Functions

1. Understanding the Problem:
The given problem asks us to find the sum of the intercepts of the line on the coordinate axes, where the range of the function is given as .

2. Analyzing the Function:
The function given is . To find the range of this function, we need to understand the nature of the function and how its behavior is influenced by the values of .

3. Understanding the Intercepts of the Line:
The equation of the line is . The intercepts of a line are given by the x-intercept and y-intercept, which can be calculated as follows:

For the x-intercept, set :

.

For the y-intercept, set :

.

4. Sum of Intercepts:
The sum of the intercepts is given by the sum of the x-intercept and y-intercept:

.

5. Relating the Range to the Intercepts:
We are given that the range of the function is , which suggests a relationship between and . We need to use this relationship to solve for the sum of intercepts. Based on the given information, the correct value is .

Final Answer:
The sum of the intercepts is .

Step 1: Differentiate
We define: Using quotient rule, Expanding the numerator: Thus, Setting ,

Step 2: Compute Maximum and Minimum Values

Step 3: Compute the sum

Step 4: Conclusion
Thus, the correct answer is:

We need to find the domain of the function:

Step 1: Condition for Logarithm
For the logarithmic function to be defined, the argument must be positive: Since is always non-negative and is only zero at , we have:

Step 2: Condition for Inverse Sine
The function is defined for . Thus, we require:

Step 3: Solve for
Rewriting the inequality: Converting to exponential form: Multiplying by 2:

Step 4: Solve for
Taking square roots: Thus, the domain of is:

We are given the function:

Step 1: Find the Domain of the Logarithm
For the logarithmic function to be defined, the argument must be positive: Rearrange the quadratic inequality: Factorizing the quadratic expression: Solving using the sign analysis method, the valid interval is:

\Step 2: Find the Range of
Since the quadratic expression is a downward-opening parabola, it attains a maximum value at its vertex. The vertex of the quadratic function is given by: Substituting : Since the logarithm base is 3, we compute: As or , the argument approaches 0, meaning approaches . Thus, the range of is: