By, resolve the following term into partial fraction, we get:23x-11x22x-19- x2

1 2 x - 1 + 4 3 + x + - 1 3 - x

By, resolve the following term into partial fraction, we get:23x-11x22x-19- x2

1 2 x - 1 + 4 3 + x + - 1 3 - x

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By, resolve the following term into partial fraction, we get:
$\frac{23 x - 11 x^{2}}{(2 x - 1) (9 - x^{2})}$

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\frac{1}{2 x - 1} + \frac{4}{3 + x} + \frac{- 1}{3 - x}

\frac{1}{2 x - 1} + \frac{4}{3 + x} \frac{- 1}{3 - x}

\frac{1}{2 x - 1} \frac{4}{3 + x} + \frac{- 1}{3 - x}

\frac{1}{2 x - 1} \frac{4}{3 + x} \frac{- 1}{3 - x}

answer is A.

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Detailed Solution

To solve the given term by partial fraction method, let us assume

\frac{23 x - 11 x^{2}}{(2 x - 1) (9 - x^{2})}

\frac{A}{2 x - 1} + \frac{B}{3 + x} + \frac{C}{3 - x}

Taking LCM on the R.H.S. and then canceling the denominator, we get
∴ 23x – 11

x^{2}

= A (3 + x) (3 - x) + B (2x - 1) (3 - x) + C (2x - 1) (3 + x)
∴ 23x – 11

x^{2}

= A (9 −

x^{2}

) + B (7x - 2

x^{2}

− 3) + C (5x + 2

x^{2}

− 3)
∴ 23x – 11

x^{2}

x^{2}

(2C - 2B - A) + x (7B + 5C) + (9A - 3B - 3C) Now, equating the coefficients of like term
2C – 2B – A = -11
7B + 5C = 23
9A – 3B – 3C = 0
Solving above three equations for three variables, we get
A = 1, B = 4, C = -1
Hence, finally we have

\frac{23 x - 11 x^{2}}{(2 x - 1) (9 - x^{2})} = \frac{1}{2 x - 1} + \frac{4}{3 + x} + \frac{- 1}{3 - x}

So, option (1) is correct.

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