(a) Express \(\frac{9x}{(2x + 1)(x^2 + 1)}\) in partial fraction
(b) If \(^{2m}P_2\) – 10 = \(^m P_2\), find the positive value of m.
Explanation
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{A}{2x + 1}\) + \(\frac{Bx + C}{x^2 + 1}\) = \(\frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(2x + 1)(x^2 + 1)}\) - - - - -- - - -(i)
Using the cover-up method, put x = \(\frac{-1}{2}\)
A = \(\frac{9x}{x^2 + 1}\) = \(\frac{9(\frac{-1}{2})}{(\frac{-1}{2})^2 + 1}\) = \(\frac{\frac{-9}{2}}{\frac{5}{4}}\) = \(\frac{-18}{5}\)
From equation (i)
9x = A(\(x^2\) + 1) + (Bx + C)(x + 1)
9x = A\(x^2\) + A + 2B\(x^2\) + Bx + 2x + C
9x = (A + B)\(x^2\) + (B + 2C)x + (A + C)
comparing coefficients(x\(^2\)
A + 2B = 0
but A = \(\frac{-18}{5}\)
- A = 2B
-(\(\frac{-18}{5}\)) = 2B
18 = 10B
B = \(\frac{18}{10}\) = \(\frac{9}{5}\).
comparing coefficients (constant terms)
A + C = 0
C = - A = -(\(\frac{-18}{5}\))
C = \(\frac{18}{5}\)
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{\frac{-18}{5}}{2x + 1}\) + \(\frac{\frac{9x}{5} + \frac{18}{5}}{x^2 + 1}\)
= \(\frac{-18}{5(2x + 1)}\) + \(\frac{9x}{5(x^2 + 1)}\) + \(\frac{18}{5(x^2 + 1)}\)
= \(\frac{18}{5}\)[\(\frac{1}{x^2 + 1}\) + \(\frac{1}{2(x^2 + 1)}\) - \(\frac{1}{2x + 1}\)]
(b) \(^{2m}P_2\) - 10 = \(^m P_2\)
\(\frac{(2m)!}{(2m - 2)!}\) - 10 = \(\frac{m!}{(m - 2)!}\)
\(\frac{(2m)(2m - 1) \times (2m - 2)!}{(2m - 2)!}\) - 10 = \(\frac{m \times (m - 1) \times (m - 2)!}{(m - 2)!}\)
⇒ 2m(2m - 1) - 10 = m(m - 1)
4m\(^2\) - 2m - 10 = m\(^2\) - m
4m\(^2\) - m\(^2\) - 2m + m - 10 = 0
3 m\(^2\) - m - 10 = 0
(3m + 5)(m - 2) = 0
m = 2( +ve value only)