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Options

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

Answer

A

Explanation

The equation of the tangent to the curve ` y=e^(x)` at ` (c,e^(c))` is
`y-e^(c )= e^(c )(x-c) " "…(i)`
The equation of the line joining `(c-1 ,e^(c-1)) ` and `(c+1, e^(c+1)) ` is
` y-e^(c-1)=(e^(c )(e-e^(-1)))/(2) (x-c+1) " "...(ii) `
On subtracting (i) from (ii), we get
` e^(c)-e^(c-1)=(e^(c)(e-e^(-1)))/(2) (x-c+1)-e^(c)(x-c) `
` rArr e^(c)-e^(c-1)=(x-c)e^(c)((e-e^(-1))/(2)-1)+e^(c)((e-e^(-1))/(2)) `
` x-c =((e-1)^(2))/(2-(e-1)^(2)) lt 0 rArr x lt c `
Hence, (i) and (ii) intersect at a point on the left of x=c.
ALITER Clearly, point P`(c,e^(c)) ` lies in between the points ` Q(c-1, e^(c-1) )` and R`(c+1,e^(c+1))` on the curve ` y=e^(x)` as shown in Fig. 4. Also, `y=e^(x)` is strictly increasing. So, the tangent at P itersects line joining Q and R on the left of point P.