Let f- -2- be given by fx- eex-e-ex- Then f is

F(x)= e^e^x+e^ e^x ⇒ f'(x)=e^x(e^e^x e^ e^x) ⇒ f'(x)=e^xe^e^x((e^e^x)^2 1) Now, as x ∈ℝ ⇒ e^x ∈ (0,∞) ⇒ e^e^x∈(1,∞) ⇒ f'(x) gt;0 So, f is strictly increasing ...

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Byju's Answer

Standard XII

Mathematics

One - One function

Let f:ℝ→ [2,∞...

Question

Let $f : R \to [2, \infty)$ be given by $f (x) = e^{e^{x}} + e^{- e^{x}}$ . Then $f$ is

one-one

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even

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bijective

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Solution

The correct option is A one-one
$f (x) = e^{e^{x}} + e^{- e^{x}} \Rightarrow f^{'} (x) = e^{x} (e^{e^{x}} - e^{- e^{x}}) \Rightarrow f^{'} (x) = \frac{e^{x}}{e^{e^{x}}} ((e^{e^{x}})^{2} - 1)$

Now, as $x \in R$
$\Rightarrow e^{x} \in (0, \infty) \Rightarrow e^{e^{x}} \in (1, \infty) \Rightarrow f^{'} (x) > 0$
So, $f$ is strictly increasing function
$\Rightarrow f$ is one-one.

By AM-GM inequality,
$\frac{e^{e^{x}} + e^{- e^{x}}}{2} \geq \sqrt{e^{e^{x}} \cdot e^{- e^{x}}}$
$\Rightarrow e^{e^{x}} + e^{- e^{x}} \geq 2$
But $e^{e^{x}} + e^{- e^{x}} \neq 2$ for any finite value of $x$ .
$∴$ Range $= (2, \infty)$
Range $\neq$ Co-domain $\Rightarrow f$ is into.
Clearly, $f$ is not even.

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Let f:R→[2,∞) be given by f(x)=eex+e−ex. Then f is

Let $f : R \to [2, \infty)$ be given by $f (x) = e^{e^{x}} + e^{- e^{x}}$ . Then $f$ is