If a line having slope 2 is a tangent to the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $x_1x_2 - y_1y_2 =$

Similar Questions

The length of the normal at any point to the curve $y=c \cosh \left(\frac{x}{c}\right)$ is

For the curve $\frac{x^n}{a^n}+\frac{y^n}{b^n}=2, (n \in N \text{ and } n > 1)$, the line $\frac{x}{a}+\frac{y}{b}=2$ is

Let $m$ be the slope of the normal $L$ drawn at $(1, 2)$ to the curve $x = t^2 - 7t + 7, y = t^2 - 4t - 10$ and $ax + by + c = 0$ be the equation of the normal $L$. If the $G$.$C$.$D$. of $(a, b, c)$ is $1$, then $m(a + b + c) =$

If $\theta$ is the acute angle between the curves $y^2 = x$ and $x^2 + y^2 = 2$, then $\tan \theta =$

Equation of the normal to the curve $y=x^2+x$ at the point $(1,2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module