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 line \(y=x\) meets \(y=k e^{x}\) for \(k \leq 0\) at

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 line

y = x

meets

y = k e^{x}

for

k \leq 0

Answer: (C)

For

k \leq 0

, the function

y = k e^{x}

is either zero (when

k = 0

) or negative (when

k < 0

).

The line

y = x

passes through the origin with positive slope for

x > 0

and negative values for

x < 0

.

When

k = 0

y = 0

intersects

y = x

only at the origin.

When

k < 0

y = k e^{x} < 0

for all

x

, while

y = x

is positive for

x > 0

. For

x < 0

, both are negative, but

k e^{x}

approaches 0 from below as

x \to - \infty

while

x \to - \infty

. They intersect at exactly one point.

Therefore, the answer is one point.

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