IB Mathematics AHL 1.15 Eigenvalues and eigenvectors AI HL Paper 2- Exam Style Questions- New Syllabus

a
Given: \( \frac{dx}{dt} = y \), \( \frac{dy}{dt} = -2x – 3y \)
Differentiate:
\( \frac{d^2x}{dt^2} = \frac{dy}{dt} \)
Substitute:
\( \frac{d^2x}{dt^2} = -2x – 3y \)
Use: \( y = \frac{dx}{dt} \)
Rewrite:
\( \frac{d^2x}{dt^2} = -2x – 3\frac{dx}{dt} \)
Rearrange:
\( \frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 2x = 0 \)
Explanation:
Differentiating the first equation and substituting into the second converts the system into the given second-order equation.
Result: Shown as required.

b
System matrix: \( \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix} \)
Characteristic equation:
\( \det \begin{pmatrix} -\lambda & 1 \\ -2 & -3 – \lambda \end{pmatrix} = 0 \)
Compute:
\( \lambda(\lambda + 3) + 2 = \lambda^2 + 3\lambda + 2 = 0 \)
Solve:
\( \lambda = \frac{-3 \pm \sqrt{9 – 8}}{2} = \frac{-3 \pm 1}{2} \)
Result:
\( \lambda = -1, -2 \)
Explanation:
The eigenvalues are found by solving the characteristic polynomial, matching the auxiliary equation.
Result: \( \lambda = -1, -2 \).

c
Eigenvalues: \( \lambda = -1, -2 \)
General solution:
\( x = Ae^{-t} + Be^{-2t} \)
Derivative:
\( \frac{dx}{dt} = -Ae^{-t} – 2Be^{-2t} \)
Initial conditions:
At \( t = 0 \), \( x = 0 \): \( A + B = 0 \)
At \( t = 0 \), \( \frac{dx}{dt} = 1 \): \( -A – 2B = 1 \)
Solve:
\( A + B = 0 \), \( -A – 2B = 1 \)
Add: \( -B = 1 \implies B = -1 \)
Then: \( A – 1 = 0 \implies A = 1 \)
Solution:
\( x = e^{-t} – e^{-2t} \)
Explanation:
The solution uses eigenvalues, with constants determined by initial conditions.
Result: \( x = e^{-t} – e^{-2t} \).

d
Function: \( x = e^{-t} – e^{-2t} \)
Maximum:
\( \frac{dx}{dt} = -e^{-t} + 2e^{-2t} = 0 \)
Solve:
\( 2e^{-2t} = e^{-t} \implies t = \ln 2 \)
Compute:
\( x = e^{-\ln 2} – e^{-2\ln 2} = \frac{1}{2} – \frac{1}{4} = 0.25 \)
Behavior:
Starts at (0, 0), peaks at \( (\ln 2, 0.25) \), decays to \( x = 0 \)

Explanation:
The maximum is found by setting the derivative to zero; the graph starts at the origin, peaks, and decays.
Result: Graph passes through (0, 0), peaks at \( (\ln 2, 0.25) \), approaches \( x = 0 \).

e

Solve: \( e^{-t} – e^{-2t} = 0.1 \)
Let \( u = e^{-t} \):
\( u – u^2 = 0.1 \implies u^2 – u + 0.1 = 0 \)
Solve:
\( u = \frac{1 \pm \sqrt{1 – 0.4}}{2} \approx 0.887, 0.113 \)
Times:
\( e^{-t} = 0.887 \implies t \approx 0.11957 \)
\( e^{-t} = 0.113 \implies t \approx 2.1830 \)
Total:
\( 2.1830 – 0.11957 \approx 2.06 \text{ days} \)
Explanation:
The time interval where \( x > 0.1 \) is found by solving the quadratic for the exponential term.
Result: 2.06 days.

f
Reason:
The concentration of mercury may not be uniform across the river due to natural variation.
Explanation:
A longer ban accounts for potential variability in mercury levels due to river currents.
Result: The concentration of mercury may not be uniform across the river due to natural variation.