| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Topic | Tangents, normals and gradients |
| Type | Increasing/decreasing intervals |
| Difficulty | Moderate -0.3 This is a straightforward calculus question requiring the quotient rule to find f'(x) and showing f'(x) < 0 for x > 1. While it involves algebraic manipulation after differentiation, it's a standard technique with no novel insight required, making it slightly easier than average. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
**Question 8:**
$$f'(x) = \frac{2x(3x^2-1)-x^2(6x)}{(3x^2-1)^2} = \frac{-2x}{(3x^2-1)^2}$$
For $x>0$, $-2x<0$ and $(\ldots)^2>0$ and $\frac{-\text{ve}}{+\text{ve}}<0$, hence decreasing function.
- **M1**: Attempt use of quotient rule, or equivalent
- **A1**: Correct unsimplified expression
- **A1**: Correct simplified expression
- **M1**: Identify that $f'(x)<0$ is required; allow 'gradient' for $f'(x)$
- **A1**: Show convincingly that the denominator is always positive and the numerator is always negative for $x>1$, and hence $f'(x)<0$
Graphical solutions could get M1 for $f'(x)<0$ is required, but need to show no stationary points to get any further credit.
**Total: [5]**8 The function f is given by $\mathrm { f } ( x ) = \frac { x ^ { 2 } } { 3 x ^ { 2 } - 1 }$, for $x > 1$. Show that f is a decreasing function.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2015 Q8 [5]}}