यदि (f(x)=\sqrt{x+1}) और (g(x)=\sqrt{2-x}) हों, तो ((fg)(x)) का अधिकतम मान क्या है?
If (f(x)=\sqrt{x+1}) and (g(x)=\sqrt{2-x}), what is the maximum value of ((fg)(x))?
Explanation opens after your attempt
Correct Answer
A. \(\frac{3}{2}\)
Concept
((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum \(\frac{9}{4}\). Therefore the maximum is \(\frac{3}{2}\).
Why this answer is correct
The correct answer is A. \(\frac{3}{2}\). ((fg)(x)=\sqrt{(x+1)(2-x)}), and the quadratic inside has maximum \(\frac{9}{4}\). Therefore the maximum is \(\frac{3}{2}\).
Exam Tip
((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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