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)=k ex-x for all real x where k is a real constant. The positive value of k for which k ex-x=0 has only one root is

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)=k ex-x for all real x where k is a real constant. The positive value of k for which k ex-x=0 has only one root is (A) 1 / e (B) 1 (C) e (D)  log e 2

Tardigrade Admin · Accepted Answer

Let f(x)=k ex-x f prime(x)=k ex-1 Substituting f prime(x)=0 ⇒ x=- logk, we get f prime prime(x)=k ex f prime prime(- log k)=1 > 0 which implies that f(x) has one minima at point x=- log k Since the equation has only one root, we get f(- log k)=0 ⇒ 1+ log k=0 ⇒ k=(1/e)

$1/ e$

$1$

$e$

$lo g_{e} 2$

1/e

1

e

loge​2

Let f(x)=kex−x 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=e1​

$1/ e$

$1$

$e$

$lo g_{e} 2$