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What is the real domain of (f(x)=\sqrt{9-x^2})?फलन (f(x)=\sqrt{9-x^2}) का वास्तविक प्रांत क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ([-3,3])([-3,3])
Step 1: The expression inside the square root must satisfy (9-x^2\geq0). Step 2: This gives (x^2\leq9), so (-3\leq x\leq3). Step 3: In square inequalities, remember both lower and upper bounds.चरण 1: वर्गमूल के अंदर (9-x^2\geq0) होना चाहिए। चरण 2: इससे (x^2\leq9) यानी (-3\leq x\leq3) मिलता है। चरण 3: वर्ग वाले असमानताओं में दोनों ओर की सीमा याद रखें।
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