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USM ITT 2010

Solusi USM IT Telkom 2010 Kode 11 – Nomor 8


Fungsi\ f(x)=sinxcosx+1\ mempunyai\ nilai\ maksimum\ ....\\(A)\quad 2\\(B)\quad 1\\(C)\quad 0\\(D)\quad 2,5\\(E)\quad 1,5
Jawaban: E

\begin{array} {rcl} f(x)&=&sinxcosx+1\\\\&=&\frac{1}{2}(2sinxcosx)+1\\\\&=&\frac{1}{2}sin2x+1\end{array}
sehingga

f_{maks}=\frac{1}{2}.1+1=1\frac{1}{2}

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