PAS MATEMATIKA BLOG: Latihan Soal 2 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{ll} 11. & Nilai yang memenuhi \\ lim_{x \to \infty} (\sqrt{4 x - 2020} - \sqrt{8 x + 2021}) adalah... . \\ \begin{array}{llll} a . & - \infty \\ b . & 0 \\ c . & 1 \\ d . & 2 \\ e . & \infty \end{array} \\ Jawab : a \\ \begin{aligned} lim_{x \to \infty} (\sqrt{4 x - 2020} - \sqrt{8 x + 2021}) \\ = lim_{x \to \infty} (\sqrt{4 x - 2020} - \sqrt{8 x + 2021}) \times \frac{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}}{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}} \\ = lim_{x \to \infty} \frac{(4 x - 2020) - (8 x + 2021)}{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}} \\ = lim_{x \to \infty} \frac{- 4 x - 4041}{\sqrt{4 x - 2020} + \sqrt{8 x + 2021}} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\ = lim_{x \to \infty} \frac{- 4 - \frac{4041}{x}}{\frac{1}{x} (\sqrt{4 x - 2020} + \sqrt{8 x + 2021})} \\ = lim_{x \to \infty} \frac{- 4 - \frac{4041}{x}}{(\sqrt{\frac{4 x}{x^{2}} - \frac{2020}{x^{2}}} + \sqrt{\frac{8 x}{x^{2}} + \frac{2021}{x^{2}}})} \\ = lim_{x \to \infty} \frac{- 4 - \frac{4041}{x}}{(\sqrt{\frac{4}{x} - \frac{2020}{x}} + \sqrt{\frac{8}{x} + \frac{2021}{x}})} \\ = \frac{- 4 - 0}{\sqrt{0 - 0} + \sqrt{0 + 0}} \\ = \frac{- 4}{0} \\ = - \infty \end{aligned} \end{array}$

$\begin{array}{ll} 12. & Nilai \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} = . . . . \\ \begin{array}{lll} a . - 1 \\ b . 1 \\ c . 2 \\ d . 4 \\ e . 8 \end{array} \\ Jawab : c \\ \begin{aligned} \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \\ = \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \times \frac{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}}{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}} \\ = \underset{x \to \infty}{Lim} \frac{4 x^{2} + 3 x - (4 x^{2} - 5 x)}{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}} \\ = \underset{x \to \infty}{Lim} \frac{3 x + 5 x}{\sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x}} \times \frac{(\frac{1}{x})}{(\sqrt{\frac{1}{x^{2}}})} \\ = \frac{3 + 5}{\sqrt{4} + \sqrt{4}} \\ = \frac{8}{4} \\ = 2 \end{aligned} \end{array}$ .

$\begin{aligned} ada cara lain yang lebih sede rhana, yaitu: \\ . & \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \\ = \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} \\ {\begin{cases} a & = 4 \\ b & = 3 \\ p & = - 4 \end{cases} \\ Jika \\ \underset{x \to \infty}{Lim} \sqrt{a x^{2} + b x + c} - \sqrt{a x^{2} + p x + q} = \frac{b - p}{2 \sqrt{a}} \\ Sehingga \\ \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} - \sqrt{4 x^{2} - 5 x} = \frac{3 - (- 5)}{2 \sqrt{4}} \\ = \frac{8}{2.2} \\ = 2 \end{aligned}$ .

$\begin{array}{ll} 13. & Nilai \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x} = . . . . \\ \begin{array}{lll} a . \infty \\ b . 1 \\ c . 2 \\ d . 4 \\ e . 8 \end{array} \\ Jawab : a \\ \underset{x \to \infty}{Lim} \sqrt{4 x^{2} + 3 x} + \sqrt{4 x^{2} - 5 x} \\ = \sqrt{\infty} + \sqrt{\infty} = \infty \end{array}$ .

$\begin{array}{ll} 14. & Nilai yang memenuhi \\ \underset{x \to \infty}{Lim} (\sqrt{3 x + 1} - \sqrt{3 x - 2}) adalah... . \\ \begin{array}{llll} a . & 0 \\ b . & 1 \\ c . & 2 \\ d . & 4 \\ e . & \infty \end{array} \\ Jawab : a \\ \begin{aligned} \underset{x \to \infty}{Lim} (\sqrt{3 x + 1} - \sqrt{3 x - 2}) \\ = \underset{x \to \infty}{Lim} (\sqrt{3 x + 1} - \sqrt{3 x - 2}) \times \frac{(\sqrt{3 x + 1} + \sqrt{3 x - 2})}{(\sqrt{3 x + 1} + \sqrt{3 x - 2})} \\ = \underset{x \to \infty}{Lim} \frac{(3 x + 1) - (3 x - 2)}{(\sqrt{3 x + 1} + \sqrt{3 x - 2})} \\ = \underset{x \to \infty}{Lim} \frac{3}{(\sqrt{3 x + 1} + \sqrt{3 x - 2})} \times \frac{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x}}} \\ = \underset{x \to \infty}{Lim} \frac{\frac{3}{\sqrt{x}}}{(\sqrt{\frac{3 x}{x} + \frac{1}{x}} + \sqrt{\frac{3 x}{x} - \frac{2}{x}})} \\ = \frac{0}{\sqrt{3 + 0} + \sqrt{3 - 0}} \\ = 0 \end{aligned} \end{array}$

$\begin{array}{ll} 15. & Nilai yang memenuhi \\ \underset{x \to \infty}{Lim} (\sqrt{4 x^{2} + 6 x + 8} - \sqrt{4 x^{2} - 8 x + 7}) adalah... . \\ \begin{array}{llll} a . & 0 \\ b . & 1 \\ c . & \frac{3}{2} \\ d . & \frac{7}{2} \\ e . & \infty \end{array} \\ Jawab : d \\ \begin{aligned} \underset{x \to \infty}{Lim} (\sqrt{4 x^{2} + 6 x + 8} - \sqrt{4 x^{2} - 8 x + 7}) \\ = \underset{x \to \infty}{Lim} (\sqrt{4 x^{2} + 6 x + 8} - \sqrt{4 x^{2} - 8 x + 7}) \\ \times \frac{(\sqrt{4 x^{2} + 6 x + 8} + \sqrt{4 x^{2} - 8 x + 7})}{(\sqrt{4 x^{2} + 6 x + 8} + \sqrt{4 x^{2} - 8 x + 7})} \\ = \underset{x \to \infty}{Lim} \frac{(4 x^{2} + 6 x + 8) - (4 x^{2} - 8 x + 7)}{\sqrt{4 x^{2} + 6 x + 8} + \sqrt{4 x^{2} - 8 x + 7}} \\ = \underset{x \to \infty}{Lim} \frac{14 x + 1}{\sqrt{4 x^{2} + 6 x + 8} + \sqrt{4 x^{2} - 8 x + 7}} \times \frac{\frac{1}{x}}{\frac{1}{x}} \\ = \underset{x \to \infty}{Lim} \frac{14 + \frac{1}{x}}{\sqrt{\frac{4 x^{2}}{x^{2}} + \frac{6 x}{x^{2}} + \frac{8}{x^{2}}} + \sqrt{\frac{4 x^{2}}{x^{2}} - \frac{8 x}{x^{2}} + \frac{7}{x^{2}}}} \\ = \frac{14 + 0}{\sqrt{4 + 0 + 0} - \sqrt{4 - 0 + 0}} \\ = \frac{14}{2 + 2} = \frac{7}{2} \end{aligned} \end{array}$

$\begin{array}{ll} 16. & Nilai yang memenuhi \\ \underset{x \to \infty}{Lim} (\sqrt{2 x^{2} + 3 x - 1} - \sqrt{x^{2} - 5 x + 3}) adalah... . \\ \begin{array}{llll} a . & 1 \\ b . & 2 \\ c . & 4 \\ d . & 8 \\ e . & \infty \end{array} \\ Jawab : e \\ \begin{aligned} \underset{x \to \infty}{Lim} (\sqrt{2 x^{2} + 3 x - 1} - \sqrt{x^{2} - 5 x + 3}) \\ = \underset{x \to \infty}{Lim} (\sqrt{2 x^{2} + 3 x - 1} - \sqrt{x^{2} - 5 x + 3}) \\ \times \frac{(\sqrt{2 x^{2} + 3 x - 1} + \sqrt{x^{2} - 5 x + 3})}{(\sqrt{2 x^{2} + 3 x - 1} + \sqrt{x^{2} - 5 x + 3})} \\ = \underset{x \to \infty}{Lim} \frac{(2 x^{2} + 3 x - 1) - (x^{2} - 5 x + 3)}{(\sqrt{2 x^{2} + 3 x - 1} + \sqrt{x^{2} - 5 x + 3})} \\ = \underset{x \to \infty}{Lim} \frac{x^{2} + 8 x - 4}{(\sqrt{2 x^{2} + 3 x - 1} - \sqrt{x^{2} - 5 x + 3})} \times \frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}} \\ = \underset{x \to \infty}{Lim} \frac{\frac{x^{2}}{x^{2}} + \frac{8 x}{x^{2}} - \frac{4}{x^{2}}}{\sqrt{\frac{2 x^{2}}{x^{4}} + \frac{3 x}{x^{4}} - \frac{1}{x^{4}}} + \sqrt{\frac{x^{2}}{x^{4}} - \frac{5 x}{x^{4}} + \frac{3}{x^{4}}}} \\ = \frac{1 + 0 - 0}{\sqrt{0 + 0 + 0} + \sqrt{0 - 0 + 0}} \\ = \frac{1}{0} \\ = \infty \end{aligned} \end{array}$

$\begin{array}{ll} 17. & Nilai yang memenuhi \\ \underset{x \to \infty}{Lim} (\sqrt{x^{2} + 3 x + 1} - \sqrt{3 x^{2} + 2 x + 5}) adalah... . \\ \begin{array}{llll} a . & - \infty \\ b . & 1 \\ c . & 2 \\ d . & 4 \\ e . & \infty \end{array} \\ Jawab : a \\ \begin{aligned} \underset{x \to \infty}{Lim} (\sqrt{x^{2} + 3 x + 1} - \sqrt{3 x^{2} + 2 x + 5}) \\ = \underset{x \to \infty}{Lim} (\sqrt{x^{2} + 3 x + 1} - \sqrt{3 x^{2} + 2 x + 5}) \\ \times \frac{(\sqrt{x^{2} + 3 x + 1} + \sqrt{3 x^{2} + 2 x + 5})}{(\sqrt{x^{2} + 3 x + 1} + \sqrt{3 x^{2} + 2 x + 5})} \\ = \underset{x \to \infty}{Lim} \frac{(x^{2} + 3 x + 1) - (3 x^{2} + 2 x + 5)}{(\sqrt{x^{2} + 3 x + 1} + \sqrt{3 x^{2} + 2 x + 5})} \\ = \underset{x \to \infty}{Lim} \frac{- 2 x^{2} + x - 4}{(\sqrt{x^{2} + 3 x + 1} - \sqrt{3 x^{2} + 2 x + 5})} \times \frac{\frac{1}{x^{2}}}{\frac{1}{x^{2}}} \\ = \underset{x \to \infty}{Lim} \frac{\frac{- 2 x^{2}}{x^{2}} + \frac{x}{x^{2}} - \frac{4}{x^{2}}}{\sqrt{\frac{x^{2}}{x^{4}} + \frac{3 x}{x^{4}} + \frac{1}{x^{4}}} + \sqrt{\frac{3 x^{2}}{x^{4}} + \frac{2 x}{x^{4}} + \frac{5}{x^{4}}}} \\ = \underset{x \to \infty}{Lim} \frac{- 2 + \frac{1}{x} - \frac{4}{x^{2}}}{\sqrt{\frac{1}{x^{2}} + \frac{3}{x^{3}} + \frac{1}{x^{4}}} + \sqrt{\frac{3}{x^{2}} + \frac{2}{x^{3}} + \frac{5}{x^{4}}}} \\ = \frac{- 2 + 0 - 0}{\sqrt{0 + 0 + 0} + \sqrt{0 + 0 + 0}} \\ = \frac{- 2}{0} \\ = - \infty \end{aligned} \end{array}$

$\begin{array}{ll} 18. & Nilai yang memenuhi \\ \underset{x \to \infty}{Lim} ((3 x - 2) - \sqrt{9 x^{2} - 2 x + 5}) adalah... . \\ \begin{array}{llll} a . & - \infty \\ b . & - \frac{5}{3} \\ c . & \frac{1}{3} \\ d . & 1 \\ e . & \infty \end{array} \\ Jawab : b \\ \begin{aligned} \underset{x \to \infty}{Lim} ((3 x - 2) - \sqrt{9 x^{2} - 2 x + 5}) \\ = \underset{x \to \infty}{Lim} (\sqrt{(3 x - 2)^{2}} - \sqrt{9 x^{2} - 2 x + 5}) \\ = \underset{x \to \infty}{Lim} (\sqrt{(9 x^{2} - 12 x + 4} - \sqrt{9 x^{2} - 2 x + 5}) \\ \underset{x \to \infty}{Lim} (\sqrt{(a x^{2} + b x + c} - \sqrt{p x^{2} + q x + r}) \\ Jika dikerjakan dengan rumus singkat \\ maka {\begin{array}{c} a = p = 3 \\ b = - 12 \\ q = - 2 \end{array} \\ = \frac{b - q}{2 \sqrt{a}} \\ = \frac{- 12 - (- 2)}{2 \sqrt{9}} \\ = \frac{- 10}{6} \\ = - \frac{5}{3} \end{aligned} \end{array}$ .

$\begin{array}{ll} 19. & Nilai yang memenuhi \\ \underset{x \to \infty}{Lim} \frac{\sqrt{4 x^{2} - 2 x} - \sqrt{x^{2} + 1}}{\sqrt{9 x^{2} - 1}} adalah... . \\ \begin{array}{llll} a . & \frac{1}{3} \\ b . & \frac{4}{9} \\ c . & \frac{1}{2} \\ d . & 1 \\ e . & \frac{3}{2} \end{array} \\ Jawab : a \\ \begin{aligned} \underset{x \to \infty}{Lim} \frac{\sqrt{4 x^{2} - 2 x} - \sqrt{x^{2} + 1}}{\sqrt{9 x^{2} - 1}} \\ = \underset{x \to \infty}{Lim} \frac{\sqrt{4 x^{2} - 2 x} - \sqrt{x^{2} + 1}}{\sqrt{9 x^{2} - 1}} \times \frac{(\sqrt{\frac{1}{x^{2}}})}{(\sqrt{\frac{1}{x^{2}}})} \\ = \underset{x \to \infty}{Lim} \frac{\sqrt{4 - \frac{2}{x}} - \sqrt{1 + \frac{1}{x^{2}}}}{\sqrt{9 - \frac{1}{x^{2}}}} \\ = \underset{x \to \infty}{Lim} \frac{\sqrt{4 - 0} - \sqrt{1 + 0}}{\sqrt{9 - 0}} \\ = \frac{2 - 1}{3} \\ = \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll} 20. & Nilai \underset{x \to \infty}{Lim} \frac{3 x^{4} + 2 x^{3} - 5 x + 2021}{2 x^{3} - 4 x^{2} + 2020} = . . . . \\ \begin{array}{lll} a . \frac{4}{9} \\ b . \frac{3}{2} \\ c . 0 \\ d . 1 \\ e . \infty \end{array} \\ Jawab : e \\ \begin{aligned} \underset{x \to \infty}{Lim} \frac{3 x^{4} + 2 x^{3} - 5 x + 2021}{2 x^{3} - 4 x^{2} + 2020} \\ = \underset{x \to \infty}{Lim} \frac{\frac{3 x^{4}}{x^{4}} + \frac{2 x^{3}}{x^{4}} - \frac{5 x}{x^{4}} + \frac{2021}{x^{4}}}{\frac{2 x^{3}}{x^{4}} - \frac{4 x^{2}}{x^{4}} + \frac{2020}{x^{4}}} \\ = \frac{3 + \frac{2}{x} - \frac{5}{x^{2}} + \frac{2021}{x^{4}}}{\frac{4}{x} - \frac{4}{x^{2}} + \frac{2020}{x^{4}}} \\ = \frac{3 + 0 - 0 + 0}{0 - 0 + 0} \\ = \infty \end{aligned} \end{array}$

PAS MATEMATIKA BLOG

Latihan Soal 2 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

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