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What is the range of \(f(x)=\sqrt{49-(x+3)^2}\)?\(f(x)=\sqrt{49-(x+3)^2}\) का range क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \([0,7]\)\([0,7]\)
Inside the square root must be non-negative: \(49-(x+3)^2\ge 0\Rightarrow (x+3)^2\le 49\). Then \(49-(x+3)^2\) can take any value from \(0\) up to \(49\). Taking square roots, \(f(x)\) takes any value from \(\sqrt{0}=0\) to \(\sqrt{49}=7\). Therefore the range is \([0,7]\).वर्गमूल के अंदर का मान non-negative होना चाहिए: \(49-(x+3)^2\ge 0\Rightarrow (x+3)^2\le 49\)। तब \(49-(x+3)^2\) का मान \(0\) से लेकर \(49\) तक बदल सकता है। वर्गमूल लेने पर \(f(x)\) के मान \(\sqrt{0}=0\) से \(\sqrt{49}=7\) तक होंगे। इसलिए range \([0,7]\) है।
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