If the tangent to the curve y = f(x) = x log_ | vedclass

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

Question

(C) The slope of the line segment joining $(1, 0)$ and $(e, e)$ is given by $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$.The derivative of the function

Vedclass · Accepted Answer

$e^{\left(\frac{1}{e - 1}\right)}$

Vedclass · Answer

(C) The slope of the line segment joining $(1, 0)$ and $(e, e)$ is given by $m = \frac{e - 0}{e - 1} = \frac{e}{e - 1}$.The derivative of the function $f(x) = x \log_{e} x$ is $f'(x) = \frac{d}{dx}(x) \cdot \log_{e} x + x \cdot \frac{d}{dx}(\log_{e} x) = 1 \cdot \log_{e} x + x \cdot \frac{1}{x} = \log_{e} x + 1$.Since the tangent at $(c, f(c))$ is parallel to the line segment,its slope $f'(c)$ must be equal to $m$:$f'(c) = \log_{e} c + 1 = \frac{e}{e - 1}$.Solving for $\log_{e} c$:$\log_{e} c = \frac{e}{e - 1} - 1 = \frac{e - (e - 1)}{e - 1} = \frac{1}{e - 1}$.Therefore,$c = e^{\frac{1}{e - 1}}$.

If the tangent to the curve $y = f(x) = x \log_{e} x$ $(x > 0)$ at a point $(c, f(c))$ is parallel to the line segment joining the points $(1, 0)$ and $(e, e)$,then $c$ is equal to:

Similar Questions

The coordinates of the point on the curve $y = x \log x$ at which the normal is parallel to the line $2x - 2y = 3$ are:

If the angle between the curves $y^2=4x$ and $y=e^{-x/2}$ is $\theta$,then $\operatorname{cosec}^2(\theta/2)=$

If the lengths of the tangent,subtangent,normal and subnormal for the curve $y=x^2+x-1$ at the point $(1,1)$ are $a, b, c$ and $d$ respectively,then their increasing order is

The angle between the curves $x^2=8y$ and $xy=8$ is

The sum of the lengths of the subtangent and the subnormal drawn at $\theta = \frac{\pi}{3}$ on the cycloid $x = a(\theta - \sin \theta)$,$y = a(1 - \cos \theta)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module