IBDP Maths AHL 5.12 Informal ideas of limit, continuity and convergence AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Test base case \( n = 1 \): M1
LHS: \( f^{(1)}(x) = \frac{1}{2} a (1 – ax)^{-3/2} \)
RHS: \( \frac{a^1 (2 \cdot 1 – 1)! (1 – ax)^{-(2 \cdot 1 + 1)/2}}{2^{2 \cdot 1} 1! (1 – 1)!} = \frac{a (1 – ax)^{-3/2}}{4} = \frac{1}{2} a (1 – ax)^{-3/2} \). A1
Assume true for \( n = k \): \( f^{(k)}(x) = \frac{a^k (2k – 1)! (1 – ax)^{-(2k + 1)/2}}{2^{2k} k! (k – 1)!} \). M1
Differentiate for \( n = k + 1 \): M1
\( f^{(k+1)}(x) = \frac{d}{dx} \left( \frac{a^k (2k – 1)! (1 – ax)^{-(2k + 1)/2}}{2^{2k} k! (k – 1)!} \right) \)
= \( \frac{(2k + 1) a^{k+1} (2k – 1)! (1 – ax)^{-(2k + 3)/2}}{2^{2k} \cdot 2k \cdot k! (k – 1)!} \). A1
= \( \frac{a^{k+1} (2k + 1)! (1 – ax)^{-(2(k+1) + 1)/2}}{2^{2(k+1)} (k+1)! k!} \). A1
Conclude: true for \( n = k + 1 \), so true for all \( n \in \mathbb{Z}^+ \). R1
[6 marks]
Maclaurin series: \( f(x) = f(0) + x f'(0) + \frac{x^2}{2} f”(0) + \dots \). M1
\( f(0) = (1 – 0)^{-1/2} = 1 \). A1
\( f'(x) = \frac{1}{2} a (1 – ax)^{-3/2}, f'(0) = \frac{1}{2} a \). A1
\( f”(x) = \frac{3}{4} a^2 (1 – ax)^{-5/2}, f”(0) = \frac{3}{4} a^2 \). A1
\( f(x) = 1 + x \cdot \frac{1}{2} a + \frac{x^2}{2} \cdot \frac{3}{4} a^2 = 1 + \frac{1}{2} ax + \frac{3}{8} a^2 x^2 + \dots \). A1
[4 marks]
For \( (1 – 2x)^{-1/2} \), set \( a = 2 \): \( 1 + x + \frac{3}{2} x^2 + \dots \). M1
For \( (1 – 4x)^{-1/2} \), set \( a = 4 \): \( 1 + 2x + 6x^2 + \dots \). A1
Multiply up to \( x^2 \): \( (1 + x + \frac{3}{2} x^2) \times (1 + 2x + 6x^2) = 1 + 3x + \frac{19}{2} x^2 + \dots \). A1
Thus, \( (1 – 2x)^{-1/2} (1 – 4x)^{-1/2} \approx \frac{2 + 6x + 19x^2}{2} \). A1
[3 marks]
For \( (1 – 2x)^{-1/2} \), \( |2x| < 1 \Rightarrow |x| < \frac{1}{2} \). M1
For \( (1 – 4x)^{-1/2} \), \( |4x| < 1 \Rightarrow |x| < \frac{1}{4} \). A1
Stricter condition: \( |x| < \frac{1}{4} \). A1
[2 marks]
Substitute \( x = \frac{1}{10} \): \( \frac{2 + 6 \cdot \frac{1}{10} + 19 \cdot \left(\frac{1}{10}\right)^2}{2} = \frac{2 + 0.6 + 0.19}{2} = \frac{279}{200} \). M1
\( (1 – \frac{2}{10})^{-1/2} (1 – \frac{4}{10})^{-1/2} = \frac{5}{2\sqrt{3}} \approx \frac{279}{200} \). A1
\( \sqrt{3} \approx \frac{500}{279} \). A1
[3 marks]
▶️ Answer/Explanation
Evaluate at \( x = 2 \): \( 1 – 2(2) = -3 \) and \( \frac{3}{4}(2 – 2)^2 – 3 = -3 \). A1
Both values are equal, hence \( f \) is continuous at \( x = 2 \). R1
Note: R1 may be awarded for justification using a graph or referring to limits. Do not award A0R1.
[2 marks]
Reflection in the \( y \)-axis: replace \( x \) with \( -x \): M1
\[ f(-x) = \left\{ \begin{array}{ll} 1 – 2(-x) = 1 + 2x, & -x \leq 2 \implies x \geq -2 \\ \frac{3}{4}((-x) – 2)^2 – 3 = \frac{3}{4}(x + 2)^2 – 3, & -x > 2 \implies x < -2 \end{array} \right. \]
Translation by \( \left( \begin{array}{c} 2 \\ 0 \end{array} \right) \): replace \( x \) with \( x – 2 \): M1
\[ g(x) = \left\{ \begin{array}{ll} 1 + 2(x – 2) = 2x – 3, & x – 2 \geq -2 \implies x \geq 0 \\ \frac{3}{4}((x – 2) + 2)^2 – 3 = \frac{3}{4}x^2 – 3, & x – 2 < -2 \implies x < 0 \end{array} \right. \] A1 (correct expressions) A1 (correct domains)
Note: Award M1 for evidence of reflecting a graph in the \( y \)-axis. Award M1 for attempting to substitute \( (x – 2) \) or translating along the positive \( x \)-axis. Award A1 for correct domains (independent of M1). Award A1 for correct expressions.
[4 marks]