IBDP Maths SL 3.7 Composite functions of the form AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Write \(3\theta\) as \(2\theta+\theta\) and use the compound-angle identity.
\(\sin 3\theta=\sin(2\theta+\theta)\)
\(\sin 3\theta=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta\)
Using \(\sin 2\theta=2\sin\theta\cos\theta\) and \(\cos 2\theta=1-2\sin^2\theta\):
\(\sin 3\theta=2\sin\theta\cos^2\theta+(1-2\sin^2\theta)\sin\theta\)
Now use \(\cos^2\theta=1-\sin^2\theta\).
\(\sin 3\theta=2\sin\theta(1-\sin^2\theta)+\sin\theta-2\sin^3\theta\)
\(\sin 3\theta=2\sin\theta-2\sin^3\theta+\sin\theta-2\sin^3\theta\)
\(\sin 3\theta=3\sin\theta-4\sin^3\theta\)
✅ Hence, \(\sin 3\theta=3\sin\theta-4\sin^3\theta\).
(b)
Let \(u=\theta+\dfrac{\pi}{6}\). Using the identity from part (a):
\(6\sin u-8\sin^3u=2(3\sin u-4\sin^3u)\)
\(6\sin u-8\sin^3u=2\sin 3u\)
Therefore:
\(y=2\sin\left(3\left(\theta+\dfrac{\pi}{6}\right)\right)\)
\(y=2\sin\left(3\theta+\dfrac{\pi}{2}\right)\)
Starting with the graph of \(y=\sin\theta\), apply the following transformations in order:
1. A horizontal stretch with scale factor \(\dfrac{1}{3}\), giving \(y=\sin 3\theta\).
2. A horizontal translation of \(\dfrac{\pi}{6}\) units to the left, giving \(y=\sin\left(3\left(\theta+\dfrac{\pi}{6}\right)\right)\).
3. A vertical stretch with scale factor \(2\).
The horizontal transformations must be applied in the stated order. The vertical stretch may be applied at any stage.
✅ Answer: Horizontal stretch by scale factor \(\dfrac{1}{3}\), translation \(\dfrac{\pi}{6}\) to the left, and vertical stretch by scale factor \(2\).
▶️ Answer/Explanation
Solution:
(a) To find \(k\), we substitute point A into the equation:
\(-2 = 1 + k\sin\left(\frac{\pi}{6}\right)\)
Since \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\):
\(-2 = 1 + \frac{k}{2}\)
\(\frac{k}{2} = -3\)
\(k = -6\)
(b) The maximum value occurs when \(\sin x\) is minimized. For \(k = -6\), the minimum of \(\sin x\) (-1) gives the maximum of \(y\):
Maximum occurs at \(x = \frac{3\pi}{2}\) (from graph or calculus)
\(y_{\text{max}} = 1 + (-6)(-1) = 7\)
Thus, \(a = \frac{3\pi}{2}\) and \(b = 7\)
Alternative Calculus Method:
Find derivative: \(y’ = -6\cos x\)
Set \(y’ = 0\): \(\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}\)
Second derivative test shows maximum at \(x = \frac{3\pi}{2}\)
Substitute back to find \(y = 7\)
Markscheme:
(a) \(-2 = 1 + k\sin\left(\frac{\pi}{6}\right)\) M1
\(-3 = \frac{1}{2}k\) A1
\(k = -6\) AG N0
(b) METHOD 1
maximum \(\Rightarrow \sin x = -1\) M1
\(a = \frac{3\pi}{2}\) A1
\(b = 1 – 6(-1) = 7\) A1 N2
METHOD 2
\(y’ = 0\) M1
\(k\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots\)
\(a = \frac{3\pi}{2}\) A1
\(b = 1 – 6(-1) = 7\) A1 N2
Note: Award A1A1 for \(\left(\frac{3\pi}{2}, 7\right)\).
[5 marks]
▶️ Answer/Explanation
Solution:
a. For a sine function \(y = a\sin(b(x+c)) + d\):
Amplitude \(a = \frac{y_{\text{max}} – y_{\text{min}}}{2} = \frac{3.5 – 0.5}{2} = 1.5\)
Vertical shift \(d = \frac{y_{\text{max}} + y_{\text{min}}}{2} = \frac{3.5 + 0.5}{2} = 2\)
Thus, \(a = 1.5\) and \(d = 2\)
b. The period can be found from the distance between max and min points:
Distance between max at \(x=1\) and min at \(x=2\) is half the period
Therefore, full period \(= 2 \times (2-1) = 2\)
Since period \(= \frac{2π}{b}\), we have \(b = \frac{2π}{2} = π\)
c. The phase shift \(c\) can be found using the maximum point:
At maximum: \(b(1 + c) = \frac{π}{2} + 2πn\) (for any integer \(n\))
Using \(b=π\) and solving for smallest \(c>0\):
\(π(1 + c) = \frac{π}{2}\) ⇒ \(1 + c = \frac{1}{2}\) ⇒ \(c = -0.5\) (invalid)
Next solution: \(π(1 + c) = \frac{5π}{2}\) ⇒ \(1 + c = \frac{5}{2}\) ⇒ \(c = 1.5\)
Thus, smallest positive \(c = 1.5\)
Markscheme:
a. \(a = 1.5\), \(d = 2\) A1A1
[2 marks]
b. \(b = \frac{2π}{2} = π\) (M1)A1
[2 marks]
c. attempt to solve an appropriate equation or apply a horizontal translation (M1)
\(c = 1.5\) A1
Note: Do not award a follow through mark for the final A1.
Award (M1)A0 for \(c = -0.5\).
[2 marks]