Question

The tangent to the curve $y=e^x$drawn at the point $(c,e^c)$ intersects the line joining the points $(c-l,e^{c-1})$ 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

The correct option is A on the left of $x=c$

${\frac{dy}{dx}}_{x=c}$ $=e^x{|}_{x=c}$ $=e^c$

Therefore equation of the tangent will be

$\frac{y-e^c}{x-c}=e^c$

$y-e^c=e^{c}x-c{e}^c$

$y-e^{c}x=e^c(1-c)$...(i)

Now the slope of the line joining $(c-1,e^{c-1})$ and $(c+1,e^{c+1})$
$=\frac{e^{c+1}-e^{c-1}}{c+1-(c-1)}$

$=\frac{e^{c+1}-e^{c-1}}{2}$

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

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

Hence

$y-\left(e^{c-1}\right)=(x-(c-1\left)\right)\frac{e^{c-1}(e^2-1)}{2}$ ...(ii)

From i, we know

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

Substituting in the given equation, we get

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

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

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

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

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

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

$=c-1-\frac{2e^{c-1}}{e^{c+1}-e^{c-1}-2e^c}$

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

$=c-k$ ...(i)

Where k is constant.

Now

$c-k<c$

Hence the intersection is towards the left of $x=c$.