Question

If \(f(x)=\sqrt{2-x^2}\) and \(g(x)=\log(1-x)\) are two real valued functions, then the domain of the function \((f+g)(x)\) is

• A. \([-2,2]\)
• B. \([- \sqrt{2},1)\)
• C. \((-\infty,1)\)
• D. \((1,2]\)

Hint:

For the sum of two functions, the domain is always the intersection of the individual domains of the functions.

Answer 1

The Correct Option is B

Step 1: Find the domain of \(f(x)=\sqrt{2-x^2}\).
For the square root function to be real valued, the expression inside the square root must be non-negative. Thus,
\[ 2-x^2\geq 0 \]
\[ x^2\leq 2 \]
\[ -\sqrt{2}\leq x\leq \sqrt{2} \]
Therefore, the domain of \(f(x)\) is
\[ [-\sqrt{2},\sqrt{2}] \]

Step 2: Find the domain of \(g(x)=\log(1-x)\).
For the logarithmic function to be defined, its argument must be positive. Hence,
\[ 1-x\gt 0 \]
\[ x\lt 1 \]
Therefore, the domain of \(g(x)\) is
\[ (-\infty,1) \]

Step 3: Find the domain of \((f+g)(x)\).
The function \((f+g)(x)\) is defined only where both \(f(x)\) and \(g(x)\) are defined simultaneously.
So, we take the intersection of the two domains:
\[ [-\sqrt{2},\sqrt{2}] \cap (-\infty,1) \]
Since \(1\lt \sqrt{2}\), the common interval becomes
\[ [-\sqrt{2},1) \]

Step 4: Final conclusion.
Hence, the domain of \((f+g)(x)\) is
\[ \boxed{[-\sqrt{2},1)} \]