In the diagram, the graphs of
$f(x) = ext{sin} x - 1$ and $g(x) = ext{cos} 2x$ are drawn for the interval
$x ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } [ -90^{ ext{o}} ; 360^{ ext{o}} ]$ - NSC Mathematics - Question 6 - 2019 - Paper 2
Question 6
In the diagram, the graphs of
$f(x) = ext{sin} x - 1$ and $g(x) = ext{cos} 2x$ are drawn for the interval
$x ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } e... show full transcript
Worked Solution & Example Answer:In the diagram, the graphs of
$f(x) = ext{sin} x - 1$ and $g(x) = ext{cos} 2x$ are drawn for the interval
$x ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } [ -90^{ ext{o}} ; 360^{ ext{o}} ]$ - NSC Mathematics - Question 6 - 2019 - Paper 2
Step 1
6.1 Write down the range of f.
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Answer
The range of the function f(x)=extsinx−1 can be determined by considering the range of extsinx, which is [−1,1]. Therefore, the range of f(x) is: [−2,0]extOR[2extand0].
Step 2
6.2 Write down the values of x in the interval x ∈ [-90°, 360°] for which graph f is decreasing.
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Answer
The function f(x) is decreasing in the intervals where its derivative is negative. For the interval [−90exto;270exto], the function f(x) is decreasing in the ranges: 90exto<x<270exto.
Step 3
6.3 Determine the value(s) of x for which PQ will be a maximum.
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Answer
To find the maximum of segment PQ, we set up the equation based on the intersection of PQ and the graphs of f and g: PQ=extcos2x−(extsinx−1)
This simplifies to: PQ=1−2extsinx−extsinx+1
Setting into formula: ext{sin} x = rac{-1}{2}
Thus, x=194.48extoextor345.52exto.
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