In the diagram below, the graphs of $f(x)= ext{cos}2x$ and $g(x)=- ext{sin}x$ are drawn for the interval $x \in [-180^{\circ}; 180^{\circ}]$ - NSC Mathematics - Question 6 - 2021 - Paper 2
Question 6
In the diagram below, the graphs of $f(x)= ext{cos}2x$ and $g(x)=- ext{sin}x$ are drawn for the interval $x \in [-180^{\circ}; 180^{\circ}]$. A and B are two points ... show full transcript
Worked Solution & Example Answer:In the diagram below, the graphs of $f(x)= ext{cos}2x$ and $g(x)=- ext{sin}x$ are drawn for the interval $x \in [-180^{\circ}; 180^{\circ}]$ - NSC Mathematics - Question 6 - 2021 - Paper 2
Step 1
6.1 Without using a calculator, determine the values of x for which cos 2x = -sin x in the interval x ∈ [-180°, 180°].
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Answer
To solve the equation extcos2x=−extsinx, we can utilize the identity of the sine function:
Rearranging the equation gives:
extcos2x+extsinx=0
Using the double angle formula for cosine, we know:
extcos2x=2extcos2x−1
Substitute this into the equation:
2extcos2x−1+extsinx=0
Substitute extsinx=1−extcos2x to obtain:
2extcos2x−extcos2x−1=0
This simplifies to:
extcos2x−1=0extorextcos2x=1
From this, we find:
extcosx=1extorextcosx=−1extorx=k⋅180extcircled0
Thus, solving for x gives:
x=150extcircled0,−30extcircled0,90extcircled0
The values of x that satisfy the condition are:
x=150extcircled0,−30extcircled0,90extcircled0.
Step 2
6.2.1 How many degrees apart are points A and B from each other?
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Answer
The distance between points A and B is given by:
AB=(−30∘)−(150∘)=−180∘
Thus, the points are 120 degrees apart.
Step 3
6.2.2 For which values of x in the given interval will f'(x) > 0?
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Answer
The derivative f′(x) corresponds to the function f(x)=extcos2x. Given that:
f′(x)=−2extsin2x
Setting this to greater than zero:
ightarrow ext{sin}2x < 0$$
This occurs for:
$$180^{\circ} < 2x < 360^{\circ}, or \ 0 < 2x < 180^{\circ}$$
Evaluating gives:
$$x \in (90^{\circ}, 180^{\circ}) \\ or \ (-90^{\circ}, 0)$$.
Step 4
6.2.3 Determine the values of k for which cos 2x + 3 = k will have no solution.
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Answer
To find the values of k, recall that for the equation to have no solution, the right-hand side must be outside of the range of the left-hand side.
Given that:
−1≤cos2x≤1
Thus,
−1+3<k>1+3
This implies:
k≤2ork≥4
So, for k<2 or k>4, there will be no solution.
![In-the-diagram-below,-the-graphs-of-$f(x)=-ext{cos}2x$-and-$g(x)=--ext{sin}x$-are-drawn-for-the-interval-$x-\in-[-180^{\circ};-180^{\circ}]$-NSC Mathematics-Question 6-2021-Paper 2.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/subject_692_paper_20438_q6_679949c8.jpg)
