IBDP Maths SL 3.7 Composite functions of the form AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ 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]
▶️ Answer/Explanation
Solution:
(a) \(f(x) = |\sin(4x)|\), \(0 \leq x \leq \pi\)
- Period of \(\sin(4x)\) is \(\frac{2\pi}{4} = \frac{\pi}{2}\)
- Absolute value reflects negative portions above \(x\)-axis
- 4 complete “humps” in \([0, \pi]\) (since \(\pi \div \frac{\pi}{2} = 2\) full periods, and absolute value creates 2 humps per period)
- Range: \([0, 1]\)
(b) \(f(x) = |\sin(\pi x)|\), \(0 \leq x \leq 4\)
- Period of \(\sin(\pi x)\) is \(\frac{2\pi}{\pi} = 2\)
- Absolute value creates “bumps” instead of waves
- 2 full periods in \([0,4]\) (\(4\div2=2\))
- Zeros at integer values (\(x=0,1,2,3,4\))
- Range: \([0, 1]\)
(c) \(f(x) = \sin(|4x|)\), \(-\pi \leq x \leq \pi\)
- Even function due to absolute value on \(x\)
- Mirror image about \(y\)-axis
- Standard sine curve for \(x \geq 0\)
- Period of \(\frac{2\pi}{4} = \frac{\pi}{2}\) for \(x \geq 0\)
- 4 complete periods in \([0,\pi]\)
- Range: \([-1, 1]\)
Key Characteristics Summary:
| Function | Period | Range | Symmetry |
|---|---|---|---|
| \(|\sin(4x)|\) | \(\frac{\pi}{2}\) | \([0,1]\) | – |
| \(|\sin(\pi x)|\) | \(1\) | \([0,1]\) | – |
| \(\sin(|4x|)\) | \(\frac{\pi}{2}\) (for \(x\geq0\)) | \([-1,1]\) | Even |
▶️ Answer/Explanation
Solution:
(a) Composition \(f \circ g(x) = e^{2\sin\left(\frac{\pi x}{2}\right)}\)
The period is determined by the inner function \(g(x)\):
Period of \(\sin\left(\frac{\pi x}{2}\right) = \frac{2\pi}{\frac{\pi}{2}} = 4\)
Thus, period of \(f \circ g = 4\)
Alternative answer in degrees: \(\frac{720^\circ}{\pi} \approx 229.18^\circ\)
(b) Solving \(e^{2\sin\left(\frac{\pi x}{2}\right)} > 4\):
Take natural log: \(2\sin\left(\frac{\pi x}{2}\right) > \ln 4\)
\(\Rightarrow \sin\left(\frac{\pi x}{2}\right) > \frac{\ln 4}{2} \approx 0.693\)
Solution intervals:
\(x \in (0.488 + 4k, 1.512 + 4k)\) for \(k \in \mathbb{Z}\)
Alternative answer in degrees: \(x \in \left(28.0^\circ + \frac{720^\circ}{\pi}k, 86.5^\circ + \frac{720^\circ}{\pi}k\right)\)
Key Steps:
- For (a), the period comes from the trigonometric component
- For (b), solve the inequality by taking natural logs
- The solution involves finding where sine exceeds \(\frac{\ln 4}{2}\)
- General solution accounts for periodicity (every 4 units)