IB Mathematics SL 5.1 Derivative interpreted as gradient function AI HL Paper 2- Exam Style Questions- New Syllabus

a
(i) Differentiate \( y = -0.1x^3 + 0.8x^2 \):
\( \frac{dy}{dx} = -0.1 \times 3x^2 + 0.8 \times 2x = -0.3x^2 + 1.6x \)
(ii) Set \( \frac{dy}{dx} = 0 \):
\( -0.3x^2 + 1.6x = 0 \implies x(-0.3x + 1.6) = 0 \implies x = 0 \text{ or } x = \frac{1.6}{0.3} = \frac{16}{3} \approx 5.3333 \)
Since \( x \in [2, 8] \), use \( x = \frac{16}{3} \).
Substitute: \( y = -0.1 \left( \frac{16}{3} \right)^3 + 0.8 \left( \frac{16}{3} \right)^2 \):
\( \left( \frac{16}{3} \right)^3 = \frac{4096}{27}, \quad \left( \frac{16}{3} \right)^2 = \frac{256}{9} \)
\( y = -0.1 \times \frac{4096}{27} + 0.8 \times \frac{256}{9} \approx -15.17037 + 22.75556 \approx 7.58519 \)
Maximum height: 7.59 m.
Explanation:
Differentiate using the power rule, solve for critical points, and evaluate \( y \) at \( x = \frac{16}{3} \), rounding to two decimal places.
Result:
(i) \( \frac{dy}{dx} = -0.3x^2 + 1.6x \)
(ii) 7.59 m

b
(i) At \( x = 4 \):
\( y = -0.1 \times 4^3 + 0.8 \times 4^2 = -0.1 \times 64 + 0.8 \times 16 = -6.4 + 12.8 = 6.4 \)
(ii) At \( x = 6 \):
\( y = -0.1 \times 6^3 + 0.8 \times 6^2 = -0.1 \times 216 + 0.8 \times 36 = -21.6 + 28.8 = 7.2 \)
Explanation:
Substitute \( x = 4 \) and \( x = 6 \) into the height function.
Result:
(i) 6.4 m
(ii) 7.2 m

c
Trapezoidal rule, 3 intervals over \( [2, 8] \), width \( h = \frac{8 – 2}{3} = 2 \).
Points: \( x = 2, 4, 6, 8 \); Heights: \( y(2) = 2.4 \), \( y(4) = 6.4 \), \( y(6) = 7.2 \), \( y(8) = 0 \).
Area:
\( \frac{h}{2} [y_0 + 2(y_1 + y_2) + y_3] = 1 \times [2.4 + 2(6.4 + 7.2) + 0] = 2.4 + 2 \times 13.6 = 2.4 + 27.2 = 29.6 \)
Explanation:
Apply the trapezoidal rule with the given heights.
Result:
29.6 m²

d
(i) Area integral:
\[ \int_2^8 (-0.1x^3 + 0.8x^2) \, dx \]
(ii) Integrate:
\( \int (-0.1x^3 + 0.8x^2) \, dx = -0.025x^4 + \frac{0.8}{3}x^3 \)
Evaluate:
\( \left[ -0.025x^4 + \frac{0.8}{3}x^3 \right]_2^8 = \left( -0.025 \times 4096 + \frac{0.8}{3} \times 512 \right) – \left( -0.025 \times 16 + \frac{0.8}{3} \times 8 \right) \approx 34.1333 – 1.7333 = 32.4 \)
Explanation:
Write the integral for the area and compute it by evaluating the antiderivative.
Result:
(i) \( \int_2^8 (-0.1x^3 + 0.8x^2) \, dx \)
(ii) 32.4 m²