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$\operatorname{In} \frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{x+p}+\frac{b}{x+q}$ then $2(a+b)=$
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$\mathrm{f}(4)$
Given :
$\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{(x+p)}+\frac{b}{(x-q)}$ ...(i)
$\begin{aligned} & \therefore \quad \frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2} \\ & =\left(x^2-5 x+6\right)+\frac{x-8}{x^2-x-2} \\ & =\left(x^2-5 x+6\right)+\frac{x-8}{(x+1)(x-2)} \\ & \Rightarrow \frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}\end{aligned}$
$=\left(x^2-5 x+6\right)+\frac{3}{x+1}-\frac{2}{x-2}$ ...(ii)
Comparing eqns. (i) and (ii), we get
$\begin{aligned} & f(x)=x^2-5 x+6, a=3, b=-2 \\ & p=1, q=2 \\ & 2(a+b)=2(3-2)=2 \times 1=2 \\ & \text { Now, } f(7)=49-35+6=20 \\ & f(6)=36-30+6=12 \\ & f(5)=25-25+6=6 \\ & f(4)=16-20+6=2 \\ & \therefore \quad 2(a+b)=f(4) .\end{aligned}$
$\frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}=f(x)+\frac{a}{(x+p)}+\frac{b}{(x-q)}$ ...(i)
$\begin{aligned} & \therefore \quad \frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2} \\ & =\left(x^2-5 x+6\right)+\frac{x-8}{x^2-x-2} \\ & =\left(x^2-5 x+6\right)+\frac{x-8}{(x+1)(x-2)} \\ & \Rightarrow \frac{x^4-6 x^3+9 x^2+5 x-20}{x^2-x-2}\end{aligned}$
$=\left(x^2-5 x+6\right)+\frac{3}{x+1}-\frac{2}{x-2}$ ...(ii)
Comparing eqns. (i) and (ii), we get
$\begin{aligned} & f(x)=x^2-5 x+6, a=3, b=-2 \\ & p=1, q=2 \\ & 2(a+b)=2(3-2)=2 \times 1=2 \\ & \text { Now, } f(7)=49-35+6=20 \\ & f(6)=36-30+6=12 \\ & f(5)=25-25+6=6 \\ & f(4)=16-20+6=2 \\ & \therefore \quad 2(a+b)=f(4) .\end{aligned}$
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