Bulan: Desember 2011

$latex [ f(x)= >\begin{cases} -x^{2}, &\text{if $x < 0$;}\ \alpha + x, &\text{if $0 \leq x \leq 1$;}\ >x^{2}, &\text{otherwise.} \end{cases} ] $

In the figure below, PQRS is rectangle. What is value of $latex a + b + c $ ?

Solusi

Berdasarkan kesebangunan segitiga-segitiga dan teorema Phytagoras, diperoleh:

$latex a^2 = 9 \cdot (9 + 16) $

$latex a^2 = 225 $

$latex a = 15 $

$latex b^2 = 9 \times 16 = 144 $

$latex b = 12 $

$latex c^2 = 16 \cdot (9 + 16) $

$latex c^2 = 400 $

$latex c = 20 $

Sehingga, $latex a + b + c = … $

Determine the value of:

$latex 100^2 – 99^2 + 98^2 – 97^2 + 96^2 – 95^2 + \ldots + 4^2 – 3^2 + 2^2 – 1^2 $

SOLUSI

Ingat kesamaan

$latex a^2 – b^2 = (a + b)(a – b) $

lebih khusus, jika $latex a – b = 1 $, maka

$latex a^2 – b^2 = a + b $, sehingga:

$latex 100^2 – 99^2 + 98^2 – 97^2 + 96^2 – 95^2 + \ldots + 4^2 – 3^2 + 2^2 – 1^2 $

$latex = (100 + 99) + (98 + 97) + (96 + 95) + \cdots + (4 + 3) + (2 + 1) $

$latex = 199 + 195 + 191 + \cdots + 7 + 3 $

yang merupakan deret hitung dengan suku pertama, $latex a = 3 $ dan beda, $latex b = 4 $.

$latex a + (n – 1)b = 199 $

$latex 3 + (n – 1)4 = 199 $

$latex 4n = 200 $

$latex n = 50 $

maka jumlah 50 suku pertama adalah:

$latex S_{n} = \frac {1}{2} \cdot n (2a + (n – 1)b) $

$latex S_{50} = \frac {1}{2} \cdot 50 (2 \cdot 3 + (50 – 1) \cdot 4) $

$latex S_{50} = 25 \cdot (6 + 196) $

Jadi, $latex S_{50} = \cdots $

Value sum the first ten terms of sequence numbers
$latex \frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3} + \frac {1}{3 \cdot 4} + \frac {1}{4 \cdot 5} + \cdots + \frac {1}{9 \cdot 10} + \frac {1}{10 \cdot 11} = … $

SOLUSI
Perhatikan pola berikut:

$latex \frac {1}{1 \cdot 2} = \frac {1}{2} $

$latex \frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3} = \frac {2}{3} $

$latex \frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3} + \frac {1}{3 \cdot 4} = \frac {3}{4} $

$latex \vdots $

Dengan demikian,

$latex \frac {1}{1 \cdot 2} + \frac {1}{2 \cdot 3} + \frac {1}{3 \cdot 4} + \frac {1}{4 \cdot 5} + \cdots + \frac {1}{9 \cdot 10} + \frac {1}{10 \cdot 11} = \frac {..}{..} $

Eleven students have a math test with the average score 65. The difference between the highest and the lowest scores is 20. If the highest and the lowest scores are ignored then average is 63. The highest score is …

Solusi
Nilai 11 siswa, misalkan x₁, x₂, …, x₁₁. Reratanya 65.
$latex \frac {x_{1} + x_{2} + … + x_{11}}{11} = 65 $
atau
$latex x_{1} + x_{2} + … + x_{11} = 65 \times 11 = 715 $ (pers. i)

Misalkan x₁ nilai tertinggi dan x₁₁ nilai terendah, selisihnya 20.
$latex x_{1} – x_{11} = 20 $ (pers. ii)

Rerata nilai tanpa menyertakan nilai tertinggi dan nilai terendah adalah 63.
$latex \frac {x_{2} + x_{3} + … + x_{10}}{9} = 63 $
atau
$latex x_{2} + x_{3} + … + x_{10} = 63 \times 9 = 567 $ (pers. iii)

Substitusi pers. iii ke pers. i
$latex x_{1} + x_{2} + … + x_{10} + x_{11} = 715 $
$latex x_{1} + x_{11} + (x_{2} + … + x_{10}) = 715 $
$latex x_{1} + x_{11} + 567 = 715 $
$latex x_{1} + x_{11} = 148 $ (pers. iv)

Berdasarkan pers. ii dan pers. iv
$latex x_{1} + x_{11} = 148 $
$latex x_{1} – x_{11} = 20 $

Eliminasi $latex x_{11} $, maka diperoleh:

$latex 2x_{1} = 168 $
$latex x_{1} = 84 $

Jadi, nilai tertinggi = 84.