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)

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