| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Determine increasing/decreasing intervals |
| Difficulty | Standard +0.3 This is a standard C3 question requiring quotient rule differentiation, solving a quadratic inequality, and applying Simpson's rule. All three parts are routine applications of techniques with no novel insight required, though the multi-step nature and algebraic manipulation place it slightly above average difficulty. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(f'(x) = \frac{2x\times(4x+1)-(x^2+3)\times4}{(4x+1)^2}\) | M1 A1 | |
| \(= \frac{4x^2+2x-12}{(4x+1)^2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(\frac{4x^2+2x-12}{(4x+1)^2} \geq 0\) | ||
| for \(x \neq -\frac{1}{4}\), \((4x+1)^2 > 0\) \(\therefore\) \(4x^2+2x-12 \geq 0\) | M1 A1 | |
| \(2(2x-3)(x+2) \geq 0\) | M1 | |
| \(x \leq -2\) or \(x \geq \frac{3}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Notes |
| \(x\): \(0, 1, 2, 3, 4, 5, 6\); \(f(x)\): \(3, \frac{4}{5}, \frac{7}{9}, \frac{12}{13}, \frac{19}{17}, \frac{28}{21}, \frac{39}{25}\) | M1 | |
| \(I \approx \frac{1}{3}\times 1\times\left[3 + \frac{39}{25} + 4\left(\frac{4}{5}+\frac{12}{13}+\frac{28}{21}\right) + 2\left(\frac{7}{9}+\frac{19}{17}\right)\right]\) | M1 | |
| \(= 6.86\) (3sf) | A1 | (10) |
# Question 7:
## Part (i)
| Answer | Mark | Notes |
|--------|------|-------|
| $f'(x) = \frac{2x\times(4x+1)-(x^2+3)\times4}{(4x+1)^2}$ | M1 A1 | |
| $= \frac{4x^2+2x-12}{(4x+1)^2}$ | A1 | |
## Part (ii)
| Answer | Mark | Notes |
|--------|------|-------|
| $\frac{4x^2+2x-12}{(4x+1)^2} \geq 0$ | | |
| for $x \neq -\frac{1}{4}$, $(4x+1)^2 > 0$ $\therefore$ $4x^2+2x-12 \geq 0$ | M1 A1 | |
| $2(2x-3)(x+2) \geq 0$ | M1 | |
| $x \leq -2$ or $x \geq \frac{3}{2}$ | A1 | |
## Part (iii)
| Answer | Mark | Notes |
|--------|------|-------|
| $x$: $0, 1, 2, 3, 4, 5, 6$; $f(x)$: $3, \frac{4}{5}, \frac{7}{9}, \frac{12}{13}, \frac{19}{17}, \frac{28}{21}, \frac{39}{25}$ | M1 | |
| $I \approx \frac{1}{3}\times 1\times\left[3 + \frac{39}{25} + 4\left(\frac{4}{5}+\frac{12}{13}+\frac{28}{21}\right) + 2\left(\frac{7}{9}+\frac{19}{17}\right)\right]$ | M1 | |
| $= 6.86$ (3sf) | A1 | **(10)** |
---7.
$$\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { 4 x + 1 } , \quad x \in \mathbb { R } , \quad x \neq - \frac { 1 } { 4 }$$
(i) Find and simplify an expression for $\mathrm { f } ^ { \prime } ( x )$.\\
(ii) Find the set of values of $x$ for which $\mathrm { f } ( x )$ is increasing.\\
(iii) Use Simpson's rule with six strips to find an approximate value for
$$\int _ { 0 } ^ { 6 } f ( x ) d x$$
\hfill \mbox{\textit{OCR C3 Q7 [10]}}