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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2007 · Paper 1 · Mathematics · Application of Derivatives

The tangent to the curve y=e^x drawn at the point (c, e^c)...

The tangent to the curve y=exy=e^x drawn at the point \left(c, e^c\right)intersectsthelinejoiningthepoints(c−1,ec−1)intersects the line joining the points \left(c-1, e^{c-1}\right) and \left(c+1, e^{c+1}\right)The tangent to the curve y=e^x drawn at the point (c, e^c) intersects the line joining the points (c-1, e^ c-1 ) andThe tangent to the curve y=e^x drawn at the point (c, e^c) intersects the line joining the points (c-1, e^ c-1 ) and

Options

  1. Aon the left of x=cx=c
  2. Bon the right of x=cx=c
  3. Cat no point
  4. Dat all points

Correct answer

A. on the left of x=cx=c

Step-by-step solution

Slope of the line joining the points (c−1,ec−1)\left(c-1, e^{c-1}\right) and (c+1,ec+1)\left(c+1, e^{c+1}\right) is equal to ec+1−ec−12>ec\frac{e^{c+1}-e^{c-1}}{2} > e^c The tangent to the curve y=e^x drawn at the point (c, e^c) intersects the line joining the points (c-1, e^ c-1 ) and
 Tangent to the curve \text{ Tangent to the curve } y=e^x  will intersect the given line to the left of the line \text{ will intersect the given line to the left of the line } x=c . \text{. } ALITER The equation of the tangent to the curve y=exy=e^x at (c,ec)\left(c, e^c\right) is y-e^c=e^c(x-c) Equation of the line joining the given points is y-e^{c-1}=\frac{e^c\left(e-e^{-1}\right)}{2}[x−(c−1)]Eliminatingx-(c-1)] EliminatingyfromEqs.(i)and(ii),wegetfrom Eqs. (i) and (ii), we get[x−(c−1)][2−(e−e−1)]=2e−1x−c=e+e−1−22−(e−e−1)<0⇒x<c.\begin{aligned} {[x-(c-1)]\left[2-\left( e-e^{-1}\right)\right] } & =2 e^{-1} \\ x-c & =\frac{e+e^{-1}-2}{2-\left(e-e^{-1}\right)}< 0 \Rightarrow x < c . \end{aligned}$
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