Mathematics

Resolve into partial fractions $$\displaystyle \frac{23x-11x^2}{(2x-1)(9-x^2)}$$


ANSWER

$$\displaystyle \frac { 1 }{ \left( 2x-1 \right) } +\frac { -1 }{ \left( 3-x \right) } +\frac { 4 }{ \left( 3+x \right) } $$


SOLUTION
We know $$\displaystyle \frac { 23x-11{ x }^{ 2 } }{ \left( 2x-1 \right) \left( 9-{ x }^{ 2 } \right)  } =\frac { 23x-11{ x }^{ 2 } }{ \left( 2x-1 \right) \left( 3-x \right) \left( 3+x \right)  } $$

Let $$\displaystyle \frac { 23x-11{ x }^{ 2 } }{ \left( 2x-1 \right) \left( x-3 \right) \left( x+3 \right)  } =\frac { A }{ \left( 2x-1 \right)  } +\frac { B }{ \left( 3-x \right)  } +\frac { C }{ \left( 3+x \right)  } $$ .....(1)

$$23x-11{ x }^{ 2 }=A\left( 3-x \right) \left( 3+x \right) +B\left( 2x-1 \right) \left( 3+x \right) +C\left( 2x-1 \right) \left( 3-x \right) \\ 23x-11{ x }^{ 2 }=A\left( { 9-x }^{ 2 } \right) +B\left( 2{ x }^{ 2 }+5x-3 \right) +C\left( -2{ x }^{ 2 }+7x-3 \right) $$
On comparing coefficients we get
$$-11=-A+2B-2C,23=5B+7C,0=9A-3B-3C$$
Solving them we get 
$$A=1,B=-1$$ and $$C=4$$
Putting this values in equation (1), we get
$$\dfrac {23x-11x^2}{(2x-1)(x-3)(x+3)}=\dfrac {1}{(2x-1)}+\dfrac {-1}{(3-x)}+\dfrac {4}{(3+x)}$$
Hence, option 'A' is correct.
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