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AP EAPCET 23-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 17 of 160

Marks: +1, -0

If

\frac{x^{3}}{(2x-1)(x+2)(x-3)}

=A+\frac{B}{2x-1}

+\frac{C}{x+2}+\frac{D}{x-3}

A=

$\frac{1}{2}$
$-\frac{1}{50}$
$-\frac{8}{25}$
$\frac{27}{25}$

Solution:

\frac{x^{3}}{(2x-1)(x+2)(x-3)}=\frac{A}{1}+\frac{B}{2x-1}

+\frac{C}{x+2}+\frac{D}{x-3}

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

\therefore A(2x-1)(x+2)(x-3)

+B(x+2)(x-3)

+C(x-3)(2x-1)+D(2x-1)(x+2)=x^{3}

Put

x=3

, we obtain

D(5)(5)=27 \Rightarrow D=\frac{27}{25}

Put

x=-2

, we obtain

C(-5)(-5)=-8 \Rightarrow C=-\frac{8}{25}

Put

x=\frac{1}{2}

, we obtain

B\left(\frac{5}{2}\right)\left(-\frac{5}{2}\right)=\frac{1}{8} \Rightarrow B=-\frac{1}{50}

Put

x=0

, we obtain

A(-1)(2)(-3)+\left(-\frac{1}{50}\right)(2)(-3)

+\left(-\frac{8}{25}\right)(-3)(-1)+\frac{27}{25}(-1)(2)=0

6A+\frac{3}{25}-\frac{24}{25}-\frac{54}{25}=0

\Rightarrow 6A=\frac{75}{25}=3

\Rightarrow A=\frac{3}{6}=\frac{1}{2}

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