IB Mathematics SL 2.2 Concept of a function, domain, range and graph AI HL Paper 2- Exam Style Questions- New Syllabus
An aircraft departs from a level runway. Designate point O as the location where the aircraft starts its ascent from the runway, with x representing the horizontal distance, in kilometers, from O. The function h describes the vertical elevation, in kilometers, of the aircraft’s nose above the runway, defined as
\[ h(x) = \frac{10}{1 + 150e^{-0.07x}} – 0.06, \quad x \geq 0 \]
(a)
(i) Calculate \(h(0)\).
(ii) Explain the significance of this value in the given scenario.
[2]
(b)
(i) Identify the horizontal asymptote of the graph of \(y = h(x)\).
(ii) Explain the meaning of this value within the context.
[2]
(c) Determine \(h'(x)\) as a function of \(x\).
[4]
A safety guideline stipulates that \(h'(x)\) should not surpass 0.2.
(d) Considering the aircraft travels at least 200 km horizontally from point O, assess whether it complies with this safety guideline.
[4]
▶️ Answer/Explanation
(a)
(i) Substituting \( x = 0 \) into the function \( h(x) \).
\[ h(0) = \frac{10}{1 + 150e^{-0.07 \times 0}} – 0.06 \]
\[ h(0) = \frac{10}{1 + 150 \times 1} – 0.06 \]
\[ h(0) = \frac{10}{151} – 0.06 \]
\[ h(0) = 0.00623 \, \text{(km)} \, (= 0.00622517) \]
(ii) This is the height of the nose of the plane above the runway when the plane is on the runway.
[2 marks]
(b)
(i) Determining the horizontal asymptote by analyzing the behavior as \( x \to \infty \).
As \( x \to \infty \), \( e^{-0.07x} \to 0 \), so:
\[ h(x) \to \frac{10}{1 + 0} – 0.06 = 10 – 0.06 = 9.94 \]
The horizontal asymptote is \( y = 9.94 \).
(ii) This is the height that the nose of the plane approaches but does not reach.
[2 marks]
(c)
Method 1 (chain rule)
Expressing \( h(x) \) for differentiation.
\[ h(x) = 10(1 + 150e^{-0.07x})^{-1} – 0.06 \]
Differentiating using the chain rule:
\[ h'(x) = -10(1 + 150e^{-0.07x})^{-2} \times 150e^{-0.07x} \times (-0.07) \]
\[ h'(x) = \frac{105e^{-0.07x}}{(1 + 150e^{-0.07x})^2} \]
Method 2 (quotient rule)
Applying the quotient rule to \( h(x) = \frac{10}{1 + 150e^{-0.07x}} – 0.06 \).
\[ h'(x) = \frac{(1 + 150e^{-0.07x})(0) – 10(150e^{-0.07x} \times -0.07)}{(1 + 150e^{-0.07x})^2} \]
\[ h'(x) = \frac{-10 \times 150e^{-0.07x} \times (-0.07)}{(1 + 150e^{-0.07x})^2} \]
\[ h'(x) = \frac{105e^{-0.07x}}{(1 + 150e^{-0.07x})^2} \]
[4 marks]
(d)
Analyzing the derivative \( h'(x) \) over the interval \( x \geq 0 \) up to 200 km.
Evaluating the maximum value of \( h'(x) \).
The maximum occurs at \( x \approx 9.90 \), where \( h'(9.90) \approx 0.175 \).
Since 0.175 is less than 0.2, and this is the maximum gradient within the first 200 km, the plane is following the safety regulation.
[4 marks]
[Total: 12 marks]