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Question

(D) Given the curve $y = x^4 - 6x^3 + 13x^2 - 10x + 5$. The slope of the tangent is given by the derivative $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$.

Vedclass · Accepted Answer

$-13$

Vedclass · Answer

(D) Given the curve $y = x^4 - 6x^3 + 13x^2 - 10x + 5$. The slope of the tangent is given by the derivative $\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$. Since the slope of the tangent 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 $2x^3 - 9x^2 + 13x - 6 = 0$. By testing integer roots,we find $(x-1)$ and $(x-2)$ are factors. Factoring gives $(x-1)(x-2)(2x-3) = 0$. The roots are $x = 1, x = 2, x = 1.5$. Since $x_1, x_2 \in \mathbb{N}$,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,$x_1x_2 - y_1y_2 = (1 	imes 2) - (3 	imes 5) = 2 - 15 = -13$.

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 $Q(x_2, y_2)$,where $x_1, x_2 \in \mathbb{N}$,then $x_1x_2 - y_1y_2 =$

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