x_1, x_2 in N. If a line having slope 2 is a | vedclass

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(A) The equation of the given curve is $y=x^4-6x^3+13x^2-10x+5$. Taking the derivative with respect to $x$, we get $\frac{dy}{dx}=4x^3-18x^2+26x-10$.

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$17$

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(A) The equation of the given curve is $y=x^4-6x^3+13x^2-10x+5$. Taking the derivative with respect to $x$, we get $\frac{dy}{dx}=4x^3-18x^2+26x-10$. Since the slope of the tangent at points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is $2$, we set $\frac{dy}{dx}=2$: $4x^3-18x^2+26x-10=2$ $4x^3-18x^2+26x-12=0$ Dividing by $2$, we get $2x^3-9x^2+13x-6=0$. Factoring the cubic equation, we find $(x-1)(2x^2-7x+6)=0$, which simplifies to $(x-1)(x-2)(2x-3)=0$. The roots are $x=1, 2, \frac{3}{2}$. Given $x_1, x_2 \in N$, we choose $x_1=1$ and $x_2=2$. For $x_1=1$, $y_1=1^4-6(1)^3+13(1)^2-10(1)+5=1-6+13-10+5=3$. For $x_2=2$, $y_2=2^4-6(2)^3+13(2)^2-10(2)+5=16-48+52-20+5=5$. Thus, $x_1x_2+y_1y_2=(1 	imes 2)+(3 	imes 5)=2+15=17$.

$x_1, x_2 \in N$. If a line having slope $2$ is a tangent to the curve $y=x^4-6x^3+13x^2-10x+5$ at points $P(x_1, y_1)$ and $Q(x_2, y_2)$, then $x_1x_2+y_1y_2=$

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