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The Area bounded by $y=e^x, 2y+2k-2kx-1-e^2=0$ is minimum for $ k=k_1$; then (where [.] denotes greatest integer function)

A. $[2k_1]\gt5$ $\qquad$ B. $|2k_1-2\pi|\lt1$ $\qquad$ C. $k_1 \in \mathbb I$ $\qquad$ D. $k_1=\frac pq; p, q$ are integers $q\not=0,1, {-1}$

my attempt

I had drawn the graphs of the exponential function and the the straightline and found that for $k\lt0$ we have $\infty$ area enclosed so $k$ must be $\gt0$. and also the line $$\frac y{\frac{e^2+1}2-k}+ {x\over1-\frac{e^2+1}{2k}}=1$$

passes through a fixed point $P\equiv\left(1,\frac{e^2+1}2\right)$

so as $k$ varies from $0$ to $\infty$ the line rotates about this fixed point. and this point P is above the graph of $y=e^x$. I reached upto this I could not make any further conclusion from here. I had a thought to find the points of intersection in terms of $k$ then integrate but to find the points of intersection seems to be impossible.

enter image description here

Graph of the question

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  • $\begingroup$ Nice question. What did you try in order to solve it? $\endgroup$
    – Dominique
    Commented Mar 14 at 15:05
  • $\begingroup$ Is this multi correct or single correct $\endgroup$
    – Sisanta Chhatoi
    Commented Mar 14 at 18:52
  • $\begingroup$ I have edited it, to tell how far I had approached. $\endgroup$
    – Skdmg
    Commented Mar 14 at 18:53
  • $\begingroup$ This is a multi correct competitive exam-based question. $\endgroup$
    – Skdmg
    Commented Mar 14 at 18:55
  • $\begingroup$ Could you provide us the graphs via geogebra or Wolfram Alpha or..., and the point $P $, please ? $\endgroup$
    – Stéphane Jaouen
    Commented Apr 11 at 9:24

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