Mathematics Multiple Choice Questions on “Integration by Partial Fractions”.
1. What form of rational function (frac{px+q}{(x-a)(x-b)}), a≠b represents?
a) (frac{A}{(x-a)})
b) (frac{B}{(x-b)})
c) (frac{A+B}{(x-a)(x-b)})
d) (frac{A}{(x-a)} + frac{B}{(x-b)})
Answer: d
Clarification: The given function (frac{px+q}{(x-a)(x-b)}), a≠b can also be written as
(frac{A}{(x-a)} + frac{B}{(x-b)}) and is further used to solve integration by partial fractions numerical.
2. Find (int frac{x^2+1}{x^2-5x+6} dx).
a) x – 5log|x-2| + 10log|x-3|+C
b) x – 3log|x-2| + 5log|x-3|+C
c) x – 10log|x-2| + 5log|x-3|+C
d) x – 5log|x-5| + 10log|x-10|+C
Answer: a
Clarification: As it is not proper rational function, we divide numerator by denominator and get
(frac{x^2+1}{x^2-5x+6} = 1-frac{5x-5}{x^2-5x+6} = 1+frac{5x-5}{(x-2)(x-3)})
Let (frac{5x-5}{(x-2)(x-3)}=frac{A}{(x-2)} + frac{B}{(x-3)})
So that, 5x–5 = A(x-3) + B(x-2)
Now, equating coefficients of x and constant on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A=-5 and B=10.
Therefore, (frac{x^2+1}{x^2-5x+6} = 1 – frac{5}{(x-2)} + frac{10}{(x-3)}).
(int frac{x^2+1}{x^2-5x+6} dx = int dx – 5int frac{dx}{(x-2)} + 10int frac{dx}{(x-3)}).
= x – 5log|x-2| + 10log|x-3|+C
3. Find (int frac{dx}{(x+1)(x+2)}).
a) (Log left|frac{x+1}{x+2}right|+ C)
b) (Log left|frac{x-1}{x+2}right|+ C)
c) (Log left|frac{x+2}{x+1}right|+ C)
d) (Log left|frac{x+1}{x-2}right|+ C)
Answer: a
Clarification: It is a proper rational function. Therefore,
(frac{1}{(x+1)(x+2)} = frac{A}{(x+1)} + frac{B}{(x+2)})
Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and the constant term, we get A+B = 0 and 2A+B = 1. Solving it we get A=1, and B=-1.
Thus, it simplifies to, (frac{1}{(x+1)} + frac{-1}{(x+2)} = int frac{dx}{(x+1)} – int frac{dx}{(x+2)}).
= log|x+1| – log|x+2| + C
= (Log left|frac{x+1}{x+2}right|+ C).
4. An improper integration fraction is reduced to proper fraction by _____
a) multiplication
b) division
c) addition
d) subtraction
Answer: b
Clarification: An improper integration factor can be reduced to proper fraction by division, i.e., if the numerator and denominator have same degree, then they must be divided in order to reduce it to proper fraction.
5. (int frac{dx}{x(x^2+1)}) equals ______
a) (log|x| – frac{1}{2} log(x^2+1)) + C
b) (log|x| + frac{1}{2} log(x^2+1)) + C
c) –(log|x| + frac{1}{2} log(x^2+1)) + C
d) (frac{1}{2} log|x| + log(x^2+1)) + C
Answer: a
Clarification: We know that (int frac{dx}{x(x^2+1)} = frac{A}{x} + frac{Bx+C}{x^2+1})
By simplifying it we get, (int frac{dx}{x(x^2+1)}=frac{(A+B) x^2+Cx+A}{x(x^2+1)})
Now equating the coefficients we get A = 0, B = 0, C=1.
(int frac{dx}{x(x^2+1)} = int frac{dx}{x} + int frac{-xdx}{(x^2+1)})
Therefore after integrating we get (log|x| – frac{1}{2} log(x^2+1)) + C.
6. (int frac{dx}{(x^2-9)}) equals ______
a) (frac{1}{6} log frac{x+3}{x-3}) + C
b) (frac{1}{6} log frac{x-3}{x+3}) + C
c) (frac{1}{5} log frac{x+3}{x-3}) + C
d) (frac{1}{3} log frac{x+3}{x-3}) + C
Answer: b
Clarification: (int frac{dx}{(x^2-9)}=frac{A}{(x-3)} + frac{B}{(x+3)})
By simplifying, it we get (frac{A(x+3)+B(x-3)}{(x^2-9)} = frac{(A+B)x+3A-3B}{(x^2-9)})
By solving the equations, we get, A+B=0 and 3A-3B=1
By solving these 2 equations, we get values of A=1/6 and B=-1/6.
Now by putting values in the equation and integrating it we get value,
(frac{1}{6} log (frac{x-3}{x+3})) + C.
7. Which form of rational function (frac{px+q}{(x-a)^2}) represents?
a) (frac{A}{(x-a)} + frac{B}{(x-a)^2})
b) (frac{A}{(x-a)^2} + frac{B}{(x-a)})
c) (frac{A}{(x-a)} – frac{B}{(x-a)^2})
d) (frac{A}{(x-a)} – frac{B}{(x-a)})
Answer: a
Clarification: It is a form of the given partial fraction (frac{px+q}{(x-a)^2}) which can also be written as
(frac{A}{(x-a)} + frac{B}{(x-a)^2}) and is further used to solve integration by partial fractions numerical.
8. (int frac{(x^2+x+1)dx}{(x+2)(x^2+1)}) equals ______
a) (frac{3}{5}log|x+2| + frac{1}{5}log|x^2+1|+frac{1}{5} tan^{-1}x+5C)
b) (frac{3}{5}log|x+2| + frac{1}{5}log|x^2+1|+frac{1}{6} tan^{-1}x+C)
c) (frac{3}{5}log|x+2| + frac{1}{6}log|x^2+1|+frac{1}{6} tan^{-1}x+C)
d) (frac{3}{5}log|x+2| + frac{1}{5}log|x^2+1|+frac{1}{5} tan^{-1}x+C)
Answer: d
Clarification: (int frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = frac{A}{(x+2)} + frac{Bx+C}{(x^2+1)})
Now equating, (x2+x+1) = A (x2+1) + (Bx+C) (x+2)
After equating and solving for coefficient we get values,
A=(frac{3}{5}), B=(frac{2}{5}), C=(frac{1}{5}), now putting these values in the equation we get,
(int frac{(x^2+x+1)dx}{(x+2)(x^2+1)} = frac{3}{5} int frac{dx}{(x+2)} + frac{1}{5} int frac{2xdx}{(x^2+1)} + frac{1}{5} int frac{dx}{(x^2+1)})
Hence it comes, (frac{3}{5} log|x+2| + frac{1}{5} log|x^2+1|+frac{1}{5}tan^{-1}x+C)
9. Identify the type of the equation (x+1)2.
a) Linear equation
b) Cubic equation
c) Identity
d) Imaginary
Answer: c
Clarification: As it represents the identity (b+a)2 it satisfies the identity (b+a)2 = (a2 + b2 +2ab) and is not linear, cubic or an imaginary equation so the correct option is Identity Equation.
10. For the given equation (x+2) (x+4) = x2 + 6x + 8, how many values of x satisfies this equation?
a) Two values of x
b) One value of x
c) All value of x
d) No value of x
Answer: c
Clarification: If we solve the L.H.S. (Left Hand Side) of the equation, we get the following value.
(x+2) (x+4) = x2 + 4x + 2x + 8 = x2 + 6x + 8.
This value is same as the R.H.S. (Right Hand Side).
So, all the values of x satisfy the equality.