IB Mathematics SL 5.3 The derivative of functions AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme (with full working)
(a)
(i) Differentiate \(y=-0.1x^{3}+0.8x^{2}\):
\[ \frac{dy}{dx}=-0.3x^{2}+1.6x. \] M1 A1
(ii) Stationary points satisfy \(\dfrac{dy}{dx}=0\):
\[ -0.3x^{2}+1.6x=0 \;\;\Rightarrow\;\; x\bigl(1.6-0.3x\bigr)=0 \;\;\Rightarrow\;\; x=0 \text{ or } x=\frac{1.6}{0.3}=\frac{16}{3}\,(=5.\overline{3}). \] Since the domain is \(2\le x\le 8\), the relevant stationary point is \(x=\dfrac{16}{3}\).
Check it is a maximum using the second derivative: \[ \frac{d^{2}y}{dx^{2}}=-0.6x+1.6,\quad \frac{d^{2}y}{dx^{2}}\Big|_{x=16/3}=-0.6\cdot\frac{16}{3}+1.6=-1.6<0 \Longrightarrow \text{local maximum}. \] Now evaluate the height: \[ y_{\max}= -0.1\!\left(\frac{16}{3}\right)^{3} + 0.8\!\left(\frac{16}{3}\right)^{2} = -\frac{1}{10}\cdot\frac{4096}{27} + \frac{8}{10}\cdot\frac{256}{9} = -\frac{4096}{270} + \frac{2048}{90} = \frac{1024}{135}\approx 7.585\ldots \] Therefore the maximum height is \(\boxed{7.59\text{ m}}\) (to 2 d.p.). M1 A1 M1 A1
Check it is a maximum using the second derivative: \[ \frac{d^{2}y}{dx^{2}}=-0.6x+1.6,\quad \frac{d^{2}y}{dx^{2}}\Big|_{x=16/3}=-0.6\cdot\frac{16}{3}+1.6=-1.6<0 \Longrightarrow \text{local maximum}. \] Now evaluate the height: \[ y_{\max}= -0.1\!\left(\frac{16}{3}\right)^{3} + 0.8\!\left(\frac{16}{3}\right)^{2} = -\frac{1}{10}\cdot\frac{4096}{27} + \frac{8}{10}\cdot\frac{256}{9} = -\frac{4096}{270} + \frac{2048}{90} = \frac{1024}{135}\approx 7.585\ldots \] Therefore the maximum height is \(\boxed{7.59\text{ m}}\) (to 2 d.p.). M1 A1 M1 A1
(b) Trapezoidal rule with three intervals on \([2,8]\)
Step size: \[ h=\frac{8-2}{3}=2. \] Function values (exact and decimal):
| \(x\) | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| \(y=-0.1x^3+0.8x^2\) | \(2.4\) | \(6.4\) | \(7.2\) | \(0\) |
Trapezoidal rule: \[ A \approx \frac{h}{2}\Big(y_0+2(y_1+y_2)+y_3\Big) = \frac{2}{2}\Big(2.4+2(6.4+7.2)+0\Big) = 1\cdot(2.4+12.8+14.4) = 29.6. \] Estimated cross-sectional area \(\;\boxed{29.6\text{ m}^2}\). A1 M1 A1
(c)
(i) The exact cross-sectional area is \[ A=\int_{2}^{8} y\,dx \;=\;\int_{2}^{8}\bigl(-0.1x^{3}+0.8x^{2}\bigr)\,dx. \] A1 A1
(ii) Evaluate: \[ \int\!\bigl(-0.1x^{3}+0.8x^{2}\bigr)\,dx = -\frac{1}{40}x^{4}+\frac{4}{15}x^{3}+C. \] Hence \[ A=\left[-\frac{1}{40}x^{4}+\frac{4}{15}x^{3}\right]_{2}^{8} =\left(-\frac{4096}{40}+\frac{2048}{15}\right)-\left(-\frac{16}{40}+\frac{32}{15}\right) =-\frac{512}{5}+\frac{2022}{15} =\frac{162}{5} = \boxed{32.4\text{ m}^2}. \] A2
Total Marks: 13