Diketahui \( f(x) = ax^2 + 2x + 4 \) dan \( g(x) = x^2+ax-2 \). Jika \( h(x) = \frac{f(x)}{g(x)} \) dengan \( h’(0) = 1 \) maka nilai \(a\) adalah…
- \( 2 \)
- \( \frac{1}{2} \)
- \( 0 \)
- \( -\frac{1}{2} \)
- \( -2 \)
(SBMPTN 2018)
Pembahasan:
Dari soal diketahui \( h(x) = \frac{f(x)}{g(x)} \) dan \( h’(0) = 1 \) maka kita peroleh berikut:
\begin{aligned} f(x) = ax^2 + 2x + 4 &\Leftrightarrow f(0) = a(0)^2+2(0)+4 = 4 \\[8pt] f(x) = ax^2 + 2x + 4 &\Leftrightarrow f'(x) = 2ax+2 \\[8pt] &\Leftrightarrow f'(0) = 2 \\[8pt] g(x) = x^2+ax-2 &\Leftrightarrow g(0) = (0)^2+a(0)-2 = -2 \\[8pt] g(x) = x^2+ax-2 &\Leftrightarrow g'(x) = 2x+a \\[8pt] &\Leftrightarrow g'(0) = a \end{aligned}
\begin{aligned} h(x) = \frac{f(x)}{g(x)} \Leftrightarrow h'(x) &= \frac{ f'(x)g(x)-f(x)g(x) }{g^2(x)} \\[8pt] \Leftrightarrow h'(0) &= \frac{ f'(0)g(0)-f(0)g(0) }{g^2(0)} \\[8pt] \Leftrightarrow 1 &= \frac{(2)(-2)-(4)(a)}{(-2)^2} \\[8pt] 1 = \frac{-4-4a}{4} \Leftrightarrow 1 &= -1-a \\[8pt] a &= -2 \end{aligned}
Jawaban E.