Covers Lessons 11–15: differentiating $\ln x$ and $\log_a x$, exponential growth and decay, optimisation with exponentials, and module synthesis.
Assessment
Select the best answer for each question.
$\frac{d}{dx}(\ln x)$ equals:
$\frac{d}{dx}(\log_{10} x)$ equals:
The half-life of a substance with decay constant $k = -0.05$ is:
The stationary point of $f(x) = xe^{-x}$ occurs at:
$\frac{d}{dx}(\ln(x^2 + 1))$ equals:
A population doubles every 4 hours. The growth constant $k$ is:
$\frac{d}{dx}(x \ln x)$ equals:
The maximum value of $f(x) = xe^{-x}$ for $x \geq 0$ is:
Short Answer
Differentiate $f(x) = \frac{\ln x}{x}$, showing all working and simplifying.
A radioactive substance decays from 80 g to 20 g in 12 days. Assuming exponential decay, find the decay constant $k$ and the half-life.
Find and classify the stationary point of $f(x) = x^2 e^{-x}$ for $x \geq 0$.
Q1: B — $\frac{d}{dx}(\ln x) = \frac{1}{x}$.
Q2: C — $\frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln 10}$.
Q3: B — $T_{1/2} = \frac{\ln 2}{|k|} = \frac{\ln 2}{0.05}$.
Q4: B — $f'(x) = e^{-x}(1-x) = 0 \Rightarrow x = 1$.
Q5: B — $\frac{d}{dx}(\ln(x^2+1)) = \frac{2x}{x^2+1}$ by chain rule.
Q6: B — $T_{double} = \frac{\ln 2}{k}$, so $k = \frac{\ln 2}{4}$.
Q7: C — Product rule: $1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Q8: C — Maximum at $x = 1$ where $f(1) = 1 \cdot e^{-1} = \frac{1}{e}$.
Q9 (3 marks): $u = \ln x$, $v = x$ [0.5]. $u' = \frac{1}{x}$, $v' = 1$ [0.5]. $f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2}$ [0.5] $= \frac{1 - \ln x}{x^2}$ [1.5].
Q10 (3 marks): $20 = 80e^{12k}$ [0.5]. $\frac{1}{4} = e^{12k}$ [0.5]. $12k = \ln(\frac{1}{4}) = -\ln 4 = -2\ln 2$ [0.5]. $k = -\frac{\ln 2}{6} \approx -0.116$ per day [0.5]. $T_{1/2} = \frac{\ln 2}{|k|} = \frac{\ln 2}{\ln 2 / 6} = 6$ days [1].
Q11 (3 marks): $f'(x) = 2xe^{-x} - x^2e^{-x}$ [0.5] $= xe^{-x}(2-x)$ [0.5]. For $x \geq 0$: $f'(x) = 0$ when $x = 0$ or $x = 2$ [0.5]. At $x = 0$: $f(0) = 0$ (minimum on domain) [0.5]. At $x = 2$: $f(2) = 4e^{-2} \approx 0.541$. Second derivative: $f''(x) = e^{-x}(x^2 - 4x + 2)$ [0.5]. $f''(2) = e^{-2}(4 - 8 + 2) = -2e^{-2} < 0$, so local maximum [0.5].