Nomor 6
Diketahui $ \sin (40^\circ +\alpha )=b \, $, dengan $ 0 < \alpha < 50^\circ $. Nilai dari $ \cos (10^\circ +\alpha )=...$
$\spadesuit \, $ Buat segitiga dari $ \sin (40^\circ +\alpha )= \frac{b}{1} = \frac{de}{mi} $
Sehingga : $ \cos (40^\circ+\alpha )= \frac{\sqrt{1-b^2}}{1} = \sqrt{1-b^2} $
$\spadesuit \, $ Konsep : $ \cos (A - B) = \cos A \cos B + \sin A \sin B $
$\spadesuit \, $ Menentukan $ \cos (10^\circ +\alpha ) $
$\begin{align} \cos (10^\circ +\alpha ) & = \cos [ (40^\circ+\alpha ) - ( 30^\circ ) ] \\ & = \cos (40^\circ+\alpha ) . \cos 30^\circ + \sin (40^\circ+\alpha ) . \sin 30^\circ \\ & = \sqrt{1-b^2} . \frac{1}{2} \sqrt{3} + b . \frac{1}{2} \\ & = \frac{1}{2} \left( \sqrt{3(1-b^2)} + b \right) \end{align}$
Jadi, nilai $ \cos (10^\circ +\alpha ) = \frac{1}{2} \left( \sqrt{3(1-b^2)} + b \right) . \heartsuit $
Sehingga : $ \cos (40^\circ+\alpha )= \frac{\sqrt{1-b^2}}{1} = \sqrt{1-b^2} $
$\spadesuit \, $ Konsep : $ \cos (A - B) = \cos A \cos B + \sin A \sin B $
$\spadesuit \, $ Menentukan $ \cos (10^\circ +\alpha ) $
$\begin{align} \cos (10^\circ +\alpha ) & = \cos [ (40^\circ+\alpha ) - ( 30^\circ ) ] \\ & = \cos (40^\circ+\alpha ) . \cos 30^\circ + \sin (40^\circ+\alpha ) . \sin 30^\circ \\ & = \sqrt{1-b^2} . \frac{1}{2} \sqrt{3} + b . \frac{1}{2} \\ & = \frac{1}{2} \left( \sqrt{3(1-b^2)} + b \right) \end{align}$
Jadi, nilai $ \cos (10^\circ +\alpha ) = \frac{1}{2} \left( \sqrt{3(1-b^2)} + b \right) . \heartsuit $
Nomor 7
Banyaknya nilai $x$ dengan $0\leq x \leq 2014\pi$ yang memenuhi persamaan $\frac{\sin 3x}{3-4\sin ^2x}=1$ adalah ...
$\clubsuit \, $ Syarat penyebut pecahan $ \frac{\sin 3x}{3-4\sin ^2x}=1 $
$ 3-4\sin ^2x \neq \rightarrow \sin ^2x \neq \frac{3}{4} \, $ ....(i)
$\clubsuit \, $ Konsep dasar
$ \sin 3x = 3\sin x - 4 \sin ^3 x $
$\clubsuit \, $ Misal $ p = \sin x \, $ , menyelesaikan persamaan
$\begin{align} \frac{\sin 3x}{3-4\sin ^2x} & =1 \\ \sin 3x & = 3-4\sin ^2x \\ 3\sin x - 4 \sin ^3 x & = 3-4\sin ^2x \\ 4 \sin ^3 x -4\sin ^2x - 3\sin x + 3 & = 0 \\ 4p^3 - 4p^2 -3p + 3 & = 0 \\ \text{dengan skema horner } \, & \, \text{ diperoleh} \\ (4p^2-3)(p-1) & = 0 \\ p^2 = \frac{3}{4} \vee p & = 1 \end{align}$
Skema hornernya :
Untuk $ p^2 = \frac{3}{4} \rightarrow \sin ^2 x = \frac{3}{4} , \, $ berdasarkan (i) maka $ \sin ^2 x = \frac{3}{4} \, $ tidak memenuhi .
Untuk $ p = 1 \rightarrow \sin x = 1 \rightarrow \sin x = \sin \frac{\pi}{2} $
solusinya :
$\begin{align} & \\ x & = \theta + k.2\pi \\ x & = \frac{\pi}{2} + k.2\pi \, \, \, \, ...(1) \end{align}$ $ \, \, \, \, \text{atau} \, \, \, \, $ $\begin{align} x & = \theta + k.2\pi \\ x & = (180^\circ - \frac{\pi}{2} ) + k.2\pi \\ x & = \frac{\pi}{2} + k.2\pi \, \, \, \, ...(2) \end{align}$
Bentuk (1) dan (2) sama, sehingga diambil salah satu saja.
$ x = \frac{\pi}{2} + k.2\pi \rightarrow x = \left( 2k+\frac{1}{2} \right)\pi \, $ ...(ii)
$\clubsuit \, $ Menentukan banyaknya solusi dari bentuk (ii) dan syarat $ 0\leq x \leq 2014\pi $
$\begin{align} 0 \leq & x \leq 2014\pi \\ 0 \leq & \left( 2k+ \frac{1}{2} \right) \pi \leq 2014 \pi \, \, \, \text{(bagi } \pi ) \\ 0 \leq & 2k+ \frac{1}{2} \leq 2014 \\ - \frac{1}{2} \leq & 2k \leq 2014 - \frac{1}{2} \, \, \, \text{(bagi 2)} \\ - \frac{1}{4} \leq & k \leq 1007 - \frac{1}{4} \\ - \frac{1}{4} \leq & k \leq 1006 + \frac{3}{4} \\ \text{karena } \, & k \, \text{ bulat, maka } \\ 0 \leq & k \leq 1006 \end{align}$
Ada 1007 bilangan bulat $ k \, $ yang memenuhi, sehingga solusi $ x \, $ berdasarkan $ x = \left( 2k+\frac{1}{2} \right)\pi \, $ juga ada 1007 solusi.
Jadi, banyaknya $ x \, $ yang memenuhi ada 1007 bilangan. $ \heartsuit $
$ 3-4\sin ^2x \neq \rightarrow \sin ^2x \neq \frac{3}{4} \, $ ....(i)
$\clubsuit \, $ Konsep dasar
$ \sin 3x = 3\sin x - 4 \sin ^3 x $
$\clubsuit \, $ Misal $ p = \sin x \, $ , menyelesaikan persamaan
$\begin{align} \frac{\sin 3x}{3-4\sin ^2x} & =1 \\ \sin 3x & = 3-4\sin ^2x \\ 3\sin x - 4 \sin ^3 x & = 3-4\sin ^2x \\ 4 \sin ^3 x -4\sin ^2x - 3\sin x + 3 & = 0 \\ 4p^3 - 4p^2 -3p + 3 & = 0 \\ \text{dengan skema horner } \, & \, \text{ diperoleh} \\ (4p^2-3)(p-1) & = 0 \\ p^2 = \frac{3}{4} \vee p & = 1 \end{align}$
Skema hornernya :
Untuk $ p^2 = \frac{3}{4} \rightarrow \sin ^2 x = \frac{3}{4} , \, $ berdasarkan (i) maka $ \sin ^2 x = \frac{3}{4} \, $ tidak memenuhi .
Untuk $ p = 1 \rightarrow \sin x = 1 \rightarrow \sin x = \sin \frac{\pi}{2} $
solusinya :
$\begin{align} & \\ x & = \theta + k.2\pi \\ x & = \frac{\pi}{2} + k.2\pi \, \, \, \, ...(1) \end{align}$ $ \, \, \, \, \text{atau} \, \, \, \, $ $\begin{align} x & = \theta + k.2\pi \\ x & = (180^\circ - \frac{\pi}{2} ) + k.2\pi \\ x & = \frac{\pi}{2} + k.2\pi \, \, \, \, ...(2) \end{align}$
Bentuk (1) dan (2) sama, sehingga diambil salah satu saja.
$ x = \frac{\pi}{2} + k.2\pi \rightarrow x = \left( 2k+\frac{1}{2} \right)\pi \, $ ...(ii)
$\clubsuit \, $ Menentukan banyaknya solusi dari bentuk (ii) dan syarat $ 0\leq x \leq 2014\pi $
$\begin{align} 0 \leq & x \leq 2014\pi \\ 0 \leq & \left( 2k+ \frac{1}{2} \right) \pi \leq 2014 \pi \, \, \, \text{(bagi } \pi ) \\ 0 \leq & 2k+ \frac{1}{2} \leq 2014 \\ - \frac{1}{2} \leq & 2k \leq 2014 - \frac{1}{2} \, \, \, \text{(bagi 2)} \\ - \frac{1}{4} \leq & k \leq 1007 - \frac{1}{4} \\ - \frac{1}{4} \leq & k \leq 1006 + \frac{3}{4} \\ \text{karena } \, & k \, \text{ bulat, maka } \\ 0 \leq & k \leq 1006 \end{align}$
Ada 1007 bilangan bulat $ k \, $ yang memenuhi, sehingga solusi $ x \, $ berdasarkan $ x = \left( 2k+\frac{1}{2} \right)\pi \, $ juga ada 1007 solusi.
Jadi, banyaknya $ x \, $ yang memenuhi ada 1007 bilangan. $ \heartsuit $
Nomor 8
Jika $\displaystyle \lim_{x \to 1} \left[ \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1}+ \frac{4(x^4-1)}{x^2-x^{-1}} \right]=...$
$\spadesuit \, $ Menentukan nilai limit masing-masing
Konsep : $ \left( \frac{a}{b} \right)^{-1} = \frac{b}{a} $
$\begin{align} & \displaystyle \lim_{x \to 1} \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4}{x(x-1)} -\frac{4-3x+x^2}{-(x-1)(x^2+x+1)} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4(x^2+x+1)}{x(x-1)(x^2+x+1)} + \frac{x(4-3x+x^2)}{(x-1)(x^2+x+1)} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4x^2+4x+4+4x-3x^2+x^3}{x(x-1)(x^2+x+1)} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{x^3+x^2+8x+4}{x^4-x} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \frac{x^4-x}{x^3+x^2+8x+4} \\ & = \frac{1^4-1}{1^3+1^2+8.1+4} = \frac{0}{14} = 0 \end{align}$
Konsep Turunan pada limit:
$ \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \rightarrow \lim_{x \to a} \frac{f(x)}{g(x)} = \displaystyle \lim_{x \to a} \frac{f^\prime (x)}{g^\prime (x)} $
diturunkan sampai hasilnya tidak sama dengan $ \frac{0}{0} $
$\begin{align} & \displaystyle \lim_{x \to 1} \frac{4(x^4-1)}{x^2-x^{-1}} = \frac{0}{0} \, \, \, \text{(diturunkan)} \\ & = \displaystyle \lim_{x \to 1} \frac{4(x^3)}{2x+\frac{1}{x^2}} \\ & = \frac{4(1^3)}{2.1+\frac{1}{1^2}} \\ & = \frac{16}{3} \end{align}$
$\spadesuit \, $ Menentukan hasil limitnya
$\begin{align} & \displaystyle \lim_{x \to 1} \left[ \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1}+ \frac{4(x^4-1)}{x^2-x^{-1}} \right] \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1} + \displaystyle \lim_{x \to 1} \frac{4(x^4-1)}{x^2-x^{-1}} \\ & = 0 + \frac{16}{3} \\ & = \frac{16}{3} \end{align}$
Jadi, nilai limitnya adalah $ \frac{16}{3} . \heartsuit $
Konsep : $ \left( \frac{a}{b} \right)^{-1} = \frac{b}{a} $
$\begin{align} & \displaystyle \lim_{x \to 1} \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4}{x(x-1)} -\frac{4-3x+x^2}{-(x-1)(x^2+x+1)} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4(x^2+x+1)}{x(x-1)(x^2+x+1)} + \frac{x(4-3x+x^2)}{(x-1)(x^2+x+1)} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4x^2+4x+4+4x-3x^2+x^3}{x(x-1)(x^2+x+1)} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \left( \frac{x^3+x^2+8x+4}{x^4-x} \right)^{-1} \\ & = \displaystyle \lim_{x \to 1} \frac{x^4-x}{x^3+x^2+8x+4} \\ & = \frac{1^4-1}{1^3+1^2+8.1+4} = \frac{0}{14} = 0 \end{align}$
Konsep Turunan pada limit:
$ \displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \rightarrow \lim_{x \to a} \frac{f(x)}{g(x)} = \displaystyle \lim_{x \to a} \frac{f^\prime (x)}{g^\prime (x)} $
diturunkan sampai hasilnya tidak sama dengan $ \frac{0}{0} $
$\begin{align} & \displaystyle \lim_{x \to 1} \frac{4(x^4-1)}{x^2-x^{-1}} = \frac{0}{0} \, \, \, \text{(diturunkan)} \\ & = \displaystyle \lim_{x \to 1} \frac{4(x^3)}{2x+\frac{1}{x^2}} \\ & = \frac{4(1^3)}{2.1+\frac{1}{1^2}} \\ & = \frac{16}{3} \end{align}$
$\spadesuit \, $ Menentukan hasil limitnya
$\begin{align} & \displaystyle \lim_{x \to 1} \left[ \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1}+ \frac{4(x^4-1)}{x^2-x^{-1}} \right] \\ & = \displaystyle \lim_{x \to 1} \left( \frac{4}{x^2-x} -\frac{4-3x+x^2}{1-x^3} \right)^{-1} + \displaystyle \lim_{x \to 1} \frac{4(x^4-1)}{x^2-x^{-1}} \\ & = 0 + \frac{16}{3} \\ & = \frac{16}{3} \end{align}$
Jadi, nilai limitnya adalah $ \frac{16}{3} . \heartsuit $
Nomor 9
Misalkan $f(0)=1$ dan $f^\prime(0)=2$. Jika $g(x)=\cos (f(x))$, maka $g^\prime(0)=...$
$\clubsuit \, $ Konsep dasar turunan
$ y = f[g(x)] \rightarrow y^\prime = f^\prime [g(x)] . g^\prime (x) $
$\clubsuit \, $ Menentukan turunan fungsi dan substitusi $ x = 0 $
$\begin{align} g(x) & =\cos (f(x)) \\ g^\prime (x) & = f^\prime (x) . [-\sin (f(x)) ] \\ x=0 \rightarrow g^\prime (0) & = - f^\prime (0) . \sin (f(0)) \\ g^\prime (0) & = - 2 . \sin (1) \\ g^\prime (0) & = - 2 \sin 1 \end{align}$
Jadi, nilai turunannya $ g^\prime (0) = - 2 \sin 1 . \heartsuit $
$ y = f[g(x)] \rightarrow y^\prime = f^\prime [g(x)] . g^\prime (x) $
$\clubsuit \, $ Menentukan turunan fungsi dan substitusi $ x = 0 $
$\begin{align} g(x) & =\cos (f(x)) \\ g^\prime (x) & = f^\prime (x) . [-\sin (f(x)) ] \\ x=0 \rightarrow g^\prime (0) & = - f^\prime (0) . \sin (f(0)) \\ g^\prime (0) & = - 2 . \sin (1) \\ g^\prime (0) & = - 2 \sin 1 \end{align}$
Jadi, nilai turunannya $ g^\prime (0) = - 2 \sin 1 . \heartsuit $
Nomor 10
Jika $\int \limits_{-1}^a \frac{x+1}{(x+2)^4} \, dx=\frac{10}{81}$ dan $a > -2$, maka $a=...$
$\spadesuit \, $ Konsep dasar integral
$ \int (ax+b)^n \, dx = \frac{1}{a}. \frac{1}{n+1}.(ax+b){n+1} + c $
$\spadesuit \, $ Menentukan integral dan nilai $ a $
$\begin{align} \int \limits_{-1}^a \frac{x+1}{(x+2)^4} \, dx & = \frac{10}{81} \\ \int \limits_{-1}^a \frac{(x+2)}{(x+2)^4} - \frac{1}{(x+2)^4} \, dx & = \frac{10}{81} \\ \int \limits_{-1}^a \frac{1}{(x+2)^3} - \frac{1}{(x+2)^4} \, dx & = \frac{10}{81} \\ \int \limits_{-1}^a (x+2)^{-3} - (x+2)^{-4} \, dx & = \frac{10}{81} \\ \left[ \frac{1}{1}. \frac{1}{-3+1}.(x+2)^{-3+1} - \frac{1}{1}. \frac{1}{-4+1}.(x+2)^{-4+1} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{1}{-2}.(x+2)^{-2} - \frac{1}{-3}.(x+2)^{-3} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{-1}{2(x+2)^2} + \frac{1}{3(x+2)^3} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{1}{3(x+2)^3} - \frac{1}{2(x+2)^2} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] - \left[ \frac{1}{3(-1+2)^3} - \frac{1}{2(-1+2)^2} \right] & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] - \left[ - \frac{1}{6} \right] & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] + \left[ \frac{1}{6} \right] & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] = \frac{10}{81} & - \frac{1}{6} \\ \text{(kedua ruas disederhanankan)} & \\ \frac{3a+4}{6(a+2)^3} & = \frac{7}{162} \end{align}$
Sehingga : $ 3a + 4 = 7 \rightarrow a = 1 $
Jadi, nilai $ a = 1. \heartsuit $
$ \int (ax+b)^n \, dx = \frac{1}{a}. \frac{1}{n+1}.(ax+b){n+1} + c $
$\spadesuit \, $ Menentukan integral dan nilai $ a $
$\begin{align} \int \limits_{-1}^a \frac{x+1}{(x+2)^4} \, dx & = \frac{10}{81} \\ \int \limits_{-1}^a \frac{(x+2)}{(x+2)^4} - \frac{1}{(x+2)^4} \, dx & = \frac{10}{81} \\ \int \limits_{-1}^a \frac{1}{(x+2)^3} - \frac{1}{(x+2)^4} \, dx & = \frac{10}{81} \\ \int \limits_{-1}^a (x+2)^{-3} - (x+2)^{-4} \, dx & = \frac{10}{81} \\ \left[ \frac{1}{1}. \frac{1}{-3+1}.(x+2)^{-3+1} - \frac{1}{1}. \frac{1}{-4+1}.(x+2)^{-4+1} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{1}{-2}.(x+2)^{-2} - \frac{1}{-3}.(x+2)^{-3} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{-1}{2(x+2)^2} + \frac{1}{3(x+2)^3} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{1}{3(x+2)^3} - \frac{1}{2(x+2)^2} \right]_{-1}^a & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] - \left[ \frac{1}{3(-1+2)^3} - \frac{1}{2(-1+2)^2} \right] & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] - \left[ - \frac{1}{6} \right] & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] + \left[ \frac{1}{6} \right] & = \frac{10}{81} \\ \left[ \frac{1}{3(a+2)^3} - \frac{1}{2(a+2)^2} \right] = \frac{10}{81} & - \frac{1}{6} \\ \text{(kedua ruas disederhanankan)} & \\ \frac{3a+4}{6(a+2)^3} & = \frac{7}{162} \end{align}$
Sehingga : $ 3a + 4 = 7 \rightarrow a = 1 $
Jadi, nilai $ a = 1. \heartsuit $
Tidak ada komentar:
Posting Komentar
Catatan: Hanya anggota dari blog ini yang dapat mengirim komentar.