PAS MATEMATIKA BLOG

Lanjutan Materi Pertidaksamaan Berkaitan Materi Pertidaksamaan Kuadrat

Sebelumya

silahkan buka di sini

A. PERTIDAKSAMAAN KUADRAT

$\begin{array}{ll} \underset{―}{Bentuk Umum} : \\ {\begin{cases} a x^{2} + b x + c < 0 \\ a x^{2} + b x + c \leq 0 \\ a x^{2} + b x + c > 0 \\ a x^{2} + b x + c \geq 0 \end{cases} \end{array}$

B. PENYELESAIAN PERTIDAKSAMAAN KUADRAT

$\begin{aligned} a x^{2} + b x + c . . . 0 \\ diubah menjadi & a x^{2} + b x + c = 0 \\ \Leftrightarrow a (x - x_{1}) (x - x_{2}) = 0 \\ \Leftrightarrow x = x_{1} atau x = x_{2} \end{aligned}$

$\begin{array}{lc} Pertidaksamaan & Himpunan Penyelesaian dengan x_{1} < x_{2} \\ a x^{2} + b x + c < 0 & {x | x_{1} < x < x_{2}, x \in R} \\ a x^{2} + b x + c \leq 0 & {x | x_{1} \leq x \leq x_{2}, x \in R} \\ a x^{2} + b x + c > 0 & {x | x < x_{1} atau x > x_{2}, x \in R} \\ a x^{2} + b x + c \geq 0 & {x | x \leq x_{1} atau x \geq x_{2}, x \in R} \end{array}$

$CONTOH SOAL$

$\begin{array}{ll} Tentukan himpunan penyelesaian \textbf{PtKSV} \\ (Pertidaksamaan Linear Satu Variabel) berikut ini! \\ a . x^{2} - 6 x + 8 < 0 \\ b . x^{2} - 6 x + 8 \leq 0 \\ c . x^{2} - 6 x + 8 > 0 \\ d . x^{2} - 6 x + 8 \geq 0 \\ Jawab : \end{array}$

$\begin{aligned} x^{2} - 6 x + 8 . . . 0 \\ diubah menjadi & x^{2} - 6 x + 8 = 0 \\ \Leftrightarrow 1 (x - 2) (x - 4) = 0 \\ \Leftrightarrow x = 2 atau x = 4 \end{aligned}$

$\begin{array}{ll}\\ &\begin{array}{|l|c|c|}\hline \textrm{Pertidaksamaan}&\textrm{HP dengan}\: \:\left (2<4 \right )&\textrm{Selang/Interval}\\\hline x^{2}-6x+8< 0&\left \{ x|2<x<4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{3-6} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{2}&&&\multicolumn{2}{l}{4}&& \end{array} \\\hline x^{2}-6x+8\leq 0&\left \{ x|2\leq x\leq 4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{3-6} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{\textcircled{2}}&&&\multicolumn{2}{l}{\textcircled{4}}&& \end{array} \\\hline x^{2}-6x+8> 0&\left \{ x|x<2\: \: \textrm{atau}\: \: x>4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{1-2}\cline{7-9} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{2}&&&\multicolumn{2}{l}{4}&& \end{array} \\\hline x^{2}-6x+8\geq 0&\left \{ x|x\leq 2\: \: \textrm{atau}\: \: x\geq 4,\: x\in \mathbb{R} \right \}&\begin{array}{ll|llll|llll}\\ &\multicolumn{2}{c}{.}&&&\multicolumn{2}{c}{.}&\\ &\multicolumn{2}{r}{.}&&&\multicolumn{2}{l}{.}&&\\\cline{1-2}\cline{7-9} &+&&-&-&&+&+&\\\hline &\multicolumn{2}{r}{\textcircled{2}}&&&\multicolumn{2}{l}{\textcircled{4}}&& \end{array} \\\hline \multicolumn{3}{|c|}{\begin{aligned}\textbf{Catatan: }&\textrm{Angka yang dilingkari termasuk}\\ &\textrm{himpunan penyelesaian} \end{aligned}}\\\hline \end{array} \end{array}$

Contoh Soal 2 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll} 6. & Penyelesaian dari pertidaksamaan \\ | \frac{3 - 2 x}{- 5} | > 5 adalah... . \\ \begin{array}{llll} a . & x < - 11 atau x > 14 \\ b . & x < - 14 atau x > 11 \\ c . & 11 < x < 14 \\ d . & - 14 < x < - 11 \\ e . & x > 14 \end{array} \\ Jawab : a \\ \begin{aligned} | \frac{3 - 2 x}{- 5} | & > 5 \\ \frac{3 - 2 x}{- 5} < & - 5 atau \frac{3 - 2 x}{- 5} > 5 \\ \frac{2 x - 3}{5} > & 5 atau \frac{2 x - 3}{5} < - 5 \\ 2 x - 3 > & 25 atau 2 x - 3 < - 25 \\ 2 x > 25 & + 3 atau 2 x < - 25 + 3 \\ x > 14 & atau x < - 11, \\ dapat juga dituliskan \\ x < - & 11 atau x > 14 \end{aligned} \end{array}$

$\begin{array}{ll} 7. & Nilai x yang memenuhi pertidaksamaan \\ | 2 - 2 | x + 1 | | > 4 adalah... . \\ \begin{array}{llll} a . & x < - 4 atau x > 2 \\ b . & x < - 3 atau x > 1 \\ c . & x < - 2 atau x > 0 \\ d . & x < - 1 atau x > 3 \\ e . & x < 0 atau x > 4 \end{array} \\ Jawab : a \\ \begin{aligned} | 2 - 2 | x + 1 | | > 4 \\ 2 - 2 | x + 1 | < - 4 & atau 2 - 2 | x + 1 | > 4 \\ - 2 | x + 1 | < - 6 & atau - 2 | x + 1 | > 2 \\ | x + 1 | > 3 & atau | x + 1 | < - 1 \\ {\begin{array}{c} (x + 1) < - 3 \\ (x + 1) > 3 \end{array} & atau {\begin{array}{c} | x + 1 | < - 1 \\ tak mungkin \end{array} \\ Selanjutnya & akan didapatkan \\ x < - 4 & atau x > 2 \end{aligned} \end{array}$

$\begin{array}{l} 8. & Nilai x yang memenuhi pertidaksamaan \\ | 3 - | x | | < 10 adalah... . \\ \begin{array}{llll} a . & x < - 14 atau x > 12 \\ b . & x < - 13 atau x > 13 \\ c . & x < - 12 atau x > 10 \\ d . & 0 < x < 10 \\ e . & - 13 < 0 < 13 \end{array} \\ Jawab : e \\ \begin{aligned} | 3 - | x | | & < 10 \\ - 10 < 3 - & | x | < 10 \\ - 13 < - & | x | < 7 \\ - 7 < & | x | \leq 13, (ingat harga | x | \geq 0) \\ 0 \leq & | x | < 13 \\ selanjutnya, \\ | x | < 13 \\ - 13 < & x < 13 \\ HP = & {x | - 13 < x < 13, x \in R} \end{aligned} \end{array}$

$\begin{array}{l} 9. & (UM UGM 05) \\ Nilai x yang memenuhi pertidaksamaan \\ | x^{2} - 3 | < 2 x adalah... . \\ \begin{array}{llll} a . & - 1 < x < 3 \\ b . & - 3 < x < 1 \\ c . & 1 < x < 3 \\ d . & - 3 < x < - 1 atau 1 < x < 3 \\ e . & x > 1 \end{array} \\ Jawab : c \\ \begin{aligned} | x^{2} - 3 | & < 2 x \\ - 2 x < (x^{2} - 3) & < 2 x \\ dipartisi men & jadi dua bagian \\ ∙ pertama \\ (x^{2} - 3) & > - 2 x \\ x^{2} + 2 x - 3 & > 0 \\ (x + 3) (x - 1) & > 0 \\ x < - 3 atau & x > 1 \\ ∙ kedua \\ (x^{2} - 3) & < 2 x \\ x^{2} - 2 x - 3 & < 0 \\ (x - 3) (x + 1) & < 0 \\ - 1 < x < 3 \\ ambil yang & memenuhi keduanya \\ berupa iris & an \\ HP = & {1 < x < 3, x \in R} \end{aligned} \end{array}$

$\begin{array}{ll} 10. & (SPMB 05) \\ Nilai x yang memenuhi pertidaksamaan \\ {| x - 2 |}^{2} < 4 | x - 2 | + 12 adalah... . \\ \begin{array}{llll} a . & {x \in R | 2 \leq x \leq 8} \\ b . & {x \in R | 4 < x < 8} \\ c . & {x \in R | - 4 < x < 8} \\ d . & {x \in R | - 2 < x < 4} \\ e . & {x \in R | 2 < x < 4} \end{array} \\ Jawab : c \\ \begin{aligned} misalkan p & = | x - 2 |, selanjutnya \\ {| x - 2 |}^{2} < & 4 | x - 2 | + 12 \\ p^{2} < & 4 p + 12 \\ p^{2} - & 4 p - 12 < 0 \\ (p - 6) & (p + 2) < 0 \\ - 2 < p & < 6, atau jika dikembalikan \\ - 2 < & | x - 2 | < 6, ingat, nilanya tidak negatif \\ 0 \leq & | x - 2 | < 6 \\ - 6 < & x - 2 < 6 \\ - 4 < & x < 8 \\ HP = & {- 4 < x < 8, x \in R} \end{aligned} \end{array}$

Contoh Soal 1 Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll} 1. & Nilai x yang memenuhi | \frac{1}{2} x + 6 | \geq 9 adalah... . \\ \begin{array}{llll} a . & - 12 < x < 6 \\ b . & - 30 \leq x \leq 6 \\ c . & x \geq 6 atau x \leq - 30 \\ d . & x < 6 atau x < - 30 \\ e . & x \leq 6 atau x \geq - 30 \end{array} \\ Jawab : c \\ \begin{aligned} | \frac{1}{2} x + 6 | & \geq 9 \\ \frac{1}{2} x + 6 & \leq - 9 atau \frac{1}{2} x + 6 \geq 9 \\ \frac{1}{2} x \leq & - 9 - 6 atau \frac{1}{2} x \geq 9 - 6 \\ \frac{1}{2} x \leq & - 15 atau \frac{1}{2} x \geq 3 \\ x \leq & - 30 atau x \geq 6 \end{aligned} \end{array}$

$\begin{array}{ll} 2. & Nilai x yang memenuhi pertidaksamaan \\ 3 | x + 1 | \leq | x - 2 | adalah... . \\ \begin{array}{llll} a . & \frac{1}{4} \leq x \leq - \frac{1}{4} \\ b . & - \frac{5}{2} \leq x \leq \frac{5}{2} \\ c . & x \leq \frac{1}{4} atau x \geq \frac{5}{2} \\ d . & x \leq - \frac{5}{2} atau x \geq \frac{1}{4} \\ e . & x \leq - \frac{5}{2} atau x \geq - \frac{1}{4} \end{array} \\ Jawab : b \\ \begin{aligned} 3 | x + 1 | & \leq | x - 2 | \\ {(3 | x + 1 |)}^{2} & \leq {(| x - 2 |)}^{2} \\ {(3 x + 3)}^{2} & \leq {(x - 2)}^{2} \\ (3 x + 3 + (x - 2)) & (3 x + 3 - (x - 2)) \leq 0 \\ (4 x + 1) (2 x + 5) & \leq 0 \\ HP = & {x | - \frac{5}{2} \leq x \leq - \frac{1}{4}, x \in R} \end{aligned} \end{array}$

$\begin{array}{l} 3. & Nilai x yang memenuhi pertidaksamaan \\ | x - 3 | < 3 adalah... . \\ \begin{array}{llll} a . & x < 3 \\ b . & - 3 < x < 3 \\ c . & x < - 3 atau x < 3 \\ d . & x > 0 atau x < 6 \\ e . & x < 0 atau x < 6 \end{array} \\ Jawab : d \\ \begin{aligned} | x - 3 | & < 3 \\ - 3 < (x - 3) & < 3 \\ - 3 + 3 < x & < 3 + 3 \\ 0 < x & < 6 \end{aligned} \end{array}$

$\begin{array}{l} 4. & Nilai x yang memenuhi pertidaksamaan \\ | x + 4 | > 8 adalah... . \\ \begin{array}{llll} a . & x > - 8 \\ b . & x < 4 atau x > 12 \\ c . & x > 4 atau x > - 12 \\ d . & x < - 4 atau x < 6 \\ e . & x > 4 atau x < - 12 \end{array} \\ Jawab : e \\ \begin{aligned} | x + 4 | & > 8 \\ (x + 4) < - 8 & atau (x + 4) > 8 \\ x < - 12 & atau x > 4 \end{aligned} \end{array}$

$\begin{array}{ll} 5. & Himpunan penyelesaian dari pertidaksamaan \\ | \frac{x + 1}{2} | > | \frac{x - 2}{3} | adalah... . \\ \begin{array}{llll} a . & HP = {x | - 7 < x < \frac{1}{5}, x \in R} \\ b . & HP = {x | x < - 7 atau x > \frac{1}{5}, x \in R} \\ c . & HP = {x | x > - 7, x \in R} \\ d . & HP = {x | - 1 < x < 2, x \in R} \\ e . & HP = {x | x < - 1 atau x > 2, x \in R} \end{array} \\ Jawab : b \\ \begin{aligned} | \frac{x + 1}{2} | & > | \frac{x - 2}{3} | \\ {(\frac{x + 1}{2})}^{2} & > {(\frac{x - 2}{3})}^{2} \\ (\frac{x + 1}{2} + \frac{x - 2}{3}) & (\frac{x + 1}{2} - \frac{x - 2}{3}) > 0 \\ (\frac{3 (x + 1) + 2 (x - 2)}{6}) & (\frac{3 (x + 1) - 2 (x - 2)}{6}) > 0 \\ (\frac{5 x - 1}{6}) & (\frac{x + 7}{6}) > 0 \\ HP = & {x | x < - 7 atau x > \frac{1}{5}, x \in R} \end{aligned} \end{array}$

Pertidaksamaan Nilai Mutlak (Kelas X Matematika Wajib)

$A. Sifat-Sifat yang Berlaku$

Untuk a > 0, maka

berlaku sifat-sifat sebagai berikut:

${\begin{cases} | x | < a & \Leftrightarrow - a < x < a \\ | x | \leq a & \Leftrightarrow - a \leq x \leq a \\ | x | > a & \Leftrightarrow x < - a atau a > a \\ | x | \geq a & \Leftrightarrow x \leq - a atau a \geq a \\ dan & perlu diingat \\ | x | & = \sqrt{x^{2}} \end{cases}$

$B. Penyelesaian Pertidaksamaan$

$a. bentuk pertama$

$| f (x) | {\begin{cases} < a & \Rightarrow - a \leq f (x) \leq a \\ \leq a & \Rightarrow - a \leq f (x) \leq a \\ > a & \Rightarrow f (x) < - a atau f (x) > a \\ \geq a & \Rightarrow f (x) \leq - a atau f (x) \geq a \end{cases}$

$CONTOH SOAL$

$\begin{array}{l} Tentukanlah himpunan penyelesaian dari \\ (1) . | 3 x - 7 | \leq 11 \\ (2) . | 3 x - 7 | > 11 \\ Jawab : \end{array}$

$\begin{aligned} (1) . | 3 x - 7 | & \leq 11 \\ - 11 \leq 3 x - 7 & \leq 11 \\ - 11 + (7) \leq 3 x - 7 + (7) & \leq 11 + (7) \\ - 4 \leq 3 x & \leq 18 (dibagi) 3 \\ - \frac{4}{3} \leq x & \leq 6 \\ HP = & {x | - \frac{4}{3} \leq x \leq 6, x \in R} \end{aligned}$

$\begin{aligned} (2) . | 3 x - 7 | & > 11 \\ (3 x - 7) < - 11 & atau (3 x - 7) > 11 \\ 3 x < - 11 + 7 & atau 3 x > 11 + 7 \\ x < - \frac{4}{3} atau & x > 6 \\ HP = & {x | x < - \frac{4}{3} atau x > 6, x \in R} \end{aligned}$

$b. bentuk kedua$

$| f (x) | {\begin{cases} < | g (x) | \\ \leq | g (x) | \\ > | g (x) | \\ \geq | g (x) | \end{cases} \Rightarrow dikuadratkan kedua ruas$

Perlu diingat di sini bahwa bentuk selanjutnya jika berupa pertidaksamaan kuadrat, maka langkah selanjutnya perlu diperhatikan langkah-langkah berikut:

$\begin{aligned} a x^{2} + b x + c & {\begin{cases} < \\ \leq \\ > \\ \geq \end{cases} 0 \\ dengan & a, b, c \in R, a \neq 0 \end{aligned}$

dan juga

$Akar-akar {\begin{cases} x_{1} & dan x_{2} dengan x_{1} \neq x_{2} \\ serta \\ x_{1} & \leq x_{2} \end{cases}$

maka dalam menentukan himpunan penyelesaiannya adalah sebagai berikut:

${\begin{cases} a x^{2} + b x + c < 0, & HP = {x | x_{1} < x < x_{2}, x \in R} \\ a x^{2} + b x + c \leq 0, & HP = {x | x_{1} \leq x \leq x_{2}, x \in R} \\ a x^{2} + b x + c > 0, & HP = {x | x < x_{1} atau x > x_{2}, x \in R} \\ a x^{2} + b x + c \geq 0, & HP = {x | x \leq x_{1} atau x \geq x_{2}, x \in R} \end{cases}$

$CONTOH SOAL$

$\begin{array}{l} Tentukanlah himpunan penyelesaian dari \\ (1) . | 3 x - 7 | \leq | x - 1 | \\ (2) . | 3 x - 7 | > | x - 1 | \\ Jawab : \end{array}$

$\begin{aligned} (1) . | 3 x - 7 | \leq | x - 1 | \\ {| 3 x - 7 |}^{2} \leq {| x - 1 |}^{2} \\ (3 x - 7)^{2} - (x - 1)^{2} & \leq 0 \\ (3 x - 7 + x - 1) & (3 x - 7 - (x - 1)) \leq 0 \\ (4 x - 8) (2 x - 6) & \leq 0 \\ (x - 2) (x - 3) & \leq 0 \\ HP = & {x | 2 \leq x \leq 3, x \in R} \end{aligned}$

$\begin{aligned} (2) . | 3 x - 7 | & > | x - 1 | \\ {| 3 x - 7 |}^{2} & > {| x - 1 |}^{2} \\ (3 x - 7)^{2} - & (x - 1)^{2} > 0 \\ (3 x - 7 + & x - 1) (3 x - 7 - (x - 1)) > 0 \\ (4 x - 8) & (2 x - 6) > 0 \\ (x - 2) & (x - 3) > 0 \\ HP = & {x | x < 2 atau x > 3, x \in R} \end{aligned}$

$c. bentuk ketiga$

Penyelesaian jenis ini mirip poin yang pertama, yaitu:

$\begin{array}{l} | f (x) | {\begin{cases} {\begin{array}{c} < g (x) \\ \leq g (x) \end{array} & maka {\begin{array}{c} - g (x) < f (x) < g (x) \\ - g (x) \leq f (x) \leq g (x) \end{array} \\ {\begin{array}{c} > g (x) \\ \geq g (x) \end{array} & maka {\begin{array}{c} {\begin{matrix} f (x) > g (x) \\ f (x) < - g (x) \end{matrix} \\ {\begin{matrix} f (x) \geq g (x) \\ f (x) \leq - g (x) \end{matrix} \end{array} \end{cases} \end{array}$

$CONTOH SOAL$

$\begin{array}{l} Tentukanlah himpunan penyelesaian dari \\ (1) . | 3 x - 7 | \leq (x - 1) \\ (2) . | 3 x - 7 | > (x - 1) \\ Jawab : \end{array}$

$\begin{aligned} (1) . - g (x) \leq f (x) & \leq g (x) \\ (a) . untuk f (x) & \geq - g (x) \\ 3 x - 7 & \geq - (x - 1) \\ 4 x & \geq 8 \\ x & \geq 2 \\ (b) . untuk f (x) & \leq g (x) \\ 3 x - 7 & \leq x - 1 \\ 2 x & \leq 6 \\ x & \leq 3 \\ HP = & {x | 2 \leq x \leq 3, x \in R} \end{aligned}$

$\begin{aligned} (2) . f (x) < - g (x) & atau f (x) > g (x) \\ (a) . untuk f (x) & < - g (x) \\ 3 x - 7 & < - (x - 1) \\ 4 x & < 8 \\ x & < 2 \\ (b) . untuk f (x) & > g (x) \\ 3 x - 7 & > x - 1 \\ 2 x & > 6 \\ x & > 3 \\ HP = & {x | x < 2 atau x > 3, x \in R} \end{aligned}$

Daftar Pustaka

Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X (Wajib). Bandung: Srikandi Empat Widya Utama.
Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata PElajaran Wajib. Solo: PT Tiga Serangkai Pustaka Mandiri.

Contoh Soal Lanjutan 4 Nilai Mutlak

$\begin{array}{ll} 21. & Jumlah akar-akar persamaan \\ x^{2} + | x | - 6 = 0, adalah . . . . \\ \begin{array}{llll} a . & - 1 \\ b . & 0 \\ c . & 1 \\ d . & 2 \\ e . & 4 \end{array} \\ Jawab : b \\ \begin{aligned} x^{2} + | x | & - 6 = 0 \\ {| x |}^{2} + | x | & - 6 = 0 \\ (| x | + 3) & (| x | - 2) = 0 \\ | x | = - 3 & atau | x | = 2 \\ tidak meme & nuhi (-3) atau x = \pm 2 \\ {\begin{cases} x_{1} & = 2 \\ x_{2} & = - 2 \end{cases} \\ x_{1} + x_{2} & = 2 + (- 2) = 0 \end{aligned} \end{array}$

$\begin{array}{ll} 22. & Penyelesaian untuk - 2 | x - 5 | = 8 adalah... . \\ \begin{array}{llll} a . & {0} \\ b . & {} \\ c . & {4, 5} \\ d . & {1, 9} \\ e . & {- 1, 9} \end{array} \\ Jawab : b \\ \begin{aligned} - 2 | x - 5 | & = 8 \\ | x - 5 | & = - 4 \\ Karena pe & rsamaan bernilai negatif, \\ maka tida & k ada nilai x yang memenuhi \\ ∴ HP & = {} \end{aligned} \end{array}$

$\begin{array}{ll} 23. & Nilai x yang memenuhi | \frac{x - 5}{2 x + 1} | = 2 adalah... . \\ \begin{array}{llll} a . & - \frac{5}{3} atau \frac{3}{5} \\ b . & \frac{5}{3} atau - \frac{3}{5} \\ c . & - \frac{7}{3} atau \frac{3}{5} \\ d . & \frac{7}{3} atau - \frac{3}{5} \\ e . & - \frac{7}{3} atau \frac{3}{7} \end{array} \\ Jawab : c \\ \begin{aligned} | \frac{x - 5}{2 x + 1} | & = 2 \\ \frac{x - 5}{2 x + 1} & = \pm 2 \\ untuk & x = 2 \\ \frac{x - 5}{2 x + 1} & = 2 \\ x - 5 & = 2 (2 x + 1) \\ x - 4 x & = 2 + 5 \\ - 3 x & = 7 \\ x & = \frac{7}{- 3} = - \frac{7}{3} \\ untuk & x = - 2 \\ x - 5 & = - 2 (2 x + 1) \\ x + 4 x & = - 2 + 5 \\ 5 x & = 3 \\ x & = \frac{3}{5} \end{aligned} \end{array}$

$\begin{array}{ll} 24. & Nilai x yang memenuhi | | 5 x - 4 | - 3 | = 2 adalah... . \\ \begin{array}{llll} a . & - 3 atau - 6 \\ b . & - 3 atau 6 \\ c . & 3 atau - 6 \\ d . & 3 atau 6 \\ e . & 6 atau 9 \end{array} \\ Jawab : b \\ \begin{aligned} | | 2 x - 3 | - 4 | & = 5 \\ | 2 x - 3 | - 4 & = \pm 5 \\ | 2 x - 3 | & = \pm 5 + 4 maka, \\ | 2 x - 3 | & = 5 + 4 = 9 (memenuhi) \\ | 2 x - 3 | & = - 5 + 4 = - 1 \\ (tidak memenuhi) \\ Selanjutnya & , \\ (2 x - 3) & = \pm 9 \\ 2 x & = \pm 9 + 3 \\ untuk & x = 9 \\ 2 x & = 9 + 3 \\ x & = \frac{12}{2} = 6 \\ untuk & x = - 9 \\ 2 x & = - 9 + 3 \\ x & = \frac{- 6}{2} = - 3 \end{aligned} \end{array}$

$\begin{array}{ll} 25. & Nilai x yang memenuhi \\ persamaan | x + 1 | - 2 | x - 3 | = 5 | x - 4 | adalah... . \\ \begin{array}{llll} a . & 1 \frac{1}{6} atau 1 \frac{2}{3} \\ b . & 1 \frac{5}{6} atau 2 \frac{2}{5} \\ c . & 1 \frac{5}{6} atau 2 \frac{2}{3} \\ d . & 2 \frac{3}{4} atau 4 \frac{2}{3} \\ e . & 3 \frac{1}{4} atau 4 \frac{1}{2} \end{array} \\ Jawab : e \\ \begin{aligned} | x + 1 | - & 2 | x - 3 | = 5 | x - 4 | \\ Perhatik & an untuk batas sesuai definisi, maka \\ x = - 1 & , x = 3, x = 4 \\ saat x & \geq 4 \\ | x + 1 | & = x + 1, | x - 3 | = x - 3, | x - 4 | = x - 4 \\ (x + 1) - & 2 (x - 3) = 5 (x - 4) \\ x - 2 x - & 5 x = - 1 - 6 - 20 \\ - 6 x & = - 27 \\ x & = \frac{27}{6} = 4 \frac{3}{6} (memenuhi) \\ saat 3 & \leq x < 4 \\ | x + 1 | & = x + 1, | x - 3 | = x - 3, | x - 4 | = 4 - x \\ (x + 1) - & 2 (x - 3) = 5 (4 - x) \\ x - 2 x + & 5 x = - 1 - 6 + 20 \\ 4 x & = 13 \\ x & = \frac{13}{4} = 3 \frac{1}{4} (memenuhi) \\ saat - & 1 \leq x < 3 \\ | x + 1 | & = x + 1, | x - 3 | = 3 - x, | x - 4 | = 4 - x \\ (x + 1) - & 2 (3 - x) = 5 (4 - x) \\ x + 2 x + & 5 x = - 1 + 6 + 20 \\ 8 x & = 25 \\ x & = \frac{25}{8} = 3 \frac{1}{8} (tidak memenuhi) \\ saat - & 1 > x \\ | x + 1 | & = - x - 1, | x - 3 | = 3 - x, | x - 4 | = 4 - x \\ (- x - 1) & - 2 (3 - x) = 5 (4 - x) \\ - x + 2 x & + 5 x = 1 + 6 + 20 \\ 6 x & = 27 \\ x & = \frac{27}{6} = 4 \frac{3}{6} (tidak memenuhi) \end{aligned} \end{array}$

Sumber Referensi

Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X. Bandung: Srikandi Empat Widya Utama.
Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.

Contoh Soal Lanjutan 3 Nilai Mutlak

$\begin{array}{l} 16. & Penyelesaian dari | 3 x - (4 x - 7) | = 6 adalah... . \\ \begin{array}{llll} a . & {- 1, - 13} \\ b . & {- 1, 13} \\ c . & {1, 13} \\ d . & {- 13, 1} \\ e . & {13} \end{array} \\ Jawab : c \\ \begin{aligned} | 3 x - (4 x - 7) | & = 6 \\ | 3 x - 4 x + 7 | & = 6 \\ | - x + 7 | & = 6 \\ (- x + 7) & = \pm 6 \\ - x & = \pm 6 - 7 \\ dikalikan & dengan - 1 \\ x & = \pm 6 + 7 \\ x_{1} & = + 6 + 7 = + 13 \\ x_{2} & = - 6 + 7 = + 1 \end{aligned} \end{array}$

$\begin{array}{l} 17. & Penyelesaian | x - 1 | = 2 x + 1 adalah... . \\ \begin{array}{llll} a . & {- 2} \\ b . & {- 2, 0} \\ c . & {- 1} \\ d . & {} \\ e . & {0} \end{array} \\ Jawab : e \\ \begin{aligned} | x - 1 | & = 2 x + 1 \\ (x - 1) & = \pm (2 x + 1) \\ untuk & {\begin{cases} x \geq 1 & maka (x - 1) = 2 x + 1 \\ x < 1 & maka (x - 1) = - (2 x + 1) \end{cases} \\ \begin{array}{cc} x \geq 1 & x < 1 \\ \begin{aligned} x - 1 & = 2 x + 1 \\ x - 2 x & = 1 + 1 \\ - x & = 2 \\ x & = - 2 \end{aligned} & \begin{aligned} x - 1 & = - (2 x + 1) \\ x - 1 & = - 2 x - 1 \\ x + 2 x & = - 1 + 1 \\ 3 x & = 0 \\ x & = 0 \end{aligned} \\ Tidak memenuhi & memenuhi \end{array} \end{aligned} \end{array}$

$\begin{array}{l} 18. & Penyelesaian | 3 a + 1 | = 2 a + 9 adalah... . \\ \begin{array}{llll} a . & {- 2} \\ b . & {8} \\ c . & {- 2, 8} \\ d . & {} \\ e . & Semua bilangan riil \end{array} \\ Jawab : c \\ \begin{aligned} | 3 a + 1 | & = 2 a + 9 \\ (3 a + 1) & = \pm (2 a + 9) \\ untuk & {\begin{cases} a \geq - \frac{1}{3} & maka (3 a + 1) = 2 a + 9 \\ a < - \frac{1}{3} & maka (3 a + 1) = - (2 a + 9) \end{cases} \\ \begin{array}{cc} a \geq - \frac{1}{3} & a < - \frac{1}{3} \\ \begin{aligned} 3 a + 1 & = 2 a + 9 \\ 3 a - 2 a & = 9 - 1 \\ a & = 8 \end{aligned} & \begin{aligned} 3 a + 1 & = - (2 a + 9) \\ 3 a + 1 & = - 2 a - 9 \\ 3 a + 2 a & = - 9 - 1 \\ 5 a & = - 10 \\ a & = \frac{- 10}{5} \\ a & = - 2 \end{aligned} \\ memenuhi & memenuhi \end{array} \end{aligned} \end{array}$

$\begin{array}{l} 19. & Penyelesaian dari | 3 x + 2 | = 4 x + 5 adalah... . \\ \begin{array}{llll} a . & - 3 \\ b . & - 1 \\ c . & - 3 dan - 1 \\ d . & {} \\ e . & 0 \end{array} \\ Jawab : b \\ \begin{array}{r} \begin{aligned} | 3 x + 2 | & = 4 x + 5 \\ (3 x + 2) & = \pm (4 x + 5) \\ untuk & {\begin{cases} x \geq - \frac{2}{3} & maka 3 x + 2 = 4 x + 5 \\ x < - \frac{2}{3} & maka 3 x + 2 = - (4 x + 5) \end{cases} \\ \begin{array}{cc} x \geq - \frac{2}{3} & x < - \frac{2}{3} \\ \begin{aligned} 3 x + 2 & = 4 x + 5 \\ 3 x - 4 x & = 5 - 2 \\ - & = 3 \\ x & = - 3 \end{aligned} & \begin{aligned} 3 x + 2 & = - (4 x + 5) \\ 3 x + 2 & = - 4 x - 5 \\ 3 x + 4 x & = - 5 - 2 \\ 7 x & = - 7 \\ x & = \frac{- 7}{7} \\ x & = - 1 \end{aligned} \end{array} \end{aligned} \end{array} \end{array}$

$\begin{array}{l} 20. & Jika | - 3 x | + 4 y^{- 1} = 6 z + 4 x, \\ maka nilai x dinyatakan dalam y dan z adalah . . . . \\ \begin{array}{llll} a . & \frac{y}{4} - 6 z \\ b . & \frac{4}{y} - 6 z \\ c . & 4 - \frac{6}{y} \\ d . & \frac{4}{7 y} + \frac{6 z}{7} \\ e . & \frac{7}{4 y} + 6 z \end{array} \\ Jawab : b \\ \begin{aligned} | - 3 x | & + 4 y^{- 1} = 6 z + 4 x \\ | - 3 x | & = 6 z + 4 x - 4 y^{- 1} \\ - 3 x & = \pm (6 z + 4 x - 4 y^{- 1}) \\ untuk & : x \geq 0 \\ - 3 x & = 6 z + 4 x - 4 y^{- 1} \\ - 3 x - 4 x & = 6 z - 4 y^{- 1} \\ - 7 x & = 6 z - 4 y^{- 1} \\ x & = \frac{6 z - 4 y^{- 1}}{- 7} \\ = \frac{4}{7 y} - \frac{6 z}{7} \\ u n t u k & : x < 0 \\ - 3 x & = - (6 z + 4 x - 4 y^{- 1}) \\ - 3 x & = - 6 z - 4 x + 4 y^{- 1} \\ - 3 x + 4 x & = - 6 z + 4 y^{- 1} \\ x & = 4 y^{- 1} - 6 z \\ = \frac{4}{y} - 6 z \end{aligned} \end{array}$

Contoh Soal Lanjutan 2 Nilai Mutlak

$\begin{array}{l} 11. & Nilai p yang memenuhi 10 - 4 | 4 - 5 p | = 26 adalah... . \\ \begin{array}{lllllll} a . & 2 atau 1 \frac{2}{3} \\ b . & 1 atau - \frac{3}{5} \\ c . & 2 atau 2 \frac{3}{5} \\ d . & - 2 \frac{3}{4} atau 1 \\ e . & - 1 atau 2 \frac{3}{5} \end{array} \\ Jawab : e \\ \begin{aligned} 10 - 4 | 4 - 5 p | & = - 26 \\ - 4 | 4 - 5 p | & = - 36 \\ | 4 - 5 p | & = 9 \\ (4 - 5 p) & = \pm 9 \\ - 5 p & = - 4 \pm 9 \\ p & = \frac{- 4 \pm 9}{- 5} \\ p & = {\begin{cases} \frac{- 4 + 9}{- 5} & = - 1 \\ atau \\ \frac{- 4 - 9}{- 5} & = \frac{13}{5} = 2 \frac{3}{5} \end{cases} \end{aligned} \end{array}$

$\begin{array}{l} 12. & Jika 3 < x < 5 maka penyelesaian untuk \\ \sqrt{x^{2} - 6 x + 9} - \sqrt{x^{2} - 10 x + 25} = . . . \\ \begin{array}{lllllll} a . & 2 x - 2 \\ b . & 2 \\ c . & 8 - 2 x \\ d . & - 2 \\ e . & 2 x - 8 \end{array} \\ Jawab : e \\ \begin{aligned} \sqrt{x^{2} - 6 x + 9} & - \sqrt{x^{2} - 10 x + 25} \\ \sqrt{(x - 3)^{2}} - \sqrt{(x - 5)^{2}} \\ = | x - 3 | - | x - 5 | \\ ingat bahwa saat \\ 3 < x < 5 & maka \\ {\begin{cases} | x - 3 | = (x - 3) \\ | x - 5 | = - (x - 5) \end{cases}, \\ sehingga \\ = | x - 3 | - | x - 5 | \\ = (x - 3) - (- (x - 5)) \\ = x - 3 + x - 5 \\ = 2 x - 8 \end{aligned} \end{array}$

$\begin{array}{l} 13. & Jika 1 < x < 5 maka penyelesaian untuk \\ \sqrt{x^{2} - 2 x + 1} + \sqrt{x^{2} - 10 x + 25} = . . . \\ \begin{array}{lllllll} a . & 2 \\ b . & 3 \\ c . & 4 \\ d . & 5 \\ e . & 6 \end{array} \\ Jawab : c \\ \begin{aligned} \sqrt{x^{2} - 2 x + 1} & + \sqrt{x^{2} - 10 x + 25} \\ = \sqrt{(x - 1)^{2}} + \sqrt{(x - 5)^{2}} \\ = | x - 1 | + | x - 5 | \\ ingat bahwa saat \\ 1 < x < 5 & maka {\begin{cases} | x - 1 | = (x - 1) \\ | x - 5 | = - (x - 5) \end{cases}, \\ sehingga \\ = | x - 1 | + | x - 5 | \\ = (x - 1) + (- (x - 5)) \\ = x - 1 + 5 - x \\ = 4 \end{aligned} \end{array}$

$\begin{array}{ll} 14. & Perhatikanlah ilustrasi grafik di bawah ini \\ \begin{matrix} \end{matrix} \end{array}$

$\begin{array}{l} . & Persamaan yang memenuhi rumus tersebut adalah... . \\ \begin{array}{llll} a . & y = | - x - 2 | \\ b . & y = - 2 x - 4 \\ c . & y = - | 2 x - 4 | \\ d . & y = | - 2 x - 4 | \\ e . & y = | - 2 x + 4 | \end{array} \\ Jawab : e \\ \begin{aligned} Dengan cara substitusi langsung \\ kita akan mendapatkan \\ ∙ untuk x = 4 menyebabkan nilai y = 4, \\ maka sampai langkah di sini hanya \\ ada 1 persamaan yang memenuhi \\ yaitu : y = | - 2 x + 4 | \end{aligned} \end{array}$

$\begin{array}{l} 15. & Gambarlah garfik untuk persamaan | x | + | y | = 4 \\ Jawab : \\ untuk | x | + | y | = 4 \\ \begin{array}{cccc} x > 0, y > 0 & x > 0, y < 0 \\ x + y = 4 & x + (- y) = 4 \\ x < 0, y > 0 & x < 0, y < 0 \\ (- x) + y = 4 & (- x) + (- y) = 4 \end{array} \end{array}$

$\begin{aligned} berikut gambar grafiknya \end{aligned}$

Contoh Soal Lanjutan Nilai Mutlak

$\begin{array}{ll} 6. & Perhatikanlah gambar berikut \end{array}$

\begin{array}{ll} Persamaan yang sesuai dengan grafik di atas adalah . . . . \\ \begin{array}{llll} a . & y = | x - 3 | \\ b . & y = - | x - 3 | \\ c . & y = | x + 3 | \\ d . & y = | x + 6 | \\ e . & y = - | x + 6 | \end{array} \\ Jawab : c \end{array}

\begin{array}{l} 7. & Diketahui suatu fungsi y = | 2 x | \\ Nilai saat - 2 adalah . . . . \\ \begin{array}{llll} a . & - 4 \\ b . & - 2 \\ c . & 0 \\ d . & 2 \\ e . & 4 \end{array} \\ Jawab : e \\ \begin{aligned} y & = | 2 x | \\ saat & x = - 2, maka \\ y & = | 2 (- 2) | \\ = | - 4 | \\ = - (- 4) \\ = 4 \end{aligned} \end{array}

\begin{array}{l} 8. & Diketahui suatu fungsi f (x) = | x - 2 | \\ Nilai f (- 2) adalah . . . . \\ \begin{array}{llll} a . & 0 \\ b . & 2 \\ c . & 4 \\ d . & 6 \\ e . & 8 \end{array} \\ Jawab : c \\ \begin{aligned} f (x) & = | x - 2 | \\ Nilai & f (- 2) berarti nilai \\ saat & x = - 2, maka \\ f (x) & = | x - 2 | \\ f (- 2) & = | - 2 - 2 | \\ = | - 4 | \\ = - (- 4) \\ = 4 \end{aligned} \end{array}

\begin{array}{l} 9. & Diketahui fungsi f (x) = | 2 - 4 x | \\ Nilai dari f (1) adalah . . . . \\ \begin{array}{llll} a . & - 1 \\ b . & 0 \\ c . & 1 \\ d . & 2 \\ e . & 4 \end{array} \\ Jawab : d \\ \begin{aligned} f (x) & = | 2 - 4 x | \\ Nilai & f (1) berarti nilai \\ saat & x = 1, maka \\ f (x) & = | 2 - 4 x) | \\ f (1) & = | 2 - 4 (1) | \\ = | - 2 | \\ = - (- 2) \\ = 2 \end{aligned} \end{array}

\begin{array}{l} 10. & Nilai k yang memenuhi | 4 k | = 16 adalah... . \\ \begin{array}{llll} a . & 2 \\ b . & 4 \\ c . & \pm 2 \\ d . & \pm 4 \\ e . & \pm 8 \end{array} \\ Jawab : d \\ \begin{aligned} | 4 k | & = 16 \\ (4 k) & = \pm 16 \\ k & = \pm \frac{16}{4} \\ = \pm 4 \end{aligned} \end{array}

Contoh Soal Nilai Mutlak (Kelas X Matematika Wajib)

$\begin{array}{ll} 1. & Penyelesaian - 5 | x - 7 | + 2 = - 13 adalah... . \\ \begin{array}{llll} a . & x = - 10 dan x = - 4 \\ b . & x = - 10 dan x = - 2 \\ c . & x = - 2 dan x = 4 \\ d . & x = 4 dan x = - 10 \\ e . & x = 4 dan x = 10 \end{array} \\ Jawab : e \\ \begin{aligned} - 5 | x - 7 | & = - 13 - 2 \\ | x - 7 | & = \frac{- 15}{- 5} \\ | x - 7 | & = 3 \\ x - 7 & = \pm 3 \\ x & = \mp 3 + 7 \\ x & = {\begin{cases} + 3 & + 7 = 10 \\ - 3 & + 7 = 4 \end{cases} \end{aligned} \end{array}$

$\begin{array}{ll} 2. & Diketahui persamaan | x - 5 | + | x + 9 | = 22 \\ dan diberikan pernyataan-pernyataan berikut \\ untuk \\ (i) x < - 9, maka - x + 5 - x + 9 = 22 \\ (ii) - 9 \leq x < 5, maka - x + 5 - x + 9 = 22 \\ (iii) x < - 9, maka - x + 5 - x - 9 = 22 \\ (iv) x \geq 5, maka x - 5 + x + 9 = 22 \\ Pernyataan yang tepa ditunjukkan oleh . . . . \\ \begin{array}{llll} a . & (i) dan (ii) \\ b . & (i) dan (iii) \\ c . & (i) dan (iv) \\ d . & (ii) dan (iii) \\ e . & (iii) dan (iv) \end{array} \\ Jawab : c \\ \begin{matrix} \end{matrix} \end{array}$

$\begin{array}{l} 3. & Nilai x yang memenuhi persamaan \\ 3 | x - 1 | - 2 | x + 1 | = 0 \\ adalah . . . . \\ \begin{array}{llll} a . & x = - \frac{1}{5} dan x = 5 \\ b . & x = - 5 dan x = - \frac{1}{5} \\ c . & x = \frac{1}{5} dan x = 5 \\ d . & x = - 5 dan x = \frac{1}{5} \\ e . & x = - 5 dan x = 5 \end{array} \\ Jawab : c \\ \begin{aligned} 3 | x - 1 | & - 2 | x + 1 | = 0 \\ 3 | x - 1 | & = 2 | x + 1 | \\ 3^{2} (x - 1)^{2} & = 2^{2} (x + 1)^{2} \\ 9 (x^{2} - 2 x + 1) & = 4 (x^{2} + 2 x + 1) \\ 9 x^{2} - 4 x^{2} & - 18 x - 8 x + 9 - 4 = 0 \\ 5 x^{2} - 26 x & + 5 = 0 \\ (5 x - 1) & (x - 5) = 0 \\ x = \frac{1}{5} & atau x = 5 \end{aligned} \end{array}$

$\begin{array}{l} 4. & Himpunan penyelesaian dari | 5 - \frac{2}{3} x | - 9 = 8 adalah... . \\ \begin{array}{llll} a . & {- 33, - 18} \\ b . & {- 18, 33} \\ c . & {33, 8} \\ d . & {8, - 33} \\ e . & {- 8, 33} \end{array} \\ Jawab : b \\ \begin{aligned} | 5 - \frac{2}{3} x | & - 9 = 8 \\ | 5 - \frac{2}{3} x | & = 8 + 9 = 17 \\ 5 - \frac{2}{3} x & = \pm 17 \\ - \frac{2}{3} x & = \pm 17 - 5 \\ x & = \frac{\pm 17 - 5}{(- \frac{2}{3})} \\ x_{1} & = (\frac{17 - 5}{- 2}) \times 3 \\ = - 18 \\ x_{2} & = (\frac{- 17 - 5}{- 2}) \times 3 \\ = 33 \end{aligned} \end{array}$

$\begin{array}{ll} 5. & Harga buku di toko "KITA" adalah 28.000, - \\ Jika harga buku tulis di toko "FAMILI" memiliki selisih 7.000, - \\ Harga buku tulis di toko "FAMILI" dapat dinyatakan dengan . . . . \\ \begin{array}{llll} a . & | x - 7000 | = 28000 \\ b . & | x + 7000 | = 28000 \\ c . & | x - 28000 | = 7000 \\ d . & | x + 28000 | = 7000 \\ e . & | x - 21000 | = 7000 \end{array} \\ Jawab : c \end{array}$

Contoh Soal Lanjutan 4 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 21.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{1-\cos ^{2}2x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{3}{2}\\\\ \textrm{b}.&\displaystyle -\frac{3}{4}\\\\ \textrm{c}.&\displaystyle -\frac{1}{2}\\\\ \textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{e}.&\displaystyle \frac{3}{4} \end{array} \end{array}$

Jawab : e

Berikut pembahasannya

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{1-\cos ^{2}2x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\tan 3x}{\sin ^{2}2x}\\ &=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{x\times \tan \textcircled{3}x}{\sin \textcircled{2}x\times \sin \textcircled{2}x}\\ &=\displaystyle \frac{1\times 3}{2\times 2}\\ &=\displaystyle \frac{3}{4} \end{aligned}$

$\begin{array}{ll}\\ 22.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle 9\\\\ \textrm{b}.&\displaystyle 7\\\\ \textrm{c}.&\displaystyle 6\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle -6 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}&\quad \textrm{perhatikan bahwa}\\ &\color{black}\textrm{misal}\: \: x=3+h,\: \: \textrm{maka}\: \: h\rightarrow 0\\ &\color{red}\textrm{selanjutnya gunakan}\: \: h=x-3,\: \: \textrm{dengan}\: \: h\rightarrow 0\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{(3+h)^{2}-9}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{h^{2}+6h+9-9}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \:\frac{h(h+6)}{\sin h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \:(h+6)\\ &=0+6=6 \end{aligned}$

atau boleh juga

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-9}{\sin (x-3)}&=\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle \frac{(x+3)(x-3)}{\sin (x-3)}\\ &=\underset{x\rightarrow \displaystyle 3}{\textrm{Lim}}\: \: \displaystyle (x+3)\\ &=3+3=6 \end{aligned}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{x-\displaystyle \frac{\pi }{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle 6\\\\ \textrm{b}.&\displaystyle 3\\\\ \textrm{c}.&\displaystyle 2\\\\ \textrm{d}.&\displaystyle -2\\\\ \textrm{e}.&\displaystyle -6 \end{array} \end{array}$

Jawab : e

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{x-\displaystyle \frac{\pi }{2}}&\quad \color{black}\textrm{perhatikan juga bahwa}\\ &x=\displaystyle \frac{\pi }{2}+h,\quad \textrm{dengan}\: \: \: h\rightarrow 0\\ &\color{red}\textrm{atau}\: \: \: h=x-\displaystyle \frac{\pi }{2}\quad \textrm{dengan}\: \: \: h\rightarrow 0\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6\left ( \displaystyle \frac{\pi }{2}+h \right )}{h}=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( \displaystyle 3\pi +6h \right )}{h}\\ &=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ( \displaystyle \pi +6h \right )}{h}=\underset{h\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{-\sin \left ( \displaystyle 6h \right )}{h}\\ &=-6 \end{aligned}$

Daftar Pustaka

1. Astuti, A. N., Miyanto, dan Ngapiningsih. 2020. Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Yogyakarta: PT. PENERBIT INTAN PARIWARA.

Contoh Soal Lanjutan 3 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 16.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \cos 4x=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -1\\\\ \textrm{b}.&-\displaystyle \frac{1}{2}\sqrt{3}\\\\ \textrm{c}.&\displaystyle -\frac{1}{2}\\\\ \textrm{d}.&\displaystyle 0\\\\ \textrm{e}.&\displaystyle 1 \end{array} \end{array}$

Jawab : e

Cukup jelas

Dengan substitusi langsung kita akan dapat mentukan hasilnya, yatu:

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \cos 4x=\cos 4(0)=\cos 0=1 \end{aligned}$

$\begin{array}{ll}\\ 17.&\textrm{Diketahui}\: \: f(x)=\begin{cases} \sin 2x, & \text{ saat } x< 0 \\ \tan x, & \text{ saat } x\geq 0 \end{cases}\\\\ &\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0^{+}}{\textrm{Lim}}\: \: f(x)=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -2\\\\ \textrm{b}.&-\displaystyle 1\\\\ \textrm{c}.&\displaystyle 0\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle \textrm{tidak ada} \end{array} \end{array}$

Jawab : c

Berikut penjelasannya

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0^{+}}{\textrm{Lim}}\: \: \displaystyle f(x)&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \tan x=\tan 0=0 \end{aligned}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{1-\cos 2x}{5\sin ^{2}x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{2}{5}\\\\ \textrm{b}.&\displaystyle \frac{2}{5}\\\\ \textrm{c}.&\displaystyle 1\\\\ \textrm{d}.&\displaystyle 2\\\\ \textrm{e}.&\displaystyle \infty \end{array} \end{array}$

Jawab : b

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 1-\cos 2x}{5\sin ^{2}x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 2\sin ^{2}x}{5\sin ^{2}x}\\ &=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{2\color{red}\sin \textcircled{1}x\times \sin \textcircled{1}x}{5\sin \color{red}\textcircled{1}x\times \sin \textcircled{1}x}\\ &=\displaystyle \frac{2\times 1\times 1}{5\times 1\times 1}\\ &=\displaystyle \frac{2}{5}\\ &\\ &\\ \color{black}\textrm{Perlu diingat}\: &\textrm{bahwa}\quad 1-\cos 2x=2\sin ^{2}x \end{aligned}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{3-\sin x}{\cos 2x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -4\\\\ \textrm{b}.&\displaystyle -2\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : b

$\begin{array}{ll}\\ 19.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: \displaystyle \frac{\sin x-2}{\cos x-3}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle -1\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : a

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: \displaystyle \frac{ \sin x-2}{\cos x-3}&\qquad\color{black}\textrm{dengan substitusi langsung didapatkan}\\ &=\underset{x\rightarrow \displaystyle \pi }{\textrm{Lim}}\: \: \displaystyle \frac{ \left (\sin \pi \right ) -2}{\left (\cos \pi \right ) -3}\\ &=\underset{x\rightarrow \displaystyle \frac{\pi }{2}}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}0-2}{-1-3}\\ &=\displaystyle \frac{-2}{-4}\\ &=\displaystyle \frac{1}{2}\\ \end{aligned}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle \frac{\pi }{6}}{\textrm{Lim}}\: \: \displaystyle \left (\cos ^{2}x-\sin ^{2}x \right )=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle -\frac{1}{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&\displaystyle 1\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : b

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle \frac{\pi }{6}}{\textrm{Lim}}\: \: \displaystyle \left ( \cos ^{2}x-\sin ^{2}x \right )&\qquad \color{black}\textrm{dengan substitusi langsung}\\ &=\cos ^{2}\left ( \displaystyle \frac{\pi }{6} \right )-\sin ^{2}\left ( \displaystyle \frac{\pi }{6} \right )\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}-\left ( \displaystyle \frac{1}{2} \right )^{2}\\ &=\displaystyle \frac{3}{4}-\frac{1}{4}=\frac{2}{4}\\ &=\displaystyle \frac{1}{2} \end{aligned}$

Daftar Pustaka

1. Tim. 2020. Modul Matematika (Peminatan Matematika dan Ilmu-Ilmu Alam Kelas XII. Tangerang: RAHMA GEMILANG.

Contoh Soal Lanjutan 2 Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 11.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{15x\tan x}{3x\sin 5x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{8}\\\\ \textrm{b}.&\displaystyle \frac{1}{6}\\\\ \textrm{c}.&\displaystyle \frac{1}{3}\\\\ \textrm{d}.&\displaystyle \frac{3}{2}\\\\ \textrm{e}.&\displaystyle 1 \end{array} \end{array}$

Jawab : e

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 15x\tan x}{3x\sin 5x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{15}x\tan \textcircled{1}x}{\color{red}\textcircled{3}x\tan \color{red}\textcircled{5}x}\\ &=\displaystyle \frac{15\times 1}{3\times 5}\\ &=\displaystyle \frac{15}{15}\\ &=1 \end{aligned}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{18x\tan x}{3x\sin 4x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{8}\\\\ \textrm{b}.&\displaystyle \frac{1}{6}\\\\ \textrm{c}.&\displaystyle \frac{1}{2}\\\\ \textrm{d}.&\displaystyle \frac{3}{2}\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : d

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 18x\tan x}{3x\sin 4x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{18}x\tan \textcircled{1}x}{\color{red}\textcircled{3}x\tan \color{red}\textcircled{4}x}\\ &=\displaystyle \frac{18\times 1}{3\times 4}\\ &=\displaystyle \frac{18}{12}\\ &=\displaystyle \frac{3}{2} \end{aligned}$

$\begin{array}{ll}\\ 13.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x\sin x}{2x^{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle 3\\\\ \textrm{d}.&\displaystyle 6\\\\ \textrm{e}.&\displaystyle 12 \end{array} \end{array}$

Jawab : d

Sebagai pembahasannya adalah sebagai berikut

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 12x\sin x}{2x^{2}}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{12}x\sin \textcircled{1}x}{\color{red}\textcircled{2}x\times \textcircled{1}x}\\ &=\displaystyle \frac{12\times 1}{2\times 1}\\ &=\displaystyle \frac{12}{2}\\ &=6 \end{aligned}$

$\begin{array}{ll}\\ 14.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin^{2} 3x}{12x^{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{3}{4}\\\\ \textrm{d}.&\displaystyle 4\\\\ \textrm{e}.&\displaystyle 16 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ \sin^{2} 3x}{12x^{2}}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\sin \textcircled{3}x\times \sin \textcircled{3}x}{\color{red}\textcircled{12}x\times \textcircled{1}x}\\ &=\displaystyle \frac{3\times 3}{12\times 1}\\ &=\displaystyle \frac{9}{12}\\ &=\displaystyle \frac{3}{4} \end{aligned}$

$\begin{array}{ll}\\ 15.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan^{2} 3x}{21x^{2}}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{3}{7}\\\\ \textrm{c}.&\displaystyle \frac{3}{4}\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle 7 \end{array} \end{array}$

Jawab : b

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ \tan^{2} 3x}{21x^{2}}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\tan \textcircled{3}x\times \tan \textcircled{3}x}{\color{red}\textcircled{21}x\times \textcircled{1}x}\\ &=\displaystyle \frac{3\times 3}{21\times 1}\\ &=\displaystyle \frac{9}{21}\\ &=\displaystyle \frac{3}{7} \end{aligned}$

Sampai di sini semoga ini semua dapat menambah khazanah dalam menyelesaikan permasalahan jika suatu ketika pembaca mendapati soal yang semisal

Semoga bermanfaat

Contoh Soal Lanjutan Limit Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{ll}\\ 6.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\tan 8x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{3}{2}\\\\ \textrm{d}.&\displaystyle 4\\\\ \textrm{e}.&\displaystyle 8 \end{array} \end{array}$

Jawab : c

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\tan 8x}&=12\times \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \left (\frac{x}{\tan 8x}\times \frac{8}{8} \right )\\ &=12\times \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{8x}{\tan 8x}\times \frac{1}{8}\\ &=\displaystyle \frac{12}{8}\times \left (\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{8x}{\tan 8x} \right )\\ &=\displaystyle \frac{12}{8}\times 1\\ &=\frac{3}{2}\\ \end{aligned}$

Selanjutnya jika ada soal menentukan nilai limit dan fungsinya berbentuk rasional dengan pembilang atau penyebut berupa sinus atau tangen (kadang keduanya muncul bersamaan yang satu diposisi pembilang dan yang satu diposisi penyebut) semisal bentuk di atas dan masing-masing berpangkat sama, maka kita kemungkinan besar tinggal mengambil koefisien dari masing-masing unsur pembilang dan penyebutnya. Demikian untuk langkah mudahnya dalam menyelesaikan soal yang ada.

Sehingga soal di atas dapat juga diselesaikan dengan cara hematnya:

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\tan 8x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{red}\textcircled{12}x}{\tan \color{red}\textcircled{8}x}\\ &=\displaystyle \frac{12}{8}\\ &=\displaystyle \frac{3}{2} \end{aligned}$

Coba perhatikan soal berikut

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{\tan 10x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{5}\\\\ \textrm{b}.&\displaystyle \frac{3}{5}\\\\ \textrm{c}.&\displaystyle \frac{5}{9}\\\\ \textrm{d}.&\displaystyle \frac{6}{5}\\\\ \textrm{e}.&\displaystyle 2 \end{array} \end{array}$

Jawab : b

Walau cukup jelas dan sebagai ulasannya adalah sebagai berikut

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin 6x}{\tan 10x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{magenta}\textcircled{6}x}{\tan \color{magenta}\textcircled{10}x}\\ &=\displaystyle \frac{6}{10}\\ &=\displaystyle \frac{3}{5} \end{aligned}$

Sebagai tambahan kita berikan contoh pula soal berikut

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{16x}{\tan 4x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle 2\\\\ \textrm{d}.&\displaystyle 4\\\\ \textrm{e}.&\displaystyle 8 \end{array} \end{array}$

Jawab : d

Cukup jelas bahwa

$\color{blue}\begin{aligned}\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{ 16x}{\tan 4x}&=\underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\color{green}\textcircled{16}x}{\tan \color{green}\textcircled{4}x}\\ &=\displaystyle \frac{16}{4}\\ &=\displaystyle 4 \end{aligned}$

$\begin{array}{ll}\\ 9.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{12x}{\sin 10x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{9}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}\\\\ \textrm{c}.&\displaystyle \frac{5}{9}\\\\ \textrm{d}.&\displaystyle \frac{6}{5}\\\\ \textrm{e}.&\displaystyle \frac{9}{5} \end{array} \end{array}$

Jawab : d

Cukup jelas

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\: \: \underset{x\rightarrow \displaystyle 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan 12x}{\sin 4x}=....\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{2}\\\\ \textrm{c}.&\displaystyle \frac{3}{2}\\\\ \textrm{d}.&\displaystyle 3\\\\ \textrm{e}.&\displaystyle 4 \end{array} \end{array}$

Jawab : d

Jawaban juga sudah cukup jelas