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. The positive value of $k$ for which $k e^{x}-x=0$ has only one root is

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.

The positive value of

k

for which

k e^{x} - x = 0

has only one root is

Answer: (A)

For

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

to have only one root, the minimum value of

f (x)

must be zero.

f^{'} (x) = k e^{x} - 1 = 0

gives

x = - ln k

(for

k > 0

f^{''} (x) = k e^{x} > 0

, so

x = - ln k

is a minimum.

At the minimum:

f (- ln k) = k e^{- l n k} - (- ln k) = k \cdot \frac{1}{k} + ln k = 1 + ln k

For exactly one root, we need

f (- ln k) = 0

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

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