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What is the range of the function \(f(x)=(x+1)^2+3\)?फलन \(f(x)=(x+1)^2+3\) का परिसर (range) क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \([3,\infty)\)\([3,\infty)\)
For every real \(x\), \((x+1)^2\ge 0\), so \(f(x)=(x+1)^2+3\ge 3\). At \(x=-1\), the squared term is zero and \(f(-1)=3\), so 3 is included in the range. Therefore, the range is \([3,\infty)\). Exam tip: a function of the form \((x-a)^2+b\) has minimum value \(b\) and range \([b,\infty)\).किसी भी वास्तविक \(x\) के लिए \((x+1)^2\ge 0\), इसलिए \(f(x)=(x+1)^2+3\ge 3\)। \(x=-1\) पर \((x+1)^2=0\) और \(f(-1)=3\), इसलिए 3 परिसर में शामिल है। अतः परिसर \([3,\infty)\) है। परीक्षा-युक्ति: \((x-a)^2+b\) के लिए न्यूनतम मान \(b\) होता है और परिसर \([b,\infty)\) होता है।
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