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})$$(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)

on the left of $x=c$

Equation of straight line joining $A(c+1,e^{c+1})$ and $B(c-1,e^{c-1})$ is

$y-e^{c+1}=\frac{e^{c+1}-e^{c-1}}{2}(x-c-1)$ (1)

Equation of tangent at $(c,e^c)$ is $y-e^c=e^c(x-c)$ (2)

Subtracting (1) from (2),we get

$e^c(e-1)=e^c[(x-c)-\frac{1}{2}(e-e^{-1})(x-c)+\frac{1}{2}(e-e^{-1})]$

$\Rightarrow \frac{1}{2}(e+e^{-1})-1=(x-c)[1-\frac{1}{2}(e-e^{-1})]$

$\Rightarrow x-c=\frac{e+e^{-1}-2}{2-e+e^{-1}}<0$

[$\because e+e^{-1}>2$ and $2+e^{-1}-e<0$]

$\Rightarrow x<c$

Thus the two lines meet to the left of $x=c$