Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem C. 873. (November 2006)

C. 873. For what real values of x will the value of the expression \sqrt{2\sin x}-\sin x be a maximum?

(5 pont)

Deadline expired on December 15, 2006.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. f(x)=\sqrt{2\sin x}-\sin x. Legyen a=\sqrt{2\sin x}, g(x)=a-\frac{1}{2}a^2=a(1-\frac{1}{2}a). Ez egy másodfokú függvény, maximumát a=1-ben veszi fel. Ekkor a=\sqrt{2\sin x}=1, vagyis sin x=1/2. Tehát f(x) értéke x=\frac{\pi}{6}+2k\pi\,\,(k\in Z) és x=\frac{5\pi}{6}+2l\pi\,\,(l\in Z) esetén lesz a legnagyobb (ez az érték 1/2).


Statistics:

345 students sent a solution.
5 points:185 students.
4 points:91 students.
3 points:12 students.
2 points:6 students.
1 point:5 students.
0 point:44 students.
Unfair, not evaluated:2 solutions.

Problems in Mathematics of KöMaL, November 2006