Pembahasan Soal Ujian Nasional (UN) SMA Matematika IPA Kode 1 tahun 2014 nomor 26 sampai 30

Nomor 26

Nilai dari

$\cos 265^o - \cos 95^o = ...$

$\spadesuit \,$ Rumus dasar :

$\cos x - \cos y = -2\sin \left( \frac{x+y}{2} \right).\sin \left( \frac{x-y}{2} \right)$

$\begin{align*} \cos 265^o - \cos 95^o & = -2\sin \left( \frac{265^o+95^o}{2} \right).\sin \left( \frac{265^o-95^o}{2} \right) \\ &= -2\sin 180^o.\sin 85^o \\ &= -2\times 0 \times \sin 85^o \\ &= 0 \end{align*}$
Jadi, nilai

$\cos 265^o - \cos 95^o = 0. \heartsuit$

Nomor 27

Nilai

$\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2+x+5} - \sqrt{x^2-2x+3} \right) =...$

$\clubsuit \,$ Rumus :

$\displaystyle \lim_{x \to \infty} \left( \sqrt{ax^2+bx+c} - \sqrt{ax^2+px+q} \right) = \frac{b-p}{2\sqrt{a}}$

$\begin{align*} \displaystyle \lim_{x \to \infty} \left( \sqrt{x^2+x+5} - \sqrt{x^2-2x+3} \right) &= \frac{b-p}{2\sqrt{a}} \\ &= \frac{1-(-2)}{2\sqrt{1}} \\ &= \frac{3}{2} \end{align*}$
Jadi, nilai

$\displaystyle \lim_{x \to \infty} \left( \sqrt{x^2+x+5} - \sqrt{x^2-2x+3} \right)=\frac{3}{2} .\heartsuit$

Nomor 28

Nilai dari

$\displaystyle \lim_{x \to 0} \frac{1-\cos x}{2x.\sin 2x} = ...$

$\spadesuit \,$ Rumus dasar :

$\cos px = 1-2\sin ^2 \frac{p}{2}x \,$ dan

$\, \displaystyle \lim_{x \to 0} \frac{\sin ax }{bx}= \frac{a}{b}$

$\spadesuit \,$ sehingga

$\cos x = 1-2\sin ^2 \frac{1}{2}x$
diperoleh:

$1-\cos x = 1- (1-2\sin ^2 \frac{1}{2}x ) = 2\sin ^2 \frac{1}{2}x = 2\sin \frac{1}{2}x \sin \frac{1}{2}x$

$\spadesuit \,$ Menghitung nilai limitnya

$\begin{align*} \displaystyle \lim_{x \to 0} \frac{1-\cos x}{2x.\sin 2x} &= \displaystyle \lim_{x \to 0} \frac{2\sin \frac{1}{2}x \sin \frac{1}{2}x}{2x.\sin 2x} \\ &= \displaystyle \lim_{x \to 0} \frac{2\sin \frac{1}{2}x }{2x} . \frac{\sin \frac{1}{2}x }{\sin 2x} \\ &= \frac{2. \frac{1}{2} }{2} . \frac{\frac{1}{2} }{2} \\ &= \frac{1}{2}. \frac{1}{4} \\ &= \frac{1}{8} \end{align*}$
Jadi, nilai

$\displaystyle \lim_{x \to 0} \frac{1-\cos x}{2x.\sin 2x} = \frac{1}{8} . \heartsuit$

Cara II :

$\spadesuit \,$ Rumus yang digunakan:

$1-\cos px = \frac{1}{2}(px)^2 \,$ dan

$\, \displaystyle \lim_{x \to 0} \frac{\sin ax }{bx}= \frac{a}{b}$

$\begin{align*} \displaystyle \lim_{x \to 0} \frac{1-\cos x}{2x.\sin 2x} &= \displaystyle \lim_{x \to 0} \frac{\frac{1}{2}(1x)^2}{2x.\sin 2x} \\ &= \displaystyle \lim_{x \to 0} \frac{1}{4} . \frac{x}{\sin 2x} \\ &= \frac{1}{4}. \frac{1}{2} \\ &= \frac{1}{8} \end{align*}$
Jadi, nilai

$\displaystyle \lim_{x \to 0} \frac{1-\cos x}{2x.\sin 2x} = \frac{1}{8} . \heartsuit$

Nomor 29

Diketahui fungsi

$g(x)= \frac{1}{3}x^3-A^2x-7$ , A konstanta. Jika

$f(x)=g(2x-1)$ dan

$f$ turun pada

$-\frac{1}{2} \leq x \leq \frac{3}{2}$ , nilai maksimum relatif

$g(x)$ adalah ....

$\clubsuit \,$ Menentukan

$f(x)$ dan turunannya :

$\begin{align*} f(x)&=g(2x-1) \\ &= \frac{1}{3}(2x-1)^3-A^2(2x-1)-7 \\ f^\prime (x) &= (2x-1)^2 . 2 - 2A^2 \\ f^\prime (x) &= 8x^2-8x+2-2A^2 \end{align*}$

$\clubsuit \, f \,$ turun pada

$-\frac{1}{2} \leq x \leq \frac{3}{2}$ , artinya

$-\frac{1}{2}$ dan

$\frac{3}{2}$ adalah akar-akar dari

$f^\prime (x) = 0 \Rightarrow 8x^2-8x+2-2A^2 = 0$ .

$x_1 . x_2 = \frac{c}{a} \Rightarrow -\frac{1}{2} . \frac{3}{2} = \frac{2-2A^2}{8} \Rightarrow A^2 = 4$

$\clubsuit \,$ Fungsi

$g(x)$ menjadi

$g(x)= \frac{1}{3}x^3-4x-7$

$\clubsuit \,$ Nilai maksimum/minimum :

$g^\prime (x) = 0$

$g^\prime (x) = 0 \Leftrightarrow x^2 - 4 = 0 \Leftrightarrow x=\pm 2$

$x=2 \Rightarrow g(2)=\frac{1}{3}.2^3-4.2-7=-\frac{37}{3}$

$x=-2 \Rightarrow g(2)=\frac{1}{3}.(-2)^3-4.(-2)-7=-\frac{5}{3}$
Jadi, nilai maksimum relatif dari

$g(x)$ adalah

$-\frac{5}{3}. \heartsuit$

Nomor 30

Hasil dari

$\int \frac{5x-1}{\left( 5x^2-2x+6 \right)^7}dx$ adalah ...

$\spadesuit \,$ Menentukan integral dengan substitusi :

$\begin{align*} \int \frac{5x-1}{\left( 5x^2-2x+6 \right)^7}dx &=\int \frac{5x-1}{\left( u \right)^7} \frac{du}{u^\prime} \, \text{(misal : } \, u=5x^2-2x+6 ) \\ \\ &= \int \frac{5x-1}{u^7} \frac{du}{10x-2} \\ &= \int \frac{5x-1}{u^7} \frac{du}{2(5x-1)} \\ &= \frac{1}{2} \int u^{-7} du \\ &= \frac{1}{2}. \frac{1}{-6} . u^{-6} + c \\ &= -\frac{1}{12} . \frac{1}{u^6} + c \\ &= -\frac{1}{12\left( 5x^2-2x+6 \right)^6} + c \end{align*}$
Jadi,