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In die diagram is die grafieke van f(x) = -3sin \frac{x}{2} en g(x) = 2cos(x - 60°) in die interval x ∈ [ -180° ; 180° ] geskes - NSC Mathematics - Question 6 - 2018 - Paper 2
Question 6
![In-die-diagram-is-die-grafieke-van---f(x)-=--3sin-\frac{x}{2}---en--g(x)-=-2cos(x---60°)-in-die-interval---x-∈-[--180°-;-180°-]-geskes-NSC Mathematics-Question 6-2018-Paper 2.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/subject_692_paper_20422_q6_679949c8.jpg)
In die diagram is die grafieke van f(x) = -3sin \frac{x}{2} en g(x) = 2cos(x - 60°) in die interval x ∈ [ -180° ; 180° ] geskes. T(δ ; g) is 'n draai punt van... show full transcript
Worked Solution & Example Answer:In die diagram is die grafieke van f(x) = -3sin \frac{x}{2} en g(x) = 2cos(x - 60°) in die interval x ∈ [ -180° ; 180° ] geskes - NSC Mathematics - Question 6 - 2018 - Paper 2
Step 2
Step 3
Bereken f(−120°) − g(−120°).
Answer
Vir f(−120°):
f(−120°) = -3sin \left( \frac{-120°}{2} \right) = -3sin(−60°) = \frac{4 + 3\sqrt{3}}{2}.
Vir g(−120°):
g(−120°) = 2cos(−120° - 60°) = 2cos(−180°) = 2(-1) = -2.
Dus, f(−120°) − g(−120°) = \frac{4 + 3\sqrt{3}}{2} - (−2) = \frac{4 + 3\sqrt{3}}{2} + 2.
Step 4
Gebruik die grafieke om die waardes van x, in die interval x ∈ [−180°; 180°], te bepaal waarvoor.
Answer
6.4.1 g(x) > 0:
Die x-intercepts van g is by x = −30° en x = 150°.
So, g(x) > 0 wanneer −30° < x < 150°.
6.4.2 f(x) < 0:
Die x-intercepts van f is by x = −120° en x = 120°.
So, f(x) < 0 wanneer −180° < x < −120° of 120° < x < 180°.