Question

Let \( f(x)=\log_5 x \, (x > 0) \) and \( g(x)=\cos^{-1}(x) \, (-1\le x \le 1) \). Then the domain of \( g \circ f \) is

• A. \( (0,1] \)
• B. \( [-1,a) \)
• C. \( [0,a) \)
• D. \( \left[\frac{1}{5},5\right] \)
• E. \( [-1,5] \)

Hint:

For composite functions, always restrict the inner function so its output fits the domain of the outer function.

Answer 1

The Correct Option is D

Note: The question in the image appears as `logx`, but based on the options, it is clear that the intended function is `f(x) = log_5(x)`.
Step 1: Understanding the Concept:
The domain of a composite function `g(f(x))` is the set of all `x` values such that `x` is in the domain of `f`, and `f(x)` is in the domain of `g`.
Step 2: Key Formula or Approach:
1. Identify the domain of the inner function, `f(x)`.
2. Identify the domain of the outer function, `g(x)`.
3. Set the output (range) of the inner function `f(x)` to be within the domain of the outer function `g(x)`.
4. Solve the resulting inequality for `x` and combine it with the domain of `f(x)`.
Step 3: Detailed Explanation:
The composite function is `g(f(x)) = g(log_5(x)) = cos^{-1}(log_5(x))`.
Condition 1: Domain of the inner function f(x).
The domain of `f(x) = log_5(x)` is given as `x>0`.
Condition 2: The output of f(x) must be in the domain of g(x).
The domain of `g(x) = cos^{-1}(x)` is `[-1, 1]`. This means the input to `g` must be between -1 and 1, inclusive. In our composite function, the input to `g` is `f(x)`.
Therefore, we must have:
\[ -1 \le f(x) \le 1 \] \[ -1 \le \log_5(x) \le 1 \] This can be split into two inequalities:
a) \(\log_5(x) \le 1\)
\(x \le 5^1\) \implies \(x \le 5\)
b) \(\log_5(x) \ge -1\)
\(x \ge 5^{-1}\) \implies \(x \ge \frac{1}{5}\)
Combining these two results, we get \(\frac{1}{5} \le x \le 5\).
Final Domain:
We must satisfy both Condition 1 (`x>0`) and Condition 2 (`\frac{1}{5} \le x \le 5`). The intersection of these two conditions is `\frac{1}{5} \le x \le 5\).
In interval notation, this is \([1/5, 5]\).
Step 4: Final Answer:
The domain of g \(\circ\) f is [1/5, 5].