Moderate -0.3 This is a straightforward application of the quotient rule to find f'(x), followed by showing the derivative is always positive. The algebra simplifies nicely (numerator becomes e^x which is always positive), making this slightly easier than average but still requiring proper technique and sign analysis.
10 Show that \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 1 + \mathrm { e } ^ { x } }\) is an increasing function for all values of \(x\).
\(f'(x) = \frac{(1+e^x)e^x - e^x e^x}{(1+e^x)^2}\) or \(f'(x) = \frac{e^x}{1+e^x} - \frac{e^{2x}}{(1+e^x)^2}\) from product rule
M1, A1
1.1a, 1.1
\(f'(x) = \frac{e^x}{(1+e^x)^2}\)
M1
2.1
\(f'(x) > 0 \Rightarrow f(x)\) is an increasing function for all \(x\)
E1
2.4
[4]
## Question 10:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{(1+e^x)e^x - e^x e^x}{(1+e^x)^2}$ or $f'(x) = \frac{e^x}{1+e^x} - \frac{e^{2x}}{(1+e^x)^2}$ from product rule | M1, A1 | 1.1a, 1.1 | Use of quotient rule or product rule (allow one error); Correct derivative |
| $f'(x) = \frac{e^x}{(1+e^x)^2}$ | M1 | 2.1 | Simplifying derivative to a form which shows it is positive |
| $f'(x) > 0 \Rightarrow f(x)$ is an increasing function for all $x$ | E1 | 2.4 | All correct and clear completion |
| **[4]** | | |
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10 Show that $\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 1 + \mathrm { e } ^ { x } }$ is an increasing function for all values of $x$.
\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q10 [4]}}