GATE 2023 ES – Question 49
Find the maximum of $f(x)=x\sqrt{4-x^2}$ on[−2,2],to two decimals.
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Correct answer: 1.9 to 2.1
Explanation
For negative x, the function is nonpositive; positive maximum lies in[0,2]. There,$f^2=x^2(4-x^2)$. Let u=x²;then $f^2=4u-u^2=4-(u-2)^2\le4$. Equality occurs at x=√2. Thus maximum is **2.00**; endpoints give0.