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If $y=\frac{a x+b}{c x+d}$ and $\frac{d x}{d y}=\frac{a d-b c}{P y^2+Q y+R}$ then $P+Q+R=$
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The correct answer is:
$(a-c)^2$
We are given that $\mathrm{y}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}}$
$\begin{aligned} & \Rightarrow c x y+d y=a x+b \\ & \Rightarrow(a-c y) x=d y-b \Rightarrow x=\frac{d y-b}{a-c y} \\ & \Rightarrow \frac{d x}{d y}=\frac{(a-c y) \cdot d-(d y-b) \cdot(-c)}{(a-c y)^2} \\ & \Rightarrow \frac{d x}{d y}=\frac{a d-c d y+c d y-b c}{(a-c y)^2}=\frac{a d-b c}{a^2+c 2 y^2-2 a c y}\end{aligned}$
Here $\mathrm{P}=\mathrm{C} 2, \mathrm{Q}=2 \mathrm{ac}, \mathrm{R}=\mathrm{Q}^2$
$\therefore \quad \mathrm{P}+\mathrm{Q}+\mathrm{R}=\mathrm{C}^2-2 \mathrm{ac}+\mathrm{a}^2=\mathrm{a}^2+\mathrm{c}^2-2 \mathrm{ac}=(\mathrm{a}-\mathrm{C})$
$\begin{aligned} & \Rightarrow c x y+d y=a x+b \\ & \Rightarrow(a-c y) x=d y-b \Rightarrow x=\frac{d y-b}{a-c y} \\ & \Rightarrow \frac{d x}{d y}=\frac{(a-c y) \cdot d-(d y-b) \cdot(-c)}{(a-c y)^2} \\ & \Rightarrow \frac{d x}{d y}=\frac{a d-c d y+c d y-b c}{(a-c y)^2}=\frac{a d-b c}{a^2+c 2 y^2-2 a c y}\end{aligned}$
Here $\mathrm{P}=\mathrm{C} 2, \mathrm{Q}=2 \mathrm{ac}, \mathrm{R}=\mathrm{Q}^2$
$\therefore \quad \mathrm{P}+\mathrm{Q}+\mathrm{R}=\mathrm{C}^2-2 \mathrm{ac}+\mathrm{a}^2=\mathrm{a}^2+\mathrm{c}^2-2 \mathrm{ac}=(\mathrm{a}-\mathrm{C})$
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