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If a line having slope 2 is a tangent to the curve $y=x^4-6 x^3+13 x^2-12 x+5$ at points $P\left(x_1, y_1\right)$ and $Q$ $\left(\mathrm{x}_2, \mathrm{y}_2\right), \mathrm{x}_1, \mathrm{x}_2 \in \mathrm{N}$ then $\mathrm{x}_1 \mathrm{x}_2-\mathrm{y}_1 \mathrm{y}_2=$
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$3$
$y=x^4-6 x^3+13 x^2-10 x+5$
$\begin{aligned} & \text { Slope of tangent }=\frac{d y}{d x}=4 x^3-18 x^2+26 x-10 \\ & \because \frac{d y}{d x}=2 \\ & 4 x^3-18 x^2+26 x-10=2 \\ & (x-1)(x-2)(2 x-3)=0 \\ & \therefore 2 \text { points can be taken as } x=1 \text { and } x=2 \\ & \text { if } x_1=2 \\ & y_1=(2)^4-6(2)^3+13(2)^2-10(2)+5=5 \\ & \text { when } x_2=1 \\ & y_2=(1)^4-6(1)^3+13(1)^2-10(1)+5=3 \\ & \therefore x_1 x_2-y_1 y_2=(1 \times 2)-(3 \times 5)=-13\end{aligned}$
$\begin{aligned} & \text { Slope of tangent }=\frac{d y}{d x}=4 x^3-18 x^2+26 x-10 \\ & \because \frac{d y}{d x}=2 \\ & 4 x^3-18 x^2+26 x-10=2 \\ & (x-1)(x-2)(2 x-3)=0 \\ & \therefore 2 \text { points can be taken as } x=1 \text { and } x=2 \\ & \text { if } x_1=2 \\ & y_1=(2)^4-6(2)^3+13(2)^2-10(2)+5=5 \\ & \text { when } x_2=1 \\ & y_2=(1)^4-6(1)^3+13(1)^2-10(1)+5=3 \\ & \therefore x_1 x_2-y_1 y_2=(1 \times 2)-(3 \times 5)=-13\end{aligned}$
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