PAS MATEMATIKA BLOG: Contoh Soal 7 Turunan Fungsi Trigonometri (Bagian 2)

$\begin{array}{ll} 31. & Fungsi f (x) = \sin x - \cos x dengan \\ 0 < x < 2 π naik pada interval . . . . \\ \begin{array}{llll} a . & 0 < x < \frac{π}{4} \\ b . & \frac{π}{4} < x < 2 π \\ c . & \frac{3 π}{4} < x < \frac{7 π}{4} \\ d . & 0 < x < \frac{3 π}{4} atau \frac{7 π}{4} < x < 2 π \\ e . & 0 < x < \frac{π}{4} atau \frac{3 π}{4} < x < 2 π \end{array} \\ Jawab : d \\ \begin{aligned} Diketahui f (x) = \sin x - \cos x \\ Fungsi f naik, jika f^{'} (x) > 0 \\ Selanjutnya \\ f^{'} (x) = \cos x + \cos x = 0 \\ \sin x = - \cos x \Leftrightarrow \frac{\sin x}{\cos x} = - 1 \\ \Leftrightarrow \tan x = - 1 \\ \Leftrightarrow \tan x = \tan \frac{3 π}{4} \\ \Leftrightarrow x = \frac{3 π}{4} \pm k . π \\ \Leftrightarrow k = 0 \Rightarrow x = \frac{3 π}{4} \\ \Leftrightarrow k = 1 \Rightarrow x = \frac{3 π}{4} \pm π = \frac{7 π}{4} \\ \Leftrightarrow k = 2 \Rightarrow x = \frac{3 π}{4} \pm 2 π = tm \\ \begin{array}{ccccccccc} + + & - - & + + \\ 0 & \frac{3 π}{4} & \frac{7 π}{4} & 2 π \end{array} \\ ambil titik uji x = \frac{1}{2} π \\ untuk x = \frac{1}{2} π \Rightarrow f^{'} (\frac{1}{2} π) \\ = \cos \frac{1}{2} π + \sin \frac{1}{2} π = 0 + 1 = 1 (positif) \\ untuk x = \frac{3}{2} π \Rightarrow f^{'} (\frac{3}{2} π) \\ = \cos \frac{3}{2} π + \sin \frac{3}{2} π = 0 - 1 = - 1 (negatif) \\ untuk x = \frac{11}{6} π \Rightarrow f^{'} (\frac{11}{6} π) \\ = \cos \frac{11}{6} π + \sin \frac{11}{6} π = \frac{1}{2} \sqrt{3} - \frac{1}{2} (positif) \end{aligned} \end{array}$

$\begin{array}{ll} 32. & Fungsi f (x) = \sin^{2} x dengan \\ 0 < x < 2 π naik pada interval . . . . \\ \begin{array}{llll} a . & \frac{π}{2} < x < π atau \frac{3 π}{2} < x < 2 π \\ b . & \frac{2 π}{3} < x < π \\ c . & 0 < x < \frac{π}{2} atau π < x < \frac{3 π}{2} \\ d . & \frac{4 π}{3} < x < 2 π \\ e . & \frac{π}{3} < x < π atau \frac{4 π}{3} < x < 2 π \end{array} \\ Jawab : c \\ \begin{aligned} Diketahui f (x) = \sin^{2} x \\ Fungsi f naik, jika f^{'} (x) > 0 \\ Selanjutnya \\ f^{'} (x) = 2 \sin x \cos x = \sin 2 x = 0 \\ \Leftrightarrow \sin 2 x = 0 \\ \Leftrightarrow \sin 2 x = \sin 0 \\ \Leftrightarrow 2 x = \pm k .2 π atau 2 x = π \pm k .2 π \\ \Leftrightarrow x = \pm k . π atau x = \frac{π}{2} \pm k . π \\ \Leftrightarrow k = 0 \Rightarrow x = 0 atau x = \frac{π}{2} \\ \Leftrightarrow k = 1 \Rightarrow x = π atau x = \frac{π}{2} + π = \frac{3 π}{2} \\ \Leftrightarrow k = 2 \Rightarrow x = 2 π atau x = \frac{π}{2} + 2 π = \frac{5}{2} π (tm) \\ \begin{array}{ccccccccc} + + & - - & + + & - - \\ 0 & \frac{π}{2} & π & \frac{3 π}{2} & 2 π \end{array} \\ ambil titik uji x = \frac{1}{6} π \\ untuk x = \frac{1}{6} π \Rightarrow f^{'} (\frac{1}{6} π) \\ = \sin 2 (\frac{1}{6} π) = \sin \frac{1}{3} π = \frac{1}{2} (positif) \\ untuk x = \frac{3}{4} π \Rightarrow f (\frac{3}{4} π) \\ = \sin 2 (\frac{3}{4} π) = - 1 (negatif) \end{aligned} \end{array}$

$\begin{array}{ll} 33. & Fungsi f (x) = \cos^{2} 2 x untuk \\ 0^{\circ} < x < 360^{\circ} turun pada interval . . . . \\ \begin{array}{llll} a . & 45^{\circ} < x < 90^{\circ} \\ b . & 135^{\circ} < x < 180^{\circ} \\ c . & 225^{\circ} < x < 270^{\circ} \\ d . & 270^{\circ} < x < 300^{\circ} \\ e . & 315^{\circ} < x < 360^{\circ} \end{array} \\ Jawab : d \\ \begin{aligned} f (x) = \cos^{2} 2 x \\ Fungsi f turun, jika f^{'} (x) < 0 \\ f^{'} (x) = 2 \cos 2 x (- \sin 2 x) (2) = - 2 \sin 4 x \\ Selanjutnya \\ \Leftrightarrow - 2 \sin 4 x = 0 \Leftrightarrow \sin 4 x = 0 \Leftrightarrow \sin 4 x = \sin 0^{\circ} \\ \Leftrightarrow {\begin{cases} 4 x = 0^{\circ} + k {.360}^{\circ} & \Rightarrow x = k {.90}^{\circ} \\ 4 x = 180^{\circ} + k {.360}^{\circ} & \Rightarrow x = 45^{\circ} + k {.90}^{\circ} \end{cases} \\ \Leftrightarrow k = 0 \Rightarrow x = 0^{\circ} atau x = 45^{\circ} \\ \Leftrightarrow k = 1 \Rightarrow x = 90^{\circ} atau x = 135^{\circ} \\ \Leftrightarrow k = 2 \Rightarrow x = 180^{\circ} atau x = 225^{\circ} \\ \Leftrightarrow k = 3 \Rightarrow x = 270^{\circ} atau x = 315^{\circ} \\ \Leftrightarrow k = 4 \Rightarrow x = 360^{\circ} atau x = 405^{\circ} (tm) \\ Gunakan titik uji pada x = 30^{\circ} \\ ∙ untuk f^{'} (30^{\circ}) = - 2 \sin 4 (30^{\circ}) = - \sqrt{3} (negatif) \\ Gunakan titik uji pada x = 60^{\circ} \\ ∙ untuk f^{'} (60^{\circ}) = - 2 \sin 4 (60^{\circ}) = \sqrt{3} (positif) \\ Gunakan titik uji pada x = 120^{\circ} \\ ∙ untuk f^{'} (120^{\circ}) = - 2 \sin 4 (120^{\circ}) = - \sqrt{3} (negatif) \\ Gunakan titik uji pada x = 150^{\circ} \\ ∙ untuk f^{'} (150^{\circ}) = - 2 \sin 4 (150^{\circ}) = \sqrt{3} (positif) \\ dan seterusnya . . . \\ \begin{array}{ccccccccccc} - - & + + & - - & + + \\ 0 & 45^{\circ} & 90^{\circ} & 135^{\circ} & 180^{\circ} \\ - - & + + & - - & + + \\ 180^{\circ} & 225^{\circ} & 270^{\circ} & 315^{\circ} & 360^{\circ} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll} 34. & (SBMPTN 2015) \\ Fungsi f (x) = 2 \sqrt{\sin^{2} x + \frac{x \sqrt{3}}{2}} \\ pada 0 < x < π turun pada interval . . . . \\ \begin{array}{llll} a . & \frac{5 π}{12} < x < \frac{11 π}{12} \\ b . & \frac{π}{12} < x < \frac{5 π}{12} \\ c . & \frac{2 π}{3} < x < \frac{5 π}{6} \\ d . & \frac{3 π}{4} < x < π \\ e . & \frac{3 π}{4} < x < \frac{3 π}{2} \end{array} \\ Jawab : c \\ \begin{aligned} Diketahui f (x) = 2 \sqrt{\sin^{2} x + \frac{x \sqrt{3}}{2}} \\ Fungsi f turun, jika f^{'} (x) < 0 \\ f^{'} (x) = \frac{\sin 2 x + \frac{1}{2} \sqrt{3}}{\sqrt{\sin^{2} x + \frac{x \sqrt{3}}{2}}} = 0 \\ \sin 2 x + \frac{1}{2} \sqrt{3} = 0 \Leftrightarrow \sin 2 x = - \frac{1}{2} \sqrt{3} \\ \Leftrightarrow \sin 2 x = \sin \frac{4 π}{3} \\ \Leftrightarrow 2 x = \frac{4 π}{3} + k .2 π atau 2 x = π - \frac{4 π}{3} + k .2 π \\ \Leftrightarrow x = \frac{2 π}{3} + k . π atau x = - \frac{π}{6} + k . π \\ \Leftrightarrow k = 0 \Rightarrow x = \frac{2 π}{3} atau x = - \frac{π}{6} (tm) \\ \Leftrightarrow k = 1 \Rightarrow x = \frac{5 π}{3} atau x = \frac{5 π}{6} \\ Gunakan titik uji pada x = \frac{π}{2} = 90^{\circ} \\ ∙ untuk f^{'} (\frac{π}{2}) = \frac{\sin 2 (\frac{π}{2}) + \frac{1}{2} \sqrt{3}}{\sqrt{\sin^{2} (\frac{π}{2}) + \frac{(\frac{π}{2}) \sqrt{3}}{2}}} = + \\ (positif) \\ Gunakan titik uji pada x = \frac{3 π}{4} = 135^{\circ} \\ ∙ untuk f^{'} (\frac{3 π}{4}) = \frac{\sin 2 (\frac{3 π}{4}) + \frac{1}{2} \sqrt{3}}{\sqrt{\sin^{2} (\frac{3 π}{4}) + \frac{(\frac{3 π}{4}) \sqrt{3}}{2}}} = - \\ (negatif) \\ \begin{array}{cccccc} + + & - - \\ 0 & \frac{2 π}{3} & \frac{5 π}{6} \end{array} \end{aligned} \end{array}$

$\begin{array}{ll} 35. & Fungsi f (x) = \sqrt{\sin^{2} x + \frac{x}{2}} \\ dengan x > 0 turun pada interval . . . . \\ \begin{array}{llll} a . & \frac{5 π}{12} < x \leq \frac{13 π}{12} \\ b . & \frac{7 π}{12} < x < \frac{11 π}{12} \\ c . & \frac{π}{12} < x < \frac{5 π}{12} \\ d . & \frac{7 π}{6} < x \leq \frac{13 π}{6} \\ e . & \frac{7 π}{6} < x \leq \frac{11 π}{6} \end{array} \\ Jawab : b \\ \begin{aligned} Diketahui f (x) = \sqrt{\sin^{2} x + \frac{x}{2}} \\ Fungsi f turun, jika f^{'} (x) < 0 \\ f^{'} (x) = \frac{\sin 2 x + \frac{1}{2}}{2 \sqrt{\sin^{2} x + \frac{x}{2}}} = 0 \\ \sin 2 x + \frac{1}{2} = 0 \Leftrightarrow \sin 2 x = - \frac{1}{2} \\ \Leftrightarrow \sin 2 x = \sin \frac{7 π}{6} \\ \Leftrightarrow 2 x = \frac{7 π}{6} + k .2 π atau 2 x = π - \frac{7 π}{6} + k .2 π \\ \Leftrightarrow x = \frac{7 π}{12} + k . π atau x = - \frac{π}{12} + k . π \\ \Leftrightarrow k = 0 \Rightarrow x = \frac{7 π}{12} atau x = - \frac{π}{12} (tm) \\ \Leftrightarrow k = 1 \Rightarrow x = \frac{19 π}{12} atau x = \frac{11 π}{12} \\ Gunakan titik uji pada x = \frac{π}{2} = 90^{\circ} \\ ∙ untuk f^{'} (\frac{π}{2}) = \frac{\sin 2 (\frac{π}{2}) + \frac{1}{2}}{\sqrt{\sin^{2} (\frac{π}{2}) + \frac{(\frac{π}{2})}{2}}} = + \\ (positif) \\ Gunakan titik uji pada x = \frac{3 π}{4} = 135^{\circ} \\ ∙ untuk f^{'} (\frac{3 π}{4}) = \frac{\sin 2 (\frac{3 π}{4}) + \frac{1}{2}}{\sqrt{\sin^{2} (\frac{3 π}{4}) + \frac{(\frac{3 π}{4})}{2}}} = - \\ (negatif) \\ \begin{array}{cccccc} + + & - - \\ 0 & \frac{7 π}{12} & \frac{11 π}{12} \end{array} \end{aligned} \end{array}$

PAS MATEMATIKA BLOG

Contoh Soal 7 Turunan Fungsi Trigonometri (Bagian 2)

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