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What is the range of the function \(f(x)=(x-3)^2+2\)?फलन \(f(x)=(x-3)^2+2\) का परास क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \([2,\infty)\)\([2,\infty)\)
Since \((x-3)^2\geq 0\), we have \(f(x)=(x-3)^2+2\geq 2\). At \(x=3\), the function attains the value 2, so its range is \([2,\infty)\). Option B gives 3, which is the x-coordinate of the vertex, not the minimum function value. Exam tip: the range of \((x-a)^2+b\) is \([b,\infty)\).क्योंकि \((x-3)^2\geq 0\), इसलिए \(f(x)=(x-3)^2+2\geq 2\)। जब \(x=3\), तब \(f(x)=2\), अतः फलन का परास \([2,\infty)\) है। विकल्प B में 3, शीर्ष-बिंदु का \(x\)-निर्देशांक है, न्यूनतम मान नहीं। परीक्षा-युक्ति: \((x-a)^2+b\) का परास हमेशा \([b,\infty)\) होता है।
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