IBDP Maths AHL 5.13 The evaluation of limits AA HL Paper 2- Exam Style Questions- New Syllabus

(a)
We need to divide 15 students into 3 groups of 5. The order of the groups does not matter (they are just “three teams”).
Number of ways = \( \frac{\binom{15}{5} \times \binom{10}{5} \times \binom{5}{5}}{3!} \)
\( = \frac{3003 \times 252 \times 1}{6} \)
\( = 126126 \).
Answer: \( \boxed{126126} \)

(b)
We integrate \( \int_0^a kte^{-3t} dt \) using integration by parts.
Let \( u = t \) so \( \frac{du}{dt} = 1 \).
Let \( \frac{dv}{dt} = e^{-3t} \) so \( v = -\frac{1}{3}e^{-3t} \).
Using \( \int u \frac{dv}{dt} dt = uv – \int v \frac{du}{dt} dt \):
\( \int_0^a kte^{-3t} dt = k \left( \left[ -\frac{t}{3}e^{-3t} \right]_0^a – \int_0^a -\frac{1}{3}e^{-3t} dt \right) \)
\( = k \left( -\frac{a}{3}e^{-3a} – 0 + \frac{1}{3} \left[ -\frac{1}{3}e^{-3t} \right]_0^a \right) \)
\( = k \left( -\frac{a}{3}e^{-3a} – \frac{1}{9}e^{-3a} + \frac{1}{9} \right) \)
Factor out \( \frac{1}{9} \):
\( = \frac{k}{9} \left( 1 – 3ae^{-3a} – e^{-3a} \right) \)
\( = \frac{k}{9} \left( 1 – (3a + 1)e^{-3a} \right) \).
Shown.

(c)
(i) Limit \( \lim_{x \to \infty} \frac{3x + 1}{e^{3x}} \). This is of the form \( \frac{\infty}{\infty} \).
Using l’Hôpital’s rule, differentiate numerator and denominator:
\( \lim_{x \to \infty} \frac{3}{3e^{3x}} = \lim_{x \to \infty} \frac{1}{e^{3x}} = 0 \).
Answer: \( \boxed{0} \)

(ii) For a valid PDF, the total area must be 1.
\( \lim_{a \to \infty} \int_0^a f(t) dt = \frac{k}{9} [1 – 0] = \frac{k}{9} \).
\( \frac{k}{9} = 1 \implies k = 9 \).
Answer: \( \boxed{k = 9} \)

(d)
The median \( m \) satisfies \( \int_0^m f(t) dt = 0.5 \).
Using the result from (b) with \( k=9 \):
\( \frac{9}{9} \left[ 1 – (3m + 1)e^{-3m} \right] = 0.5 \)
\( 1 – (3m + 1)e^{-3m} = 0.5 \)
\( (3m + 1)e^{-3m} = 0.5 \)
Using a GDC to solve for \( m \):

\( m \approx 0.5594 \).
Answer: \( \boxed{0.559 \text{ minutes}} \) (or 33.6 seconds)