IBDP Maths SL 3.8 Solving trigonometric equations AA HL Paper 2- Exam Style Questions- New Syllabus

(a)

The standard period of \( \tan(kx) \) is \( \frac{\pi}{|k|} \).
Here \( k = 2 \), so the period is \( \frac{\pi}{2} \).

\( \boxed{\frac{\pi}{2}} \)

(b)

Step 1: Set up equations using given points.
Substitute \( x = \frac{\pi}{12} \), \( f(x) = 5 \):
\( a \tan\left(2 \cdot \frac{\pi}{12}\right) + b = 5 \)
\( a \tan\left(\frac{\pi}{6}\right) + b = 5 \)
\( a \cdot \frac{1}{\sqrt{3}} + b = 5 \)    (1)

Substitute \( x = \frac{\pi}{3} \), \( f(x) = 7 \):
\( a \tan\left(2 \cdot \frac{\pi}{3}\right) + b = 7 \)
\( a \tan\left(\frac{2\pi}{3}\right) + b = 7 \)
\( a \cdot (-\sqrt{3}) + b = 7 \)    (2)

Step 2: Solve the system.
Subtract (1) from (2):
\( (-\sqrt{3}a + b) – \left( \frac{a}{\sqrt{3}} + b \right) = 7 – 5 \)
\( -\sqrt{3}a – \frac{a}{\sqrt{3}} = 2 \)
Multiply by \( \sqrt{3} \):
\( -3a – a = 2\sqrt{3} \)
\( -4a = 2\sqrt{3} \)
\( a = -\frac{\sqrt{3}}{2} \)

Substitute into (1):
\( -\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} + b = 5 \)
\( -\frac{1}{2} + b = 5 \)
\( b = \frac{11}{2} \)

\( \boxed{a = -\frac{\sqrt{3}}{2}}, \quad \boxed{b = \frac{11}{2}} \)