HOTS Zone : Nilai Mutlak
Table of Contents
Tipe:
No.
Nilai maksimum dari \({\sqrt{\left|x-y\right|}+\sqrt{\left|y-z\right|}+\sqrt{\left|z-x\right|}}\) dengan- \(\sqrt{6063}\)
- \(2\sqrt{2021}\)
- \(\sqrt{4042}+\sqrt{2021}\)
- \(2\sqrt{2021}+\sqrt{4042}\)
- \(2021\sqrt{2}+2021\)
ALTERNATIF PENYELESAIAN
Misalkan x ≥ y ≥ z dan \({P=\sqrt{\left|x-y\right|}+\sqrt{\left|y-z\right|}+\sqrt{\left|z-x\right|}}\). P bisa kita tulis menjadi,
\({P=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}}\)
Nilai P akan semakin besar jika nilai x semakin besar dan nilai z semakin kecil. Didapatx = 2021 dan z = 0 . Substitusikan ke P menjadi
\({P=\sqrt{2021-y}+\sqrt{y}+\sqrt{2021}}\)
Dengan menggunakan ketaksamaan Cauchy-Schwarz didapat,
\(\begin{aligned} \left(\sqrt{2021-y}\cdot1+\sqrt{y}\cdot1\right)^2&\leq \left(\left(\sqrt{2021-y}\right)^2+\left(\sqrt{y}\right)^2\right)\left(1^2+1^2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq \left(2021-y+y\right)\left(2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq 4042\\ \sqrt{2021-y}+\sqrt{y}&\leq \sqrt{4042}\\ \sqrt{2021-y}+\sqrt{y}+\sqrt{2021}&\leq \sqrt{4042}+\sqrt{2021} \end{aligned}\)
\({P=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}}\)
Nilai P akan semakin besar jika nilai x semakin besar dan nilai z semakin kecil. Didapat
\({P=\sqrt{2021-y}+\sqrt{y}+\sqrt{2021}}\)
Dengan menggunakan ketaksamaan Cauchy-Schwarz didapat,
\(\begin{aligned} \left(\sqrt{2021-y}\cdot1+\sqrt{y}\cdot1\right)^2&\leq \left(\left(\sqrt{2021-y}\right)^2+\left(\sqrt{y}\right)^2\right)\left(1^2+1^2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq \left(2021-y+y\right)\left(2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq 4042\\ \sqrt{2021-y}+\sqrt{y}&\leq \sqrt{4042}\\ \sqrt{2021-y}+\sqrt{y}+\sqrt{2021}&\leq \sqrt{4042}+\sqrt{2021} \end{aligned}\)
Jadi, nilai maksimum dari \({\sqrt{\left|x-y\right|}+\sqrt{\left|y-z\right|}+\sqrt{\left|z-x\right|}}\) dengan x,y,z ∈ [0, 2021] adalah \(\sqrt{4042}+\sqrt{2021}\).
JAWAB: C
JAWAB: C
No.
Didefinisikan fungsi nilai mutlak$|x|=\begin{cases}x,&x\geq0\\-x,&x\lt0\end{cases}$
Diberikan bilangan real
ALTERNATIF PENYELESAIAN
Agar nilai x76 − x16 maksimum, maka x76 harus terbesar dan x16 harus terkecil. Dengan kata lain,
x76 = x77 = ⋯ = x100
x1 = x2 = ⋯ = x16
Untuk x17 sampai x75 harus bernilai 0. Misalx76 = a > 0, dan x16 = b < 0 . Substitusikan ke persamaan,
25a − 16b = 1
25a + 16b = 0
Didapata = $\dfrac1{50}$ dan b = $-\dfrac1{32}$.
$\dfrac1{50}-\left(-\dfrac1{32}\right)=\dfrac{41}{800}$
41 + 800 =841
x76 = x77 = ⋯ = x100
x1 = x2 = ⋯ = x16
Untuk x17 sampai x75 harus bernilai 0. Misal
25a − 16b = 1
25a + 16b = 0
Didapat
$\dfrac1{50}-\left(-\dfrac1{32}\right)=\dfrac{41}{800}$
41 + 800 =
Jadi, m + n = 841.
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