Question

Let $f:R\to R$ be an odd function and suppose $f\left(e^x\right)=e^{f\left(x\right)}$ for all $xϵR$. Then

• A. $f\left(\frac{1}{e}\right)<f\left(1\right)<f\left(e\right)$
• B. $f\left(\frac{1}{e}\right)>f\left(1\right)>f\left(e\right)$
• C. $f\left(1\right)<f\left(\frac{1}{e}\right)<f\left(e\right)$
• D. $f\left(1\right)>f\left(\frac{1}{e}\right)>f\left(e\right)$

Answer 1

The correct option is A $f\left(\frac{1}{e}\right)<f\left(1\right)<f\left(e\right)$

We know that $f:R\to R$ is an odd function. hence $f(-x)=-f\left(x\right)$

Or $f\left(x\right)+f(-x)=0$ ....$\left(1\right)$

Let's put $x=0$ in the eq.$\left(1\right)$, we get $f\left(0\right)+f\left(0\right)=0$

$\Rightarrow f\left(0\right)=0$

Now the given condition is $f\left(e^x\right)=e^{f\left(x\right)}$ ....$\left(2\right)$

Now let's put $x=0$ in the equation $\left(2\right)$, we get $\to f\left(e^0\right)=e^{f\left(0\right)}$

From above calculations we know that $f\left(0\right)=0$,

$\Rightarrow f\left(e^0\right)=e^0$

$\because e^0=1$, Hence $\Rightarrow f\left(1\right)=1$ ....$\left(3\right)$

Now let's put $x=1$ in eq$\left(2\right)$, we get $\Rightarrow f\left(e\right)=e^{f\left(1\right)}$

$\because f\left(1\right)=1$, hence $f\left(e\right)=e$.....$\left(4\right)$

Now if we out $x=-1$ in the equation $\left(2\right)$, we get $\Rightarrow f\left(\frac{1}{e}\right)=e^{f(-1)}$

As the function $f$ is odd function, so $f(-x)=-f\left(x\right)$ or $f(-1)=-f\left(1\right)=-1$

Hence $f\left(\frac{1}{e}\right)=e^{-1}=\frac{1}{e}$

$\Rightarrow f\left(\frac{1}{e}\right)=\frac{1}{e}$ ...$\left(5\right)$

Now from eq. $\left(3\right),\left(4\right)$ and $\left(5\right)$ we have,

$\Rightarrow f\left(1\right)=1$

$\Rightarrow f\left(e\right)=e$

$\Rightarrow f\left(\frac{1}{e}\right)=\frac{1}{e}$

As we know that, $\frac{1}{e}<1<e$

Hence $f\left(\frac{1}{e}\right)<f\left(1\right)<f\left(e\right)$

Correct option is $A$.