The tangent to the curve `y=e^x` drawn at the point `(c,e^c)` intersects the line joining `(c-1,e^(c-1)) and (c+1,e^(c+1))` (a) on the left of `n=c` (b) on the right of `n=c` (c) at no points (d) at all points

Options

A. On the left of `x=c`
B. On the right of `x=c`
C. At no point
D. At all points

Explanation

Slope of the tangent to `y=e^(x)` at `(c, e^(c))` is given by
`" "m_(1)=((dy)/(dx))_((c,e^(c))"")= e^c`
Therefore, the equation of the tangent at `(1, e)` is
`" "y-e= e(x-1)`
Also the slope of the line joining the points `(c-1, e^(c-1)) and (c+1, e^(c+1))` is
`" "m_(2)= (e^(c+1)-e^(c-1))/((c+1)-(c-1))`
`" "= (e^(c+1)-e^(c-1))/(2)`
`" "=e^c((e-e^(-1))/(2))`
We observe `m_2 gt m_1`
Thus, the tangent to the curve `y= e^(x)` will intersect the given line to the left of the line `x=c` as shown in the following figure.
Hence the correct answer is (a).