IB Mathematics SL 2.5 Modelling with the various functions AI HL Paper 2- Exam Style Questions- New Syllabus

(a)

Using technology to solve \( 1.5^x = 6 – \frac{3}{x} \):

\( x \approx 0.638 \quad \text{and} \quad x \approx 4.10 \).

\( \boxed{x \approx 0.638 \text{ and } x \approx 4.10} \).

(b)

(i) The shaded area (under \( f(x) \) between intersection points):

\( \boxed{\int_{0.638}^{4.10} 1.5^x \, dx} \).

(ii) Evaluating the integral:

\( \int_{0.638}^{4.10} 1.5^x \, dx \approx 9.81 \).

\( \boxed{9.81} \).

(iii) The area between curves is:

\( \int_{0.638}^{4.10} \big[ g(x) – f(x) \big] \, dx = \int_{0.638}^{4.10} \left(6 – \frac{3}{x} – 1.5^x\right) dx \approx 5.38 \).

\( \boxed{5.38} \).

(c)

Parallel tangents ⇒ equal derivatives:

\( f'(x) = 1.5^x \ln(1.5) \), \( g'(x) = \frac{3}{x^2} \).

Set \( 1.5^x \ln(1.5) = \frac{3}{x^2} \).

Using technology (e.g., graphing or solver):

\( k \approx 1.86 \).

\( \boxed{k \approx 1.86} \).