If a continuous function $f$ defined on the real line $\mathbf{R}$, assumes positive and negative values in $\mathbf{R}$ then the equation $f(x)=0$ has a root in $\mathbf{R}$. For example, if it is known that a continuous function $f$ on $\mathbf{R}$ is positive at some point and its minimum value is negative then the equation $f(x)=0$ has a root in $\mathbf{R}$. Consider $f(x)=k e^{x}-x$ for all real $x$ where $k$ is a real constant. For $k>0$, the set of all values of $k$ for which $k e^{x}-x=0$ has two distinct roots is

(B) $\left(\frac{1}{e}, 1\right)$

(A) $\left(0, \frac{1}{e}\right)$

(C) $\left(\frac{1}{e}, \infty\right)$

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Application of Derivatives

If a continuous function

f

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

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 e^{x} - x

for all real

x

where

k

is a real constant.

For

k > 0

, the set of all values of

k

for which

k e^{x} - x = 0

has two distinct roots is

Answer: (B)

From the previous question, we found that the minimum value of

f (x) = k e^{x} - x

f (- ln k) = 1 + ln k

.

For two distinct roots, the minimum value must be negative:

1 + ln k < 0 ⟹ ln k < - 1 ⟹ k < \frac{1}{e}

Since

k > 0

(given), the set of values is

k \in (0, \frac{1}{e})

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Single correct option