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If (f(x)=\sqrt{x+1}) and (g(x)=\sqrt{2-x}), what is the maximum value of ((fg)(x))?यदि (f(x)=\sqrt{x+1}) और (g(x)=\sqrt{2-x}) हों, तो ((fg)(x)) का अधिकतम मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\frac{3}{2})(\frac{3}{2})
((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum (\frac{9}{4}). Therefore the maximum is (\frac{3}{2}).((fg)(x)=\sqrt{(x+1)(2-x)}) और अंदर का द्विघात अधिकतम (\frac{9}{4}) है। इसलिए अधिकतम (\frac{3}{2}) है।
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