IB Mathematics SL 5.2 Increasing and decreasing function AI SL Paper 1- Exam Style Questions- New Syllabus
Question
Consider the graph of the following function:
\[f(x)=\dfrac{16}{x}+4\ln|x|\,,\quad \text{for } x\neq 0.\]
(a) Write down the equation of the vertical asymptote of \(f(x)\). [1]
(b) Find \(f'(x)\). [3]
(c) Write down the interval in which \(f(x)\) is increasing. [2]
▶️Answer/Explanation
Markscheme
(a)
\(x=0\). A1
Reason: as \(x\to 0^\pm\), \(\dfrac{16}{x}\) and \(\ln|x|\) diverge, so there is a vertical asymptote at \(x=0\).
Reason: as \(x\to 0^\pm\), \(\dfrac{16}{x}\) and \(\ln|x|\) diverge, so there is a vertical asymptote at \(x=0\).
(b)
Differentiate term-by-term:
\(\displaystyle \frac{d}{dx}\!\left(\frac{16}{x}\right) = -\frac{16}{x^2}\) A1
\(\displaystyle \frac{d}{dx}\!\big(4\ln|x|\big) = \frac{4}{x}\) (valid for \(x\ne 0\)). A1
Therefore \(\displaystyle f'(x) = -\frac{16}{x^2}+\frac{4}{x} = \frac{4(x-4)}{x^2}\). A1
\(\displaystyle \frac{d}{dx}\!\left(\frac{16}{x}\right) = -\frac{16}{x^2}\) A1
\(\displaystyle \frac{d}{dx}\!\big(4\ln|x|\big) = \frac{4}{x}\) (valid for \(x\ne 0\)). A1
Therefore \(\displaystyle f'(x) = -\frac{16}{x^2}+\frac{4}{x} = \frac{4(x-4)}{x^2}\). A1
(c)
\(f(x)\) is increasing where \(f'(x)>0\). Using \(f'(x)=\dfrac{4(x-4)}{x^2}\), note \(x^2>0\) for \(x\ne 0\).
Hence \(f'(x)>0 \iff x-4>0 \iff x>4\).
Interval: \(\boxed{(4,\infty)}\). A1 A1
Hence \(f'(x)>0 \iff x-4>0 \iff x>4\).
Interval: \(\boxed{(4,\infty)}\). A1 A1
Total Marks: 6
Question
Ava manages a company whose profit per year was found to be changing at a rate of \( \frac{dP}{dt} = 3t^2 – 8t \), where \( P \) is the company’s profit in thousands of dollars and \( t \) is the time since the company was founded, measured in years.
(a) Determine whether the profit is increasing or decreasing when \( t = 2 \). [2]
One year after the company was founded, the profit was 4 thousand dollars.
(b) Find an expression for \( P(t) \), when \( t \geq 0 \). [4]
▶️ Answer/Explanation
Markscheme
(a)
Evaluate the rate of change at \( t = 2 \):
\[ \begin{aligned} \frac{dP}{dt} &= 3t^2 – 8t \\ \text{At } t = 2: \quad \frac{dP}{dt} &= 3 \times 2^2 – 8 \times 2 \\ &= 3 \times 4 – 16 = 12 – 16 = -4 \end{aligned} \]
Since \( \frac{dP}{dt} = -4 < 0 \), the profit is decreasing. M1 A1
[2 marks]
Evaluate the rate of change at \( t = 2 \):
\[ \begin{aligned} \frac{dP}{dt} &= 3t^2 – 8t \\ \text{At } t = 2: \quad \frac{dP}{dt} &= 3 \times 2^2 – 8 \times 2 \\ &= 3 \times 4 – 16 = 12 – 16 = -4 \end{aligned} \]
Since \( \frac{dP}{dt} = -4 < 0 \), the profit is decreasing. M1 A1
[2 marks]
(b)
Integrate \( \frac{dP}{dt} \):
\[ \begin{aligned} \frac{dP}{dt} &= 3t^2 – 8t \\ P(t) &= \int (3t^2 – 8t) \, dt \\ &= 3 \cdot \frac{t^3}{3} – 8 \cdot \frac{t^2}{2} + c \\ &= t^3 – 4t^2 + c \end{aligned} \] A1 A1
Use the condition \( P(1) = 4 \):
\[ \begin{aligned} P(1) &= 1^3 – 4 \times 1^2 + c = 4 \\ 1 – 4 + c &= 4 \\ c &= 7 \end{aligned} \] (M1)
Thus, \( P(t) = t^3 – 4t^2 + 7 \). A1
[4 marks]
Integrate \( \frac{dP}{dt} \):
\[ \begin{aligned} \frac{dP}{dt} &= 3t^2 – 8t \\ P(t) &= \int (3t^2 – 8t) \, dt \\ &= 3 \cdot \frac{t^3}{3} – 8 \cdot \frac{t^2}{2} + c \\ &= t^3 – 4t^2 + c \end{aligned} \] A1 A1
Use the condition \( P(1) = 4 \):
\[ \begin{aligned} P(1) &= 1^3 – 4 \times 1^2 + c = 4 \\ 1 – 4 + c &= 4 \\ c &= 7 \end{aligned} \] (M1)
Thus, \( P(t) = t^3 – 4t^2 + 7 \). A1
[4 marks]
Total Marks: 6