Diketahui $f(x)=x^2+4x+4$ dan $g(x)=x^2-4x+4$. Rumus $\sqrt{f(g(x))}+\sqrt{g(f(x))}=$ ...
A. $2x^2+16$
B. $2x^2+10$
C. $2x^2+8$
D. $2x^2-8$
E. $2x^2-10$
Jawab; C
$f(x)=x^2+4x+4=(x+2)^2$
$g(x)=x^2-4x+4=(x-2)^2$
$\sqrt{f(g(x))}+\sqrt{g(f(x))}=\sqrt{f((x-2)^2)}+\sqrt{g((x+2)^2)}$
$=\sqrt{[(x-2)^2+2]^2}+\sqrt{[(x+2)^2-2]^2}$
$=[(x-2)^2+2]+[(x+2)^2-2]$
$=[x^2-4x+4+2]+[x^2+4x+4-2]$
$=2x^2+8$
Sumber Soal; Mandiri Matematika Jilid 1 untuk SMA/MA Kelas X Kelompok Wajib, Halaman 93 No. 87
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