Question

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 $(c+1,e^{c+1})$

• A. on the left of x = c
• B. on the right of x = c
• C. at no point
• D. at all points

Answer 1

Answer: (A)

The equation of tangent to the curve $y=e^x$ at $(c,e^c)$ is $y-e^c=e^c(x-c)$ .......(1)
and equation of line joining $\left(c-1,e^{c-1}\right)$ and $\left(c+1,e^{c+1}\right)$
$y-e^{c-1}=\frac{e^{c+1}-e^{c-1}}{\left(c+1\right)-\left(c-1\right)}\left[x-\left(c-1\right)\right]$
$\Rightarrow y-e^{c-1}=\frac{e^c\left(e-e^{-1}\right)}{2}\left[x-c+1\right]$ ......(2)
Subtracting equation (1) from (2), we get
$e^c-e^{c-1}=e^c\left(x-c\right)\left[\frac{e-e^{-1}-2}{2}\right]+e^c\left(\frac{e-e^{-1}}{2}\right)$
$\Rightarrow x-c=\frac{\frac{\left[1-e^{-1}-\left(\frac{e-e^{-1}}{2}\right)\right]}{e-e^{-1}-2}}{2}=\frac{2-e-e^{-1}}{e-e^{-1}-2}$
$=\frac{e+e^{-1}-2}{2-\left(e-e^{-1}\right)}$
$=\frac{\frac{e+e^{-1}}{2}-1}{1-\frac{e-e^{-1}}{2}}=\frac{+ve}{-ve}=-ve$
$\Rightarrow x-c<0\Rightarrow x<c$
$\therefore$ The two lines meet on the left of line x = c.