IBDP Maths AHL 5.19Maclaurin series AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) \( f'(x) = -2 \sin 2x e^{\cos 2x} = 0 \) M1.
\( \sin 2x = 0 \) A1.
\( x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi \) A1.
Coordinates: \( (-\pi, e) \), \( \left( -\frac{\pi}{2}, \frac{1}{e} \right) \), \( (0, e) \), \( \left( \frac{\pi}{2}, \frac{1}{e} \right) \), \( (\pi, e) \) A2.
[5 marks]
(b) \( f”(x) = 4 e^{\cos 2x} (\sin^2 2x – \cos 2x) \) M1.
At \( x = -\pi \), \( f”(-\pi) = -4e \) (local maximum) A1.
At \( x = -\frac{\pi}{2} \), \( f”(-\frac{\pi}{2}) = 4e^{-1} \) (local minimum) A1.
At \( x = 0 \), \( f”(0) = -4e \) (local maximum) A1.
At \( x = \frac{\pi}{2} \), \( f”(\frac{\pi}{2}) = 4e^{-1} \) (local minimum).
At \( x = \pi \), \( f”(\pi) = -4e \) (local maximum).
Conclusion: Two local maxima at \( x = -\pi, 0, \pi \), one local minimum at \( x = -\frac{\pi}{2} \) A1.
[4 marks]
(c) Sketch shows maxima at \( x = 0 \) (\( y = e \)) and \( x = \pi \) (\( y = e \)), minimum at \( x = \frac{\pi}{2} \) (\( y = \frac{1}{e} \)) M1.
Correct shape reflecting decrease from \( x = 0 \) to \( x = \frac{\pi}{2} \), increase from \( x = \frac{\pi}{2} \) to \( x = \pi \) A1.
Accurate labeling of key points A1.
For larger interval: 
[3 marks]
(d) (i) \( \cos 2x = 1 – 2x^2 + \frac{2x^4}{3} \) M1 A1.
(ii) \( e^{\cos 2x – 1} = 1 – 2x^2 + \frac{8x^4}{3} \) M1 A1.
(iii) \( f(x) = e \left( 1 – 2x^2 + \frac{8x^4}{3} \right) \) M1 A1.
[6 marks]
(e) Using first two non-zero terms: \( f(x) \approx e (1 – 2x^2) \) M1.
\( \int_{0}^{\frac{1}{10}} e (1 – 2x^2) \, dx = e \left[ x – \frac{2x^3}{3} \right]_{0}^{\frac{1}{10}} = e \left( \frac{1}{10} – \frac{2}{3000} \right) = \frac{298e}{3000} = \frac{149e}{1500} \) M1 A1.
[3 marks]
▶️ Answer/Explanation
(a) For \( n = 1 \):
LHS: \( f^{(1)}(x) = \frac{1}{2} a (1 – ax)^{-3/2} \) M1.
RHS: \( \frac{a^1 (2 \cdot 1 – 1)! (1 – ax)^{-3/2}}{2^{2 \cdot 1} 1! (1-1)!} = \frac{a (1) (1 – ax)^{-3/2}}{4 \cdot 1 \cdot 1} = \frac{a}{2} (1 – ax)^{-3/2} \) A1.
True for \( n = 1 \).
Assume true for \( n = k \):
\( f^{(k)}(x) = \frac{a^k (2k-1)! (1 – ax)^{-\frac{2k+1}{2}}}{2^{2k} k! (k-1)!} \) M1.
For \( n = k+1 \), differentiate:
\( f^{(k+1)}(x) = \frac{d}{dx} \left( \frac{a^k (2k-1)! (1 – ax)^{-\frac{2k+1}{2}}}{2^{2k} k! (k-1)!} \right) \).
= \( \frac{(2k+1)}{2} \cdot (-a) \cdot \frac{a^k (2k-1)! (1 – ax)^{-\frac{2k+3}{2}}}{2^{2k} k! (k-1)!} \) M1.
Multiply numerator and denominator by \( 2k \):
= \( \frac{(2k+1) a^{k+1} (2k-1)! (1 – ax)^{-\frac{2k+3}{2}}}{2^{2k+1} k! (k-1)! \cdot 2k} \).
= \( \frac{a^{k+1} (2k+1)! (1 – ax)^{-\frac{2(k+1)+1}{2}}}{2^{2(k+1)} (k+1)! k!} \) A1.
True for \( n = k+1 \). Since true for \( n = 1 \), true for all \( n \in \mathbb{Z}^+ \) A1.
[6 marks]
(b) Maclaurin series: \( f(x) = f(0) + x f'(0) + \frac{x^2}{2} f”(0) + \dots \).
\( f(0) = (1 – 0)^{-1/2} = 1 \) M1.
\( f'(x) = \frac{1}{2} a (1 – ax)^{-3/2} \), so \( f'(0) = \frac{1}{2} a \) A1.
\( f”(x) = \frac{3}{4} a^2 (1 – ax)^{-5/2} \), so \( f”(0) = \frac{3}{4} a^2 \).
Series: \( f(x) = 1 + x \cdot \frac{1}{2} a + \frac{x^2}{2} \cdot \frac{3}{4} a^2 = 1 + \frac{1}{2} ax + \frac{3}{8} a^2 x^2 \) A1.
[3 marks]
(c) For \( (1 – 2x)^{-1/2} \), use part (b) with \( a = 2 \):
\( (1 – 2x)^{-1/2} = 1 + \frac{1}{2} \cdot 2x + \frac{3}{8} \cdot 4x^2 = 1 + x + \frac{3}{2} x^2 + \dots \) M1.
For \( (1 – 4x)^{-1/2} \), use part (b) with \( a = 4 \):
\( (1 – 4x)^{-1/2} = 1 + \frac{1}{2} \cdot 4x + \frac{3}{8} \cdot 16x^2 = 1 + 2x + 6x^2 + \dots \) M1.
Multiply up to \( x^2 \): \( (1 + x + \frac{3}{2} x^2) \cdot (1 + 2x + 6x^2) = 1 + (2x + x) + (6x^2 + 2x \cdot x + \frac{3}{2} x^2) = 1 + 3x + \frac{19}{2} x^2 \).
Thus, \( (1 – 2x)^{-1/2} (1 – 4x)^{-1/2} \approx \frac{2 + 6x + 19x^2}{2} \) A1.
[3 marks]
(d) For \( (1 – 2x)^{-1/2} \), \( |2x| < 1 \Rightarrow |x| < \frac{1}{2} \).
For \( (1 – 4x)^{-1/2} \), \( |4x| < 1 \Rightarrow |x| < \frac{1}{4} \) M1.
Stricter condition: \( |x| < \frac{1}{4} \) A1.
[2 marks]
(e) Substitute \( x = \frac{1}{10} \) into \( \frac{2 + 6x + 19x^2}{2} \):
\( \frac{2 + 6 \cdot \frac{1}{10} + 19 \cdot \left(\frac{1}{10}\right)^2}{2} = \frac{2 + 0.6 + 0.19}{2} = \frac{2.79}{2} = \frac{279}{200} \) M1.
\( (1 – 2 \cdot \frac{1}{10})^{-1/2} (1 – 4 \cdot \frac{1}{10})^{-1/2} = \left(1 – \frac{1}{5}\right)^{-1/2} \left(1 – \frac{2}{5}\right)^{-1/2} = \frac{5}{2\sqrt{3}} \approx \frac{279}{200} \).
Solve: \( \frac{5}{2\sqrt{3}} = \frac{279}{200} \Rightarrow \sqrt{3} = \frac{5 \cdot 200}{2 \cdot 279} = \frac{500}{279} \) A1.
Answer: \( \frac{500}{279} \) A1.
[3 marks]