The angle between the tangent lines to the gr | vedclass

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

Question

(D) Given $f(x) = \int_{2}^{x} (2t - 5) \, dt$. Evaluating the integral: $f(x) = [t^2 - 5t]_{2}^{x} = (x^2 - 5x) - (4 - 10) = x^2 - 5x + 6$. The graph

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Given $f(x) = \int_{2}^{x} (2t - 5) \, dt$. Evaluating the integral: $f(x) = [t^2 - 5t]_{2}^{x} = (x^2 - 5x) - (4 - 10) = x^2 - 5x + 6$. The graph cuts the $x$-axis where $f(x) = 0$. $x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0$,so $x = 2$ and $x = 3$. The points of intersection are $(2, 0)$ and $(3, 0)$. The slope of the tangent is $f'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5$. At $x = 2$,$m_1 = 2(2) - 5 = -1$. At $x = 3$,$m_2 = 2(3) - 5 = 1$. The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. $	an 	heta = \left| \frac{-1 - 1}{1 + (-1)(1)} ight| = \left| \frac{-2}{0} ight| = \infty$. Thus,$	heta = \frac{\pi}{2}$.

The angle between the tangent lines to the graph of the function $f(x) = \int_{2}^{x} (2t - 5) \, dt$ at the points where the graph cuts the $x$-axis is

Similar Questions

Let the curve be represented by $x=2(\cos t+t \sin t)$ and $y=2(\sin t-t \cos t)$. Then the normal at any point '$t$' of the curve is at a distance of . . . . . . units from the origin.

The chord of the curve $y=x^{2}+2ax+b$ joining the points where $x=\alpha$ and $x=\beta$ is parallel to the tangent to the curve at abscissa $x$ equal to:

Let $a$ be a fixed positive real number and $n$ be an arbitrary constant. For the curve $y = \frac{x^n}{a^{n-1}}$, if the length of the subnormal at any point $(\alpha, \beta)$ is proportional to $a^2$, then $n =$

If the equation of the tangent at $(2,3)$ on the curve $y^2 = ax^3 + b$ is $y = 4x - 5$,then the value of $a^2 + b^2$ is:

If the tangent to the curve $y^{2} = x^{3}$ at $(m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{3})$, then the value of $mM$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module