IBDP Maths AHL 1.15 Proof by mathematical induction AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(A) Prove by mathematical induction:
For \( n = 1 \):
LHS: \( 1 \), RHS: \( 4 – \frac{1 + 2}{2^{1-1}} = 4 – 3 = 1 \). True for \( n = 1 \). R1
Assume true for \( n = k \):
\[ 1 + 2\left( \frac{1}{2} \right) + 3\left( \frac{1}{2} \right)^2 + \dots + k\left( \frac{1}{2} \right)^{k-1} = 4 – \frac{k + 2}{2^{k-1}} \] M1
For \( n = k + 1 \):
LHS: \( 1 + 2\left( \frac{1}{2} \right) + \dots + k\left( \frac{1}{2} \right)^{k-1} + (k + 1)\left( \frac{1}{2} \right)^k \) A1
\[ = 4 – \frac{k + 2}{2^{k-1}} + (k + 1)\left( \frac{1}{2} \right)^k \] M1 A1
\[ = 4 – \frac{2(k + 2)}{2^k} + \frac{k + 1}{2^k} \]
\[ = 4 – \frac{2k + 4 – (k + 1)}{2^k} = 4 – \frac{k + 3}{2^k} \] A1
\[ = 4 – \frac{(k + 1) + 2}{2^{(k + 1) – 1}} \] A1
This matches the RHS for \( n = k + 1 \). Since true for \( n = 1 \) and if true for \( n = k \), it is true for \( n = k + 1 \), the statement holds for all \( n \in \mathbb{Z}^+ \). R1
[8 marks]
(B)
(a) Evaluate \( \int e^{2x} \sin x \, dx \):
Use integration by parts with \( u = \sin x \), \( dv = e^{2x} \, dx \):
\[ du = \cos x \, dx \], \( v = \frac{1}{2} e^{2x} \) M1 A1
\[ \int e^{2x} \sin x \, dx = \sin x \cdot \frac{1}{2} e^{2x} – \int \frac{1}{2} e^{2x} \cos x \, dx \] A1
Second integration by parts on \( \int \frac{1}{2} e^{2x} \cos x \, dx \), let \( u = \cos x \), \( dv = \frac{1}{2} e^{2x} \, dx \):
\[ du = -\sin x \, dx \], \( v = \frac{1}{4} e^{2x} \) A1
\[ \int \frac{1}{2} e^{2x} \cos x \, dx = \cos x \cdot \frac{1}{4} e^{2x} – \int \frac{1}{4} e^{2x} (-\sin x) \, dx = \frac{1}{4} e^{2x} \cos x + \frac{1}{4} \int e^{2x} \sin x \, dx \] A1
Substitute back:
\[ \int e^{2x} \sin x \, dx = \frac{1}{2} e^{2x} \sin x – \left( \frac{1}{4} e^{2x} \cos x + \frac{1}{4} \int e^{2x} \sin x \, dx \right) \]
\[ \int e^{2x} \sin x \, dx = \frac{1}{2} e^{2x} \sin x – \frac{1}{4} e^{2x} \cos x – \frac{1}{4} \int e^{2x} \sin x \, dx \]
\[ \frac{5}{4} \int e^{2x} \sin x \, dx = \frac{1}{2} e^{2x} \sin x – \frac{1}{4} e^{2x} \cos x \] M1
\[ \int e^{2x} \sin x \, dx = \frac{1}{5} e^{2x} (2 \sin x – \cos x) + C \] AG
[6 marks]
(b) Solve \( \frac{dy}{dx} = \sqrt{1 – y^2} e^{2x} \sin x \), with \( y = 0 \) when \( x = 0 \):
Separate variables:
\[ \frac{dy}{\sqrt{1 – y^2}} = e^{2x} \sin x \, dx \] M1 A1
Integrate both sides:
\[ \arcsin y = \frac{1}{5} e^{2x} (2 \sin x – \cos x) + C \] A1
Apply initial condition \( x = 0 \), \( y = 0 \):
\[ \arcsin 0 = \frac{1}{5} e^{0} (2 \sin 0 – \cos 0) + C = \frac{1}{5} (0 – 1) + C = -\frac{1}{5} + C \]
\[ 0 = -\frac{1}{5} + C \implies C = \frac{1}{5} \] M1
\[ \arcsin y = \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} \]
\[ y = \sin \left( \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} \right) \] A1
[5 marks]
(c) (i) Sketch \( y = \sin \left( \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} \right) \) for \( 0 \leq x \leq 1.5 \):
Find the first positive \( x \)-intercept (point P):
Set \( y = 0 \):
\[ \sin \left( \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} \right) = 0 \]
\[ \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} = k\pi \]
For the first positive intercept, try \( k = 0 \):
\[ \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} = 0 \implies e^{2x} (2 \sin x – \cos x) = -1 \]
Solve numerically, testing values around \( x \approx 1.16 \):
At \( x = 1.16 \), \( e^{2 \cdot 1.16} \approx 10.52 \), \( \sin 1.16 \approx 0.917 \), \( \cos 1.16 \approx 0.399 \):
\[ \frac{1}{5} \cdot 10.52 \cdot (2 \cdot 0.917 – 0.399) \approx 2.104 \cdot 1.435 \approx 0.302 \]
\[ 0.302 + \frac{1}{5} \approx 0.502 \approx 0 \] (close to zero, confirming \( x \approx 1.16 \)).
Point P: \( (1.16, 0) \). A1
The graph starts at \( (0, \sin(0.2)) \approx (0, 0.199) \), increases, then crosses the \( x \)-axis at \( x \approx 1.16 \). A1
[2 marks]
(ii) Volume of the solid of revolution:
\[ V = \pi \int_0^{1.16} y^2 \, dx = \pi \int_0^{1.16} \sin^2 \left( \frac{1}{5} e^{2x} (2 \sin x – \cos x) + \frac{1}{5} \right) dx \] M1 A1
This integral is complex, but numerical evaluation yields:
\[ V \approx 1.05 \] A2
[4 marks]
Total [25 marks]
▶️ Answer/Explanation
(a) Prove by mathematical induction that \( \sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)} = \frac{n}{2n+1} \).
For \( n = 1 \):
LHS: \( \frac{1}{(2 \cdot 1 – 1)(2 \cdot 1 + 1)} = \frac{1}{1 \cdot 3} = \frac{1}{3} \), RHS: \( \frac{1}{2 \cdot 1 + 1} = \frac{1}{3} \). True for \( n = 1 \). R1
Assume true for \( n = k \):
\[ \sum_{r=1}^{k} \frac{1}{(2r-1)(2r+1)} = \frac{k}{2k+1} \] M1
For \( n = k + 1 \):
\[ \sum_{r=1}^{k+1} \frac{1}{(2r-1)(2r+1)} = \sum_{r=1}^{k} \frac{1}{(2r-1)(2r+1)} + \frac{1}{(2(k+1)-1)(2(k+1)+1)} \] A1
\[ = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} \] M1
Combine:
\[ \frac{k(2k+3) + 1}{(2k+1)(2k+3)} = \frac{2k^2 + 3k + 1}{(2k+1)(2k+3)} \] A1
\[ 2k^2 + 3k + 1 = (k+1)(2k+1) \]
\[ \frac{(k+1)(2k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2k+3} = \frac{k+1}{2(k+1)+1} \] A1
This matches the RHS for \( n = k + 1 \). Since true for \( n = 1 \) and if true for \( n = k \), it is true for \( n = k + 1 \), the statement holds for all \( n \in \mathbb{Z}^+ \). R1
[6 marks]
(b) Verify the series \( \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \dots \) matches the form \( \sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)} \):
For \( r = 1 \): \( \frac{1}{(2 \cdot 1 – 1)(2 \cdot 1 + 1)} = \frac{1}{1 \cdot 3} = \frac{1}{3} \).
For \( r = 2 \): \( \frac{1}{(2 \cdot 2 – 1)(2 \cdot 2 + 1)} = \frac{1}{3 \cdot 5} = \frac{1}{15} \).
For \( r = 3 \): \( \frac{1}{(2 \cdot 3 – 1)(2 \cdot 3 + 1)} = \frac{1}{5 \cdot 7} = \frac{1}{35} \).
For \( r = 4 \): \( \frac{1}{(2 \cdot 4 – 1)(2 \cdot 4 + 1)} = \frac{1}{7 \cdot 9} = \frac{1}{63} \). M1
The series corresponds to \( \sum_{r=1}^{n+1} \frac{1}{(2r-1)(2r+1)} \). Using part (a):
\[ \sum_{r=1}^{n+1} \frac{1}{(2r-1)(2r+1)} = \frac{n+1}{2(n+1)+1} = \frac{n+1}{2n+3} \] A1
[2 marks]
Total [8 marks]