Let's analyze the function
f(x)=ex+e−x where
f:ℝ⟶ℝ.
First, we check if the function is one-one (injective).
An injective function means that if
f(a)=f(b), then
a=b. Let's assume
f(a)=f(b).
So, we have:
ea+e−a=eb+e−bLet's rewrite the equation as:
ea+=eb+ Given that the exponential function
ex is strictly increasing, and
e−x is strictly decreasing, the sum is minimized when
x=0. This involves checking various values, but combining these two observations, we can see that the function is symmetric and not strictly monotonic. Therefore, it is not one-one.
Next, we check if the function is onto (surjective).
A surjective function means that for every
y∈ℝ, there exists an
x∈ℝ such that
f(x)=y. Here, considering
f(x)=ex+e−x, let's find the range.
Define a new function
g(x)=ex+e−x. We use the fact that:
g(0)=2When
x⟶∞, then
ex⟶∞ and
e−x⟶0, so
g(x)⟶∞.
When
x⟶−∞, then
ex⟶0 and
e−x⟶∞, so
g(x)⟶∞.
But at no point does the function
ex+e−x produce negative values. Therefore, the range of
f(x)≥2. Hence, the function is not surjective.
Given that the function is neither injective nor surjective, it cannot be bijective.
Thus, the correct answer is:
Option D: not bijective