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If a continous founction of defined on the real line R, assumes positive and negative values in R, then the equation f(x)=0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum values is negative, then the equation `f(x)=0` has a root in R. Considetr `f(x)=ke^(x)-x` for all real x where k is real constant.
The positive value of k for which `ke^(x)-x=0` has only root is

A

`1/e`

B

1

C

e

D

`log_(e)2`

Text Solution

Verified by Experts

The correct Answer is:
A
Let `f(x)ke^(x)-x`
f'(x)=ke^(x)-1=0`
`impliesx=-ln k`
f''(x)=ke^(x)`
`therefore[f''(x)]_(x=-lnk)=1gt0`
Hence, `f(-lnk)=1+lnk`
For one root of given question
`1+lnk =0`
`impliesk =1/e`
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