PAS MATEMATIKA BLOG: Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

$\begin{array}{ll} 51. & Turunan pertama dari fungsi \\ g (x) = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} adalah . . . . \\ \begin{array}{llll} a . & \frac{1}{\cos^{2} x} - \frac{1}{\sin^{2} x} \\ b . & \frac{1}{\cos^{2} x} + \frac{1}{\sin^{2} x} \\ c . & \frac{1}{\sin^{2} x \cos^{2} x} \\ d . & \frac{- 1}{\sin^{2} x \cos^{2} x} \\ e . & \sin^{2} x \cos^{2} x \end{array} \\ Jawab : a \\ \begin{aligned} Diket & ahui \\ g (x) & = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \\ = \frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x} = \frac{1}{\sin x \cos x} \\ maka \\ g^{'} (x) & = \frac{0. (\sin x \cos x) - 1. (\cos^{2} x - \sin^{2} x)}{(\sin x \cos x)^{2}} \\ = \frac{\sin^{2} x - \cos^{2} x}{\sin^{2} x \cos^{2} x} \\ = \frac{1}{\cos^{2} x} - \frac{1}{\sin^{2} x} \end{aligned} \end{array}$

$\begin{array}{ll} 52. & Diketahui h (x) = \cos (\frac{3}{x}), \\ maka \frac{d h}{d x} \\ \begin{array}{llll} a . & - 3 \sin \frac{3}{x} \\ b . & - \frac{3}{x^{2}} \sin \frac{3}{x} \\ c . & - \frac{3}{x} \sin \frac{3}{x} \\ d . & \frac{3}{x^{2}} \sin \frac{3}{x} \\ e . & \frac{3}{x} \sin \frac{3}{x} \end{array} \\ Jawab : d \\ \begin{aligned} \cos \frac{3}{x} & = - \sin \frac{3}{x} (\frac{0. (x) - 3.1}{x^{2}}) \\ = \frac{- (- 3)}{x^{2}} \sin \frac{3}{x} \\ = \frac{3}{x^{2}} \sin \frac{3}{x} \end{aligned} \end{array}$

$\begin{array}{ll} 53. & Turunan pertama dari \tan (\cos x), \\ terhadap x adalah . . . . \\ \begin{array}{llll} a . & - \sec^{2} (\cos x) \sin x \\ b . & \sec^{2} (\cos x) \sin x \\ c . & \sec^{2} (\sin x) \cos x \\ d . & \sin x \\ e . & - \sin x \end{array} \\ Jawab : a \\ \begin{aligned} Misal & kan \\ y & = \tan x (\cos x) \\ y^{'} & = \sec^{2} (\cos x) \times (- \sin x) \\ = - \sec^{2} (\cos x) . \sin x \end{aligned} \end{array}$

$\begin{array}{ll} 54. & (UN 2005) Turunan pertama dari \\ f (x) = \sqrt[3]{\cos^{2} (3 x^{2} + 5 x)} adalah . . . . \\ \begin{array}{llll} a . & \frac{2}{3} \cos^{.^{- \frac{1}{3}}} (3 x^{2} + 5 x) \sin (3 x^{2} + 5 x) \\ b . & \frac{2}{3} (6 x + 5) \cos^{.^{- \frac{1}{3}}} (3 x^{2} + 5 x) \\ c . & - \frac{2}{3} \cos^{.^{- \frac{1}{3}}} (3 x^{2} + 5 x) \sin (3 x^{2} + 5 x) \\ d . & - \frac{2}{3} (6 x + 5) \tan (3 x^{2} + 5 x) \sqrt[3]{\cos^{2} (3 x^{2} + 5 x)} \\ e . & \frac{2}{3} (6 x + 5) \tan (3 x^{2} + 5 x) \sqrt[3]{\cos^{2} (3 x^{2} + 5 x)} \end{array} \\ Jawab : d \\ \begin{aligned} Misal & kan \\ f (x) & = \sqrt[3]{\cos^{2} (3 x^{2} + 5 x)} \\ f^{'} (x) & = \cos^{.^{\frac{2}{3}}} (3 x^{2} + 5 x) \\ = \frac{2}{3} \cos^{.^{- \frac{1}{2}}} (3 x^{2} + 5 x) \times (- \sin (3 x^{2} + 5 x)) \\ \times (6 x + 5) \\ = - \frac{2}{3} (6 x + 5) \cos^{.^{- \frac{1}{3}}} (3 x^{2} + 5 x) \sin (3 x^{2} + 5 x) \\ = - \frac{2}{3} (6 x + 5) \cos^{.^{\frac{2}{3}}} (3 x^{2} + 5 x) \\ \times \cos^{- 1} (3 x^{2} + 5 x) \times \sin (3 x^{2} + 5 x) \\ = - \frac{2}{3} (6 x + 5) \tan (3 x^{2} + 5 x) \sqrt[3]{\cos^{2} (3 x^{2} + 5 x)} \end{aligned} \end{array}$ .

$\begin{array}{ll} 55. & Persamaan garis singgung pada kurva \\ y = 3 \sin x pada titik yang berabsis \frac{π}{3} \\ adalah . . . . \\ \begin{array}{llll} a . & y = \frac{2}{3} (x - \frac{π}{3}) - \frac{2 \sqrt{2}}{3} \\ b . & y = \frac{2}{3} (x - \frac{π}{3}) + \frac{2 \sqrt{2}}{3} \\ c . & y = \frac{3}{2} (x - \frac{π}{3}) - \frac{3 \sqrt{3}}{2} \\ d . & y = \frac{3}{2} (x - \frac{π}{3}) + \frac{3 \sqrt{3}}{2} \\ e . & y = \frac{3}{2} (x - \frac{π}{3}) - \frac{3 \sqrt{2}}{2} \end{array} \\ Jawab : d \\ \begin{aligned} y = 3 \sin x, saat x_{0} = \frac{π}{3} \\ y_{0} = 3 \sin (\frac{π}{3}) = 3 (\frac{1}{2} \sqrt{3}) = \frac{3 \sqrt{3}}{2} \\ kita cari gradien m saat y^{'}, yaitu : \\ m = y^{'} = 3 \cos x, saat x_{0} = \frac{π}{3} \\ m = 3 \cos (\frac{π}{3}) = 3 (\frac{1}{2}) = \frac{3}{2} \\ Persamaan garis singgungnya adalah : \\ y = m (x - x_{0}) + y_{0} \\ y = \frac{3}{2} (x - \frac{π}{3}) + \frac{3 \sqrt{3}}{2} \end{aligned} \end{array}$

$\begin{array}{ll} 56. & Kurva y = \sin x + \cos x untuk \\ 0 < x < π memotong sumbu X \\ di titik A. Persamaan garis \\ singgung di titik A adalah . . . . \\ \begin{array}{llll} a . & y = - \sqrt{2} (x - \frac{π}{4}) \\ b . & y = - \sqrt{2} (x - \frac{π}{2}) \\ c . & y = - \sqrt{2} (x - \frac{3 π}{4}) \\ d . & y = \sqrt{2} (x - \frac{π}{4}) \\ e . & y = \sqrt{2} (x - \frac{3 π}{4}) \end{array} \\ Jawab : c \\ \begin{aligned} Kurva memotong sumbu X \\ di titik A, berarti y = 0 \\ \sin x + \cos x = 0 \\ \sin x = - \cos x \\ \frac{\sin x}{\cos x} = - 1 \Leftrightarrow \tan x = - 1 \\ \tan x = \tan (\frac{3 π}{4}) \Rightarrow x = \frac{3 π}{4} \\ Jadi, titik A-nya : (\frac{3 π}{4}, 0) \\ dan nilai gradien m = y^{'}, yaitu : \\ m = \cos x - \sin x \\ m = \cos (\frac{3 π}{4}) - \sin (\frac{3 π}{4}) \\ m = - \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{2} = - \sqrt{2} \\ Persamaan garis singgung di A : \\ y = m (x - x_{0}) + y_{0} \\ \Leftrightarrow y = - \sqrt{2} (x - \frac{3 π}{4}) + 0 \\ \Leftrightarrow y = - \sqrt{2} (x - \frac{3 π}{4}) \end{aligned} \end{array}$

$\begin{array}{ll} 57. & Persamaan garis singgung pada \\ kurva y = \sec^{2} x pada titik yang \\ berabsis \frac{π}{3} adalah . . . . \\ \begin{array}{llll} a . & y = 8 \sqrt{3} (x - \frac{π}{3}) - 4 \\ b . & y = 8 \sqrt{3} (x - \frac{π}{3}) + 4 \\ c . & y = - 8 \sqrt{3} (x - \frac{π}{3}) - 4 \\ d . & y = - 8 \sqrt{3} (x - \frac{π}{3}) + 4 \\ e . & y = 4 \sqrt{3} (x - \frac{π}{3}) - 4 \end{array} \\ Jawab : b \\ \begin{aligned} y = \sec^{2} x, saat x_{0} = \frac{π}{3} \\ y_{0} = \sec^{2} (\frac{π}{3}) = (2)^{2} = 4 \\ kita cari gradien m saat y^{'}, yaitu : \\ m = y^{'} = 2 \sec^{2} x \tan x, saat x_{0} = \frac{π}{3} \\ m = 2 \sec^{2} (\frac{π}{3}) \tan (\frac{π}{3}) \\ = 2 (4) \sqrt{3} = 8 \sqrt{3} \\ Persamaan garis singgungnya adalah : \\ y = m (x - x_{0}) + y_{0} \\ y = 8 \sqrt{3} (x - \frac{π}{3}) + 4 \end{aligned} \end{array}$

$\begin{array}{ll} 58. & Kurva berikut yang memiliki \\ garis singgung dengan gradien \\ 4 \sqrt{3} adalah . . . . \\ \begin{array}{llll} a . & y = 2 \sin x pada titik (\frac{π}{3}, \sqrt{3}) \\ b . & y = \cos 2 x pada titik (\frac{π}{12}, \frac{1}{2}) \\ c . & y = \tan x pada titik (π, 0) \\ d . & y = 2 \sec x pada titik (\frac{π}{3}, 2) \\ e . & y = \cot x pada titik (\frac{π}{4}, 1) \end{array} \\ Jawab : d \\ \begin{array}{cll} a & y = 2 \sin x & m = 2 \cos \frac{π}{3} \\ y^{'} = 2 \cos x & m = 2. \frac{1}{2} = 1 \\ b & y = \cos 2 x & m = - 2 \sin 2 (\frac{π}{12}) \\ y^{'} = - 2 \sin 2 x & m = - 2. \frac{1}{2} = - 1 \\ c & y = \tan x & m = \sec^{2} (π) \\ y^{'} = \sec^{2} x & m = (- 1)^{2} = 1 \\ d & y = 2 \sec x & m = 2 \sec (\frac{π}{3}) \tan (\frac{π}{3}) \\ y^{'} = 2 \sec x \tan x & m = 2.2 . \sqrt{3} = 4 \sqrt{3} \\ e & y = \cot x & m = - \csc^{2} (\frac{π}{4}) \\ y^{'} = - \csc^{2} x & m = - {(\sqrt{2})}^{2} = - 2 \end{array} \end{array}$

$\begin{array}{ll} 59. & Persamaan garis singgung pada \\ kurva y = \sec x di titik yang \\ berabsis \frac{π}{4} adalah . . . . \\ \begin{array}{llll} a . & y = \sqrt{3} x - \frac{\sqrt{3} π}{4} + \sqrt{3} \\ b . & y = \sqrt{3} x + \frac{\sqrt{3} π}{4} + \sqrt{3} \\ c . & y = \sqrt{2} x - \frac{\sqrt{2} π}{4} + \sqrt{2} \\ d . & y = \sqrt{2} x + \frac{\sqrt{2} π}{4} + \sqrt{2} \\ e . & y = \sqrt{2} x - \frac{\sqrt{2} π}{4} + \sqrt{3} \end{array} \\ Jawab : c \\ \begin{aligned} y = \sec x, saat x_{0} = \frac{π}{4} \\ y_{0} = \sec (\frac{π}{4}) = \sqrt{2} \\ kita cari gradien m saat y^{'}, yaitu : \\ m = y^{'} = \sec x \tan x, saat x_{0} = \frac{π}{4} \\ m = \sec (\frac{π}{4}) \tan (\frac{π}{4}) = \sqrt{2} .1 = \sqrt{2} \\ Persamaan garis singgungnya adalah : \\ y = m (x - x_{0}) + y_{0} \\ \Leftrightarrow y = \sqrt{2} (x - \frac{π}{4}) + \sqrt{2} \\ \Leftrightarrow y = \sqrt{2} x - \frac{\sqrt{2} π}{4} + \sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll} 60. & Persamaan garis singgung pada \\ kurva y = \sin x + \cos x di titik yang \\ berabsis \frac{π}{2} akan memotong sumbu \\ Y dengan ordinatnya berupa . . . . \\ \begin{array}{llll} a . & \frac{π}{2} + 1 \\ b . & \frac{π}{2} - 1 \\ c . & 1 - \frac{π}{2} \\ d . & 2 + \frac{π}{2} \\ e . & 2 - \frac{π}{2} \end{array} \\ Jawab : a \\ \begin{aligned} y = \sin x + \cos x, saat x_{0} = \frac{π}{2} \\ y_{0} = \sin (\frac{π}{2}) + \cos (\frac{π}{2}) = 1 + 0 = 1 \\ kita cari gradien m saat y^{'}, yaitu : \\ m = \cos x - \sin x \\ m = \cos (\frac{π}{2}) - \sin (\frac{π}{2}) \\ m = 0 - 1 = - 1 \\ Persamaan garis singgungnya adalah : \\ y = m (x - x_{0}) + y_{0} \\ \Leftrightarrow y = - 1 (x - \frac{π}{2}) + 1 \\ \Leftrightarrow y = - x + \frac{π}{2} + 1 \\ Ordinat garis singgungnya saat \\ memotong sumbu-Y adalah : x = 0, \\ maka \\ y = - 0 + \frac{π}{2} + 1 = \frac{π}{2} + 1 \end{aligned} \end{array}$

PAS MATEMATIKA BLOG

Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

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