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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2020 · 6 September · Evening Shift · Mathematics · Application of Derivatives

If the tangent to the curve, y = f(x) = x _ e x, (x > 0) at a point...

If the tangent to the curve,y=f(x)=xlog⁡ey = f\left(x\right) = x{\log}_{e}x, \left(x > 0\right)at a point \left(c, f\left(c\right)\right)isparalleltotheline−segmentjoiningthepoints(1,0)and(e,e)is parallel to the line-segment joining the points \left(1, 0\right) and \left(e, e\right),then cc is equal to :

Options

  1. Ae−1e\dfrac{e - 1}{e}
  2. B{e}^{\left(\dfrac{1}{e - 1}\right)}
  3. C{e}^{\left(\dfrac{1}{1 - e}\right)}
  4. D1e−1\dfrac{1}{e - 1}

Correct answer

B. {e}^{\left(\dfrac{1}{e - 1}\right)}

Step-by-step solution

f′(c)=1+lnc=ee−1f ' \left(c\right) = 1 + lnc = \dfrac{e}{e - 1}

lnc=1e−1lnc = \dfrac{1}{e - 1}

c=e(1e−1)c = {e}^{\left(\dfrac{1}{e - 1}\right)}

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