HOTS Zone : Suku Banyak (Polinom) [3]
Tipe:
No.
Diberikan polinomALTERNATIF PENYELESAIAN
Perhatikan bahwa P(k) = 2k untuk k ∈ {1, 2, 3, 4}, , jadi bisa kita tulis:
\(\begin{aligned} P(x)&=(x-1)(x-2)(x-3)(x-4)+2x\\ P(5)&=(5-1)(5-2)(5-3)(5-4)+2(5)\\ &=(4)(3)(2)(1)+10\\ &=\color{blue}\boxed{\boxed{\color{black}34}} \end{aligned}\)
\(\begin{aligned} P(x)&=(x-1)(x-2)(x-3)(x-4)+2x\\ P(5)&=(5-1)(5-2)(5-3)(5-4)+2(5)\\ &=(4)(3)(2)(1)+10\\ &=\color{blue}\boxed{\boxed{\color{black}34}} \end{aligned}\)
Jadi, P(5) = 34 .
No.
Banyaknya polinomial P(x) yang memenuhi- Tak hingga
- 1
- 0
- 28
- 27
ALTERNATIF PENYELESAIAN
Untuk x = 28,
\(\begin{aligned} 28P(27)&=(28-28)P(28)\\ P(27)&=0 \end{aligned}\)
Karena $P(x-1)=\dfrac{x-28}xP(x)$, makaP(27) = P(26) = ⋯ = P(1) = P(0) = 0. Sehingga,
P(x) = ax (x − 1)(x − 2)⋯(x − 27)
AgarP(28) = 28! , maka a = 1 .
Hanya ada 1 polinom.
\(\begin{aligned} 28P(27)&=(28-28)P(28)\\ P(27)&=0 \end{aligned}\)
Karena $P(x-1)=\dfrac{x-28}xP(x)$, maka
Agar
Hanya ada 1 polinom.
Jadi, bBanyaknya polinomial P(x) yang memenuhi P(28) = 28! dan xP(x − 1) = (x − 28)P(x) adalah 1.
JAWAB: B
JAWAB: B
No.
Diberikan polinomial p(x) berderajat 2017 yang mempunyai akar-akar positif berbeda dan hasil kalinya 2017. Didefinisikan- $\sqrt{2017}$
- $-\sqrt{2017}$
- 2017
- −2017
- $\dfrac{\sqrt{2017}}{2017}$
ALTERNATIF PENYELESAIAN
Misal k1, k2, ..., k2017 > 0 merupakan akar-akar dari p(x).
k1k2⋯k2017 = 2017
p(x) = a(x − k1)(x − k2)⋯(x − k2017)
\(\begin{aligned} q(x)&=p\left(x^2\right)\\ &=a\left(x^2-k_1\right)\left(x^2-k_2\right)\cdots\left(x^2-k_{2017}\right)\\ &=a\left(x+\sqrt{k_1}\right)\left(x-\sqrt{k_1}\right)\left(x+\sqrt{k_2}\right)\left(x-\sqrt{k_2}\right)\cdots\left(x+\sqrt{k_{2017}}\right)\left(x-\sqrt{k_{2017}}\right)\\ \end{aligned}\)
Akar-akar dari q(x) adalah
$-\sqrt{k_1},-\sqrt{k_2},...,-\sqrt{k_{2017}},\sqrt{k_1},\sqrt{k_2},...,\sqrt{k_{2017}}$
\(\begin{aligned} b_1b_2\cdots b_{2017}&=\left(-\sqrt{k_1}\right)\left(-\sqrt{k_2}\right)\cdots\left(-\sqrt{k_{2017}}\right)\\ &=-\sqrt{k_1k_2\cdots k_{2017}}\\ &=\color{blue}\boxed{\boxed{\color{black}-\sqrt{2017}}} \end{aligned}\)
\(\begin{aligned} q(x)&=p\left(x^2\right)\\ &=a\left(x^2-k_1\right)\left(x^2-k_2\right)\cdots\left(x^2-k_{2017}\right)\\ &=a\left(x+\sqrt{k_1}\right)\left(x-\sqrt{k_1}\right)\left(x+\sqrt{k_2}\right)\left(x-\sqrt{k_2}\right)\cdots\left(x+\sqrt{k_{2017}}\right)\left(x-\sqrt{k_{2017}}\right)\\ \end{aligned}\)
Akar-akar dari q(x) adalah
$-\sqrt{k_1},-\sqrt{k_2},...,-\sqrt{k_{2017}},\sqrt{k_1},\sqrt{k_2},...,\sqrt{k_{2017}}$
\(\begin{aligned} b_1b_2\cdots b_{2017}&=\left(-\sqrt{k_1}\right)\left(-\sqrt{k_2}\right)\cdots\left(-\sqrt{k_{2017}}\right)\\ &=-\sqrt{k_1k_2\cdots k_{2017}}\\ &=\color{blue}\boxed{\boxed{\color{black}-\sqrt{2017}}} \end{aligned}\)
Jadi, $b_1b_2\cdots b_{2017}=-\sqrt{2017}$.
JAWAB: B
JAWAB: B
Post a Comment