HOTS Zone : AM GM HM QM
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.
Diberikan bilangan real x dan y sehingga memenuhi persamaanALTERNATIF PENYELESAIAN
\(\begin{aligned}
4x^2+y^2&=4x^2-2y+7\\
4x^2-4x+y^2+2y&=7\\
(2x-1)^2+(y+1)^2&=3^2
\end{aligned}\)
Misal
\(2x-1=a\longrightarrow x=\dfrac{a+1}2\)
\(y+1=b\longrightarrow y=b-1\)
\(\begin{aligned} 5x+6y&=5\cdot\dfrac{a+1}2+6(b-1)\\[3.8pt] &=\dfrac{5a+5}2+6b-6\\[3.8pt] &=\dfrac{5a+12b-7}2 \end{aligned}\)
Misal
\(2x-1=a\longrightarrow x=\dfrac{a+1}2\)
\(y+1=b\longrightarrow y=b-1\)
\(\begin{aligned} 5x+6y&=5\cdot\dfrac{a+1}2+6(b-1)\\[3.8pt] &=\dfrac{5a+5}2+6b-6\\[3.8pt] &=\dfrac{5a+12b-7}2 \end{aligned}\)
\(\begin{aligned}
(5a+12b)^2&\leq(5^2+12^2)(a^2+b^2)\\
&\leq(13^2)(3^2)\\
&\leq39^2
\end{aligned}\)
\(\begin{array}{rcccl} -39&\leq&5a+12b&\leq&39\\ -39&\leq&5(2x-1)+12(y+1)&\leq&39\\ -39&\leq&10x-5+12y+12&\leq&39\\ -39&\leq&10x+12y+7&\leq&39\\ -46&\leq&10x+12y&\leq&32\\ -23&\leq&5x+6y&\leq&16 \end{array}\)
\(\begin{aligned} |A|+|B|&=|16|+|-23|\\ &=\boxed{\boxed{39}} \end{aligned}\)
\(\begin{array}{rcccl} -39&\leq&5a+12b&\leq&39\\ -39&\leq&5(2x-1)+12(y+1)&\leq&39\\ -39&\leq&10x-5+12y+12&\leq&39\\ -39&\leq&10x+12y+7&\leq&39\\ -46&\leq&10x+12y&\leq&32\\ -23&\leq&5x+6y&\leq&16 \end{array}\)
\(\begin{aligned} |A|+|B|&=|16|+|-23|\\ &=\boxed{\boxed{39}} \end{aligned}\)
Jadi, |A| + |B| = 39 .
No.
Jika a, b, c bilangan real positif, buktikan bahwa \[\frac1{\left(\dfrac{b}a\right)^2+\dfrac{2c}a}+\frac1{\left(\dfrac{c}b\right)^2+\dfrac{2a}b}+\frac1{\left(\dfrac{a}b\right)^2+\dfrac{2b}c}\ge1\]ALTERNATIF PENYELESAIAN
\begin{aligned}
\frac1{\left(\dfrac{b}a\right)^2+\dfrac{2c}a}+\frac1{\left(\dfrac{c}b\right)^2+\dfrac{2a}b}+\frac1{\left(\dfrac{a}b\right)^2+\dfrac{2b}c}&=\frac{a^2}{b^2+2ac}+\frac{b^2}{c^2+2ab}+\frac{c^2}{a^2+2bc}\\
&\ge\frac{(a+b+c)^2}{b^2+2ac+c^2+2ab+a^2+2bc}&{\color{red}\text{Ketaksamaan Sedrakyan}}\\[3.7pt]
&=1
\end{aligned}
Jadi, terbukti bahwa \(\frac1{\left(\dfrac{b}a\right)^2+\dfrac{2c}a}+\frac1{\left(\dfrac{c}b\right)^2+\dfrac{2a}b}+\frac1{\left(\dfrac{a}b\right)^2+\dfrac{2b}c}\ge1\).
No.
Diberikan bilangan real positif x, y, z yang memenuhi$x+\dfrac1y=1$ dan $y+\dfrac1z=5$
Nilai minimum dari $z+\dfrac1x$ adalah ....
- $\dfrac15$
- $\dfrac25$
- 1
- 2
- $\dfrac52$
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(x+\dfrac1y\right)\left(y+\dfrac1z\right)\left(z+\dfrac1x\right)&=x+y+z+\dfrac1x+\dfrac1y+\dfrac1z+xyz+\dfrac1{xyz}\\[4pt]
\left(1\right)\left(5\right)\left(z+\dfrac1x\right)&=1+5+z+\dfrac1x+xyz+\dfrac1{xyz}\\[4pt]
4\left(z+\dfrac1x\right)&=6+xyz+\dfrac1{xyz}
\end{aligned}\)
dengan ketaksamaan AM-GM bisa dibuktikan bahwa $xyz+\dfrac1{xyz}\geq2$, sehingga
\(\begin{aligned} 4\left(z+\dfrac1x\right)&\geq6+2=8\\ \left(z+\dfrac1x\right)&\geq2 \end{aligned}\)
dengan ketaksamaan AM-GM bisa dibuktikan bahwa $xyz+\dfrac1{xyz}\geq2$, sehingga
\(\begin{aligned} 4\left(z+\dfrac1x\right)&\geq6+2=8\\ \left(z+\dfrac1x\right)&\geq2 \end{aligned}\)
Jadi, nilai minimum dari $z+\dfrac1x$ adalah 2.
JAWAB: D
JAWAB: D
No.
Jika $\dfrac{12}p+\dfrac{48}q=1$, dengan p dan q bilangan real positif, maka nilai terkecil yang mungkin dariALTERNATIF PENYELESAIAN
\(\begin{aligned}
(p+q)\left(\dfrac{12}p+\dfrac{48}q)&=(p+q)\left(\dfrac{\left(\sqrt{12}\right)^2}p+\dfrac{left(\sqrt{48}\right)^2}q)\\[4pt]
(p+q)(1)&\geq\left(\sqrt{12}+\sqrt{48}\right)^2\\
p+q&\geq\left(2\sqrt3+4\sqrt3\right)^2\\
&\geq\left(6\sqrt3\right)^2\\
&\geq \color{blue}\boxed{\boxed{\color{black}108}}
\end{aligned}\)
Jadi, nilai terkecil yang mungkin dari p + q adalah 108.
No.
Diberikan a, b adalah bilangan real positif. Jika nilai a, b yang menyebabkan bentuk $$\left(a+\dfrac4a\right)\left(b+\dfrac5b\right)$$ minimum adalah am dan bm, nilai minimum dari bentuk di atas adalah K, dan nilai dari $a_m+b_m+K=p+q\sqrt{r}$, makaALTERNATIF PENYELESAIAN
\(\begin{aligned}
a+\dfrac4a&\geq2\sqrt{a\cdot\dfrac4a}\\
&\geq4
\end{aligned}\)
Kesamaan terjadi saat $a=\dfrac4a$ ataua = 2.
\(\begin{aligned} b+\dfrac5b&\geq2\sqrt{b\cdot\dfrac5b}\\ &\geq2\sqrt5 \end{aligned}\)
Kesamaan terjadi saat $b=\dfrac5b$ atau $b=\sqrt5$.
\(\begin{aligned} \left(a+\dfrac4a\right)\left(b+\dfrac5b\right)&\geq4\cdot2\sqrt5\\[4pt] &\geq8\sqrt5 \end{aligned}\)
$2+\sqrt5+8\sqrt5=2+9\sqrt5$
p = 2,q = 9, r = 5
p + q + r = 2 + 9 + 5 =16
Kesamaan terjadi saat $a=\dfrac4a$ atau
\(\begin{aligned} b+\dfrac5b&\geq2\sqrt{b\cdot\dfrac5b}\\ &\geq2\sqrt5 \end{aligned}\)
Kesamaan terjadi saat $b=\dfrac5b$ atau $b=\sqrt5$.
\(\begin{aligned} \left(a+\dfrac4a\right)\left(b+\dfrac5b\right)&\geq4\cdot2\sqrt5\\[4pt] &\geq8\sqrt5 \end{aligned}\)
$2+\sqrt5+8\sqrt5=2+9\sqrt5$
p = 2,
p + q + r = 2 + 9 + 5 =
Jadi, p + q + r = 16 .
No.
Diberikan bilangan realALTERNATIF PENYELESAIAN
\(\begin{aligned}
ab&\leq\left(\dfrac{a+b}2\right)^2\\[4pt]
&\leq\left(\dfrac82\right)^2\\[4pt]
&\leq16
\end{aligned}\)
\(\begin{aligned} P&=\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\\[8pt] &=1+\dfrac1a+\dfrac1b+\dfrac1{ab}\\[4pt] &=1+\dfrac{a+b+1}{ab}\\[4pt] &=1+\dfrac{8+1}{ab}\\[4pt] &=1+\dfrac9{ab}\\[4pt] &\geq1+\dfrac9{16}=\dfrac{25}{16} \end{aligned}\)
25 − 16 =9
\(\begin{aligned} P&=\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\\[8pt] &=1+\dfrac1a+\dfrac1b+\dfrac1{ab}\\[4pt] &=1+\dfrac{a+b+1}{ab}\\[4pt] &=1+\dfrac{8+1}{ab}\\[4pt] &=1+\dfrac9{ab}\\[4pt] &\geq1+\dfrac9{16}=\dfrac{25}{16} \end{aligned}\)
25 − 16 =
Jadi, m − n = 9.
No.
Misalkan m merupakan nilai terbesar dari z yang memenuhi- 11
- 12
- 17
- 14
- 16
ALTERNATIF PENYELESAIAN
\(\begin{aligned} x^2+y^2+z^2&=(x+y+z)^2-2(xy+xz+yz)\\ &=5^2-2(3)\\ &=19\\ x^2+y^2&=19-z^2 \end{aligned}\)
dengan CSI didapat,
\(\begin{aligned} \left(x^2+y^2\right)\left(1^2+1^2\right)&\geq(x+y)^2\\ \left(19-z^2\right)\left(2\right)&\geq(5-z)^2\\ 38-2z^2&\geq25-10z+z^2\\ 3z^2-10z-13&\leq0\\ (3z-13)(z+1)&\leq0\\ \end{aligned}\)
$-1\leq z\leq\dfrac{13}3$
13 + 3 = 16
Jadi, p + q = 16.
JAWAB: E
JAWAB: E
No.
Diketahui a, b, c adalah bilangan riil dengan $\dfrac{a}{c}=8$. Nilai terkecil dari $\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}$ adalah ....- 1
- 4
- 8
- 16
- 65
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}&\geq2\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{c^2}}\\
&=2\sqrt{\dfrac{a^2}{c^2}}\\
&=2\left(\dfrac{a}{c}\right)\\
&=2\cdot8\\
&=\color{blue}\boxed{\boxed{\color{black}16}}
\end{aligned}\)
Jadi, nilai terkecil dari $\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}$ adalah 16.
JAWAB: D
JAWAB: D
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