IB Mathematics SL 3.5 Definition of cos , sin and tan angles AA SL Paper 2- Exam Style Questions- New Syllabus

(a) Determine the period:
For a tangent function of the form \( \tan(kx) \), the period is \( \frac{\pi}{k} \).
Here, \( k = 2 \).
Therefore, the period is \( \frac{\pi}{2} \).
Answer: \( \displaystyle \boxed{\frac{\pi}{2}} \)

(b) Find \( a \) and \( b \):
Substitute the given points into \( f(x) = a \tan(2x) + b \).

For \( \left( \frac{\pi}{12}, 5 \right) \):
\( a \tan\left( 2 \cdot \frac{\pi}{12} \right) + b = 5 \)
\( a \tan\left( \frac{\pi}{6} \right) + b = 5 \)
Using exact values, \( \tan\left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \):
\( \frac{a}{\sqrt{3}} + b = 5 \) — (1)

For \( \left( \frac{\pi}{3}, 7 \right) \):
\( a \tan\left( 2 \cdot \frac{\pi}{3} \right) + b = 7 \)
\( a \tan\left( \frac{2\pi}{3} \right) + b = 7 \)
Using exact values, \( \tan\left( \frac{2\pi}{3} \right) = -\sqrt{3} \):
\( -a\sqrt{3} + b = 7 \) — (2)
Subtract equation (2) from equation (1):
\( \left( \frac{a}{\sqrt{3}} + b \right) – ( -a\sqrt{3} + b ) = 5 – 7 \)
\( \frac{a}{\sqrt{3}} + a\sqrt{3} = -2 \)
Multiply by \( \sqrt{3} \):
\( a + 3a = -2\sqrt{3} \)
\( 4a = -2\sqrt{3} \implies a = -\frac{\sqrt{3}}{2} \)
Substitute \( a = -\frac{\sqrt{3}}{2} \) into (1):
\( \frac{-\frac{\sqrt{3}}{2}}{\sqrt{3}} + b = 5 \)
\( -\frac{1}{2} + b = 5 \implies b = 5.5 = \frac{11}{2} \)
Answer: \( a = \boxed{-\frac{\sqrt{3}}{2}}, \quad b = \boxed{\frac{11}{2}} \)