Cosine of the angle of intersection of curves | vedclass

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

Question

(B) First,find the point of intersection by setting the curves equal: $3^{x-1} \log x = x^x - 1$. At $x = 1$,$3^0 \log 1 = 1^1 - 1$,which gives $1 	i

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) First,find the point of intersection by setting the curves equal: $3^{x-1} \log x = x^x - 1$. At $x = 1$,$3^0 \log 1 = 1^1 - 1$,which gives $1 	imes 0 = 0$. Thus,the curves intersect at $(1, 0)$.For the first curve $y = 3^{x-1} \log x$,the derivative is $\frac{dy}{dx} = 3^{x-1} \log 3 \log x + 3^{x-1} \cdot \frac{1}{x}$.At $x = 1$,$m_1 = 3^0 \log 3 \log 1 + 3^0 \cdot \frac{1}{1} = 0 + 1 = 1$.For the second curve $y = x^x - 1$,let $u = x^x$,then $\log u = x \log x$. Differentiating gives $\frac{1}{u} \frac{du}{dx} = \log x + 1$,so $\frac{du}{dx} = x^x (1 + \log x)$.Thus,$\frac{dy}{dx} = x^x (1 + \log x)$.At $x = 1$,$m_2 = 1^1 (1 + \log 1) = 1(1 + 0) = 1$.Since $m_1 = m_2 = 1$,the tangents are parallel,meaning the angle of intersection $	heta = 0$.Therefore,$\cos 	heta = \cos 0 = 1$.

Cosine of the angle of intersection of curves $y = 3^{x - 1} \log x$ and $y = x^x - 1$ is

Similar Questions

For the function $f(x) = x^2 - 6x + 8$ where $2 \le x \le 4$,the value of $x$ for which $f'(x)$ vanishes is:

If $f(x) = \sqrt{1 + \cos^2(x^2)}$,then $f'\left(\frac{\sqrt{\pi}}{2}\right)$ is

If $y = \cos^{-1}(\tanh x) + \sinh(\sin 6x)$, then $\frac{dy}{dx} =$

$\frac{d}{dx} \left( \frac{\sec x + \tan x}{\sec x - \tan x} \right) = $

If $y = t^{4/3} - 3t^{-2/3}$,then $\frac{dy}{dt} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module