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What is the minimum value of the function \(f(x)=2\sqrt{x-1}+3\)?फलन \(f(x)=2\sqrt{x-1}+3\) का न्यूनतम मान क्या है?
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Correct answer:सही उत्तर: 33
For the square root to be defined, \(x-1\ge 0\), so the domain is \(x\ge 1\). Since \(\sqrt{x-1}\ge 0\), we have \(2\sqrt{x-1}+3\ge 3\). At \(x=1\), \(f(1)=3\); hence the minimum value is 3. Exam tip: For an expression of the form \(a\sqrt{g(x)}+b\) with \(a>0\), check the minimum when \(\sqrt{g(x)}=0\).वर्गमूल के लिए \(x-1\ge 0\), इसलिए फलन का परिभाषा-क्षेत्र \(x\ge 1\) है। चूँकि \(\sqrt{x-1}\ge 0\), अतः \(2\sqrt{x-1}+3\ge 3\)। जब \(x=1\), तब \(f(1)=3\), इसलिए न्यूनतम मान 3 है। परीक्षा-युक्ति: \(a\sqrt{g(x)}+b\) में यदि \(a>0\) हो, तो न्यूनतम मान सामान्यतः \(\sqrt{g(x)}=0\) पर जाँचें।
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