The positive value of k for which ke^x − x = 0 has only one root is (A) 1/e

Question

If a continuous function f defined on the real line R, assumes positive and negative value 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 value is negative then the equation f(x) = 0 has a root in R.

Consider f(x) = kex − x for all real x where k is a real constant.

The positive value of k for which kex − x = 0 has only one root is

(A) 1/e

(B) 1

(C) e

(D) loge2

Answer 1

Answer is (A) 1/e

Let f(x) = kex - x. Then

f'(x) =kex - 1

Substituting f'(x) = 0 ⇒ x = - logk, we get

f(x) = kex

f(-logk) = 1 > 0

which implies that f(x) has one minima at point

x = - logk

Since the equation has only one root, we get

f(-logk) = 0

⇒ 1 + logk = 0 ⇒ k = 1/e