Question

The domain of the function \(f(x)=\sqrt{x-1+\sqrt{6-x}\) is}

• A. \([0,\infty)\)
• B. \((-\infty,6)\)
• C. \([1,6]\)
• D. None of these

Hint:

For domain questions involving square roots, make every expression inside the square root non-negative and then take the common interval.

Answer 1

The Correct Option is C

Concept:
The given function is: \[ f(x)=\sqrt{x-1}+\sqrt{6-x} \] For a square root function to be real-valued, the expression inside every square root must be non-negative. That means: \[ \text{Quantity inside square root}\geq 0 \]

Step 1: Apply the condition on the first square root.
The first square root is: \[ \sqrt{x-1} \] For this square root to be real: \[ x-1\geq 0 \] Solving: \[ x\geq 1 \] So, the first condition is: \[ x\in[1,\infty) \]

Step 2: Apply the condition on the second square root.
The second square root is: \[ \sqrt{6-x} \] For this square root to be real: \[ 6-x\geq 0 \] Solving: \[ -x\geq -6 \] Multiplying by \(-1\), the inequality sign changes: \[ x\leq 6 \] So, the second condition is: \[ x\in(-\infty,6] \]

Step 3: Find the common interval.
Both square roots are present in the same function, so both conditions must be satisfied at the same time. Therefore: \[ x\geq 1 \] and \[ x\leq 6 \] Combining both: \[ 1\leq x\leq 6 \] Hence, the domain is: \[ [1,6] \]

Step 4: Check the options.
Option (A) \([0,\infty)\) is not correct because values greater than \(6\) make \(\sqrt{6-x}\) imaginary. Option (B) \((-\infty,6)\) is not correct because values less than \(1\) make \(\sqrt{x-1}\) imaginary, and it also excludes \(6\). Option (C) \([1,6]\) satisfies both conditions. Option (D) is not correct because option (C) is correct. Hence, the correct answer is: \[ \boxed{(C)\ [1,6]} \]