Bx+C)/(x^2+1) decomposition of (3x+2)/((x+1)(x^2+1)), A is found by setting x=-1:
A
A=(-3+2)/(1+1)=-1/2
B
A=1/2
C
A=-1
D
A=1
39
Which of these is a repeated linear factor?
A
(x-1)(x-2)
B
(x-1)^2
C
x^2+1
D
(x+1)(x+2)(x+3)
40
5x-1 / (x^2+x-2) = 5x-1/((x-1)(x+2)). If partial fraction = A/(x-1)+B/(x+2), then A:
A
A=(5-1)/(1+2)=4/3
B
A=(5-1)/3=4/3
C
A=3
D
A=4
41
For B in Q41: B = ?
A
B=(5(-2)-1)/((-2)-1)=(-11)/(-3)=11/3
B
B=11
C
B=-11/3
D
B=3/11
42
The partial fraction of x/(x^2-1) = 1/2 × 1/(x-1) + ?
A
1/2 × 1/(x+1)
B
−1/2 × 1/(x+1)
C
1/(x+1)
D
1
43
Partial fractions technique always requires the rational expression to be:
A
Proper
B
Improper
C
Equal degree
D
Irrational
44
For 1/((x-a)(x-b)) where a≠b:
A
1/(a-b) × (1/(x-a) - 1/(x-b))
B
1/(a-b) × (1/(x-b) - 1/(x-a))
C
1/(a+b) × ...
D
(1/(x-a)+1/(x-b))/(a-b)
45
The partial fraction expansion helps evaluate which type of integral?
A
Only polynomial integrals
B
Rational function integrals
C
Only trigonometric integrals
D
Only exponential integrals
46
For (2x^2+3)/(x^3+x), factoring gives:
A
x(x^2+1)
B
x(x+1)(x-1)
C
x^2(x+1)
D
x(x^2-1)
47
For (2x^2+3)/(x(x^2+1))=A/x+(Bx+C)/(x^2+1). A equals:
A
3
B
2
C
1
D
0
48
After finding A=3, equating: Bx^2+Cx+Ax^2+A=2x^2+3. Coefficient of x^2: A+B=2. So B=?
A
-1
B
2
C
1
D
0
49
And C from coefficient of x in (2x^2+3)/(x(x^2+1)) = 3/x+(-x+C)/(x^2+1). Coefficient of x: C=0.
A
C=0
B
C=1
C
C=-1
D
C=3
50
The expression (x^2+x+1)/(x(x^2+1)) is decomposed. A=?
A
A=1
B
A=0
C
A=2
D
A=3
51
Partial fractions are based on the theorem that:
A
Every rational function can be written as sum of simpler fractions
B
All polynomials are irreducible
C
Only real-valued fractions exist
D
Polynomials always factor completely over integers
52
For (x-1)/(x+1), this is improper since:
A
Degrees are equal
B
Numerator degree > denominator degree
C
Denominator is zero
D
It has no real roots
53
Long division of (x-1)/(x+1) gives:
A
1 - 2/(x+1)
B
1 + 2/(x+1)
C
x - 1/(x+1)
D
x + 1/(x+1)
54
For 4/(x^2-4), the simplest decomposition is:
A
1/(x-2) - 1/(x+2)
B
2/(x-2) + 2/(x+2)
C
1/(x-2) + 1/(x+2)
D
4/(x-2)(x+2)
55
Partial fractions require the denominator to have:
A
Only one factor
B
Fully factored form
C
Common factors with numerator
D
Degree ≥ 3
56
For (x^2-3x+2)/((x-1)(x+1)(x-2)), there is a common factor:
A
(x-1) in numerator and denominator
B
(x-2) only
C
(x+1)
D
No common factor
57
After cancellation, (x^2-3x+2)/((x-1)(x+1)(x-2))=
A
(x-2)/((x+1)(x-2))
B
1/(x+1)
C
(x-2)/((x+1))
D
1/((x-1)(x+1))
58
For 1/(x-1)^2, the partial fraction is:
A
A/(x-1)^2 only
B
A/(x-1)+B/(x-1)^2
C
A/(x-1)-B/(x-1)^2
D
Not decomposable
59
For x/((x-1)(x-2)^2), the form is:
A
A/(x-1)+B/(x-2)+C/(x-2)^2
B
A/(x-1)+B/(x-2)^2
C
A/(x-2)+B/(x-1)
D
A/x+B/(x-1)+C/(x-2)
60
For A/(x-1)+B/(x-2)+C/(x-2)^2 where numerator=x. At x=1: A=?
A
A=1/(-1)=-1
B
A=1
C
A=-1
D
A=1/1=1
61
The method of equating coefficients in partial fractions is called:
A
Cover-up method
B
Identity method (comparing coefficients)
C
Heaviside method
D
Substitution method
62
The cover-up (Heaviside) method works directly for:
A
Repeated factors
B
Distinct linear factors
C
Irreducible quadratics
D
All types
63
(3x-2)/((x+1)(2x-1)). At x=-1: A(2(-1)-1)=3(-1)-2. A=?
A
A=5/3
B
A=-5/3
C
A=5
D
A=1
64
(3x-2)/((x+1)(2x-1)). At x=1/2: B=?
A
B=(-1/2)/(3/2)=-1/3
B
B=1/3
C
B=-1/3
D
B=1/2
65
For x/(x^2+x-2)=x/((x+2)(x-1)). Partial fractions:
A
A/(x+2)+B/(x-1)
B
(Ax+B)/(x^2+x-2)
C
A/(x+2)(x-1)
D
Impossible
66
A for x/((x+2)(x-1)) at x=-2: A=?
A
A=(-2)/((-2)-1)=2/3
B
A=-2/3
C
A=2/3
D
A=-3
67
B for x/((x+2)(x-1)) at x=1: B=?
A
B=1/(1+2)=1/3
B
B=1
C
B=-1/3
D
B=2/3
68
The number of partial fraction terms equals:
A
Number of factors in denominator (counting multiplicity)
B
Degree of denominator
C
Degree of numerator
D
Number of roots of numerator
69
For complex partial fractions (with imaginary roots), we prefer:
A
Complex decomposition
B
Real decomposition using quadratic factors
C
Only real linear factors
D
Not applicable
70
The expression 1/x^2 is already in partial fraction form as:
A
1/(x-1)+1/(x+1)
B
1/x-1/(x+1)
C
Cannot be simplified
D
1/x^2 itself
71
For (2x^2-1)/((x-1)(x+1)(x-2)), degree of numerator is:
A
2
B
3
C
1
D
0
72
Degree of denominator (x-1)(x+1)(x-2) is:
A
2
B
3
C
1
D
4
73
Since deg(num)=2 < deg(den)=3, the fraction is:
A
Improper
B
Proper
C
Equal
D
Undefined
74
The partial fraction of 2/(x^2-3x+2) = 2/((x-1)(x-2)):
A
2/(x-1)-2/(x-2)
B
2/(x-2)-2/(x-1)
C
1/(x-1)-1/(x-2)
D
1/(x-2)-1/(x-1)
75
In telescoping series, partial fractions of 1/n(n+1) = 1/n - ?
A
1/(n+1)
B
1/(n-1)
C
1/n
D
n/(n+1)
76
The partial fraction of (x+3)/((x+1)(x+2)):
A
A/(x+1)+B/(x+2)
B
(Ax+B)/(x+1)(x+2)
C
A/(x+1)^2
D
Not possible
77
A for (x+3)/((x+1)(x+2)) at x=-1:
A
A=2
B
A=-2
C
A=1
D
A=3
78
B for (x+3)/((x+1)(x+2)) at x=-2:
A
B=1
B
B=-1
C
B=2
D
B=-2
79
So (x+3)/((x+1)(x+2)) = 2/(x+1) - 1/(x+2). Check at x=0:
A
(0+3)/((0+1)(0+2)) = 3/2 and 2-1/2=3/2 ✓
B
3/2 and 2+1/2 ✗
C
3 and 3/2 ✗
D
Undefined
80
Partial fraction of 1/(x^2+4) (irreducible quadratic) is:
A
A/(x+2)+B/(x-2)
B
(Ax+B)/(x^2+4)
C
1/(x+2i)(x-2i) (complex)
D
Not possible in reals
81
But 1/(x^2+4) is already in its simplest real partial fraction form:
A
False, it can be simplified
B
True, since denominator is irreducible
C
Only if A=0
D
Only if B=1
82
The number of unknowns in partial fraction of 5x/((x-1)^2(x+2)) is:
A
2
B
3
C
4
D
1
83
For 5x/((x-1)^2(x+2)), at x=-2 (for C):
A
C=(-10)/((-2-1)^2)=(-10)/9
B
C=10/9
C
C=-10/9
D
C=10
84
Partial fractions are a special case of:
A
Matrix decomposition
B
Algebraic fraction decomposition into simpler terms
C
Polynomial factorization
D
Completing the square
85
The expression (2x+1)/((x-1)^2) decomposed. B (the numerator of 1/(x-1)^2) found by setting x=1:
A
B=3
B
B=2
C
B=1
D
B=-1
86
For A in (2x+1)/((x-1)^2)=A/(x-1)+B/(x-1)^2, differentiate or equate x^1 coefficient. A=?
A
A=2
B
A=1
C
A=3
D
A=0
87
The identity (2x+1)=A(x-1)+B with B=3, A=2 checks at x=0:
A
2(0)+1=2(-1)+3 → 1=1 ✓
B
1=1 ✗
C
0=1 ✗
D
Cannot check
88
For integration of rational functions, partial fractions convert to:
A
Products
B
Simpler fractions (like A/(x-a)) easy to integrate
C
Derivatives
D
Series
89
∫(1/(x(x+1)))dx after partial fractions (1/x-1/(x+1)) =
A
ln|x|-ln|x+1|+C
B
ln|x|+ln|x+1|+C
C
-ln|x|+ln|x+1|+C
D
ln|x+1|-ln|x|+C
90
The expansion of (2x+3)/((x^2+1)(x+2)): number of constants needed:
A
2
B
3
C
4
D
1
91
Which factor in the denominator needs a linear numerator in partial fractions?
A
Linear factor (x-a)
B
Repeated linear factor (x-a)^2
C
Irreducible quadratic (ax^2+bx+c)
D
All factors
92
The expression (x^3+1)/(x^2-1) is:
A
Proper
B
Improper
C
Neither
D
Undefined
93
After dividing x^3+1 by x^2-1:
A
x + (x+1)/(x^2-1)
B
x + x/(x^2-1)
C
x + 1/(x^2-1)
D
x + (x^2+1)/(x^2-1)
94
Partial fractions of (x+1)/(x^2-1)=(x+1)/((x-1)(x+1)):
A
1/(x-1) (after cancellation)
B
A/(x-1)+B/(x+1)
C
Not possible
D
(x+1)/(x-1)(x+1)
95
The partial fraction method works over:
A
Only integers
B
Only reals
C
Any field (reals, rationals, etc.)
D
Only complex numbers
96
For (x-2)/((x^2-4)) = (x-2)/((x-2)(x+2)):
A
1/(x-2)
B
1/(x+2)
C
(x-2)/(x+2)
D
Cannot simplify
97
The partial fraction technique is also used in:
A
Finding Laplace transform inverses
B
Computing derivatives only
C
Simplifying square roots
D
Solving differential equations by separation
98
For (ax+b)/((x-p)(x-q)) with p≠q, partial fractions give A/(x-p)+B/(x-q) where A=
A
(ap+b)/(p-q)
B
(aq+b)/(q-p)
C
(ap+b)/(q-p)
D
(aq+b)/(p-q)
99
The sum of partial fractions must equal:
A
The original rational expression
B
Zero
C
One
D
The denominator only
0
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