Question

If \( \frac{x+1}{x^3(x-1)} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x-1} \) then

• A. a = b = c = -d
• B. a = b = 2c = -d
• C. a = 2b = c = -d
• D. a = b = 2c = d

Hint:

When dealing with repeated linear factors in partial fractions like \(x^3\), the "cover-up" method only works for the highest power (finding 'c' by setting x=0). For the other coefficients (a and b), comparing coefficients or substituting other convenient values (like x=-1) is necessary.

Answer 1

The Correct Option is B

We start with the partial fraction decomposition identity:
\( x+1 = a x^2(x-1) + b x(x-1) + c(x-1) + d x^3 \).
To find the coefficients, we can substitute strategic values for x.
Let \(x = 1\): \(1+1 = d(1)^3 \implies 2 = d\).
Let \(x = 0\): \(0+1 = c(0-1) \implies 1 = -c \implies c = -1\).
Now, we compare the coefficients of the powers of x by expanding the identity.
\( x+1 = a(x^3 - x^2) + b(x^2 - x) + c(x-1) + dx^3 \).
\( x+1 = (a+d)x^3 + (-a+b)x^2 + (-b+c)x - c \).
Comparing the coefficient of \(x^3\): \(a+d = 0\). Since \(d=2\), we have \(a = -2\).
Comparing the coefficient of \(x^2\): \(-a+b = 0\). Since \(a=-2\), we have \(-(-2)+b=0 \implies b=-2\).
So we have the values: \(a = -2\), \(b = -2\), \(c = -1\), and \(d = 2\).
Now we check the given options with these values.
Option (B) states \(a = b = 2c = -d\).
\(a = -2\) and \(b = -2\), so \(a=b\) is true.
\(2c = 2(-1) = -2\), so \(a=b=2c\) is true.
\(-d = -(2) = -2\), so \(a=b=2c=-d\) is also true.
The relationship holds.