If `f: (e,oo) rarr R & f(x)=log[log (logx)],` then f is -
(a)f is one-one and onto
(b)f is one-one but onto
(c)f is onto but not one-one
(d)the range of f is equal to its domain

Options

A. f is one-one and onto
B. f is one-one but onto
C. f is onto but not one-one
D. the range of f is equal to its domain

Answer

To determine the properties of the function \( f(x) = \log(\log(\log x)) \) where the domain is \( (e, \infty) \), we will analyze whether the function is one-one (injective) and onto (surjective).

### Step 1: Determine the domain of \( f(x) \)

The function \( f(x) = \log(\log(\log x)) \) requires that the argument of each logarithm is positive.

1. **Innermost logarithm**: For \( \log x \) to be defined and positive, we need \( x > e \).
2. **Next logarithm**: For \( \log(\log x) \) to be defined and positive, we need \( \log x > 1 \), which implies \( x > e^1 = e \).
3. **Outermost logarithm**: For \( \log(\log(\log x)) \) to be defined, we need \( \log(\log x) > 0 \), which implies \( \log x > 1 \) or \( x > e \).

Thus, the domain of \( f(x) \) is \( (e, \infty) \).

### Step 2: Determine the range of \( f(x) \)

Next, we will analyze the range of \( f(x) \).

1. As \( x \) approaches \( e \) from the right, \( \log x \) approaches \( 1 \), and thus \( \log(\log x) \) approaches \( 0 \). Therefore, \( \log(\log(\log x)) \) approaches \( \log(0) \), which approaches \( -\infty \).
2. As \( x \) approaches \( \infty \), \( \log x \) approaches \( \infty \), and thus \( \log(\log x) \) also approaches \( \infty \). Therefore, \( \log(\log(\log x)) \) approaches \( \infty \).

Thus, the range of \( f(x) \) is \( (-\infty, \infty) \).

### Step 3: Determine if \( f(x) \) is one-one

To determine if \( f(x) \) is one-one, we can check if it is a strictly increasing function.

1. The derivative \( f'(x) \) can be calculated using the chain rule:
\[ f'(x) = \frac{1}{\log(\log x)} \cdot \frac{1}{\log x} \cdot \frac{1}{x} \]
Since \( x > e \), \( \log x > 1 \) and \( \log(\log x) > 0 \). Therefore, \( f'(x) > 0 \) for all \( x \in (e, \infty) \).

Since \( f'(x) > 0 \), \( f(x) \) is strictly increasing, which implies that it is one-one.

### Step 4: Determine if \( f(x) \) is onto

Since the range of \( f(x) \) is \( (-\infty, \infty) \) and the co-domain is also \( \mathbb{R} \), we conclude that \( f(x) \) is onto.

### Conclusion

Since \( f(x) \) is both one-one and onto, the correct option is:

**(a) f is one-one and onto.**

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