# Question 6:

## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F'(x) = k(ax^2 - x^3)$ | M1 | |
| $F''(x) = k(2xa - 3x^2)$ | M1 | |
| $kx(2a - 3x) = 0$ | | |
| $a = \frac{3}{2} \times \frac{8}{3}$ or $2\times4 - 3\times\frac{8}{3} = 0$ | M1d | Dep on 1st B |
| $a = 4$* | A1cso* | |

## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F(2) = \frac{4}{15} \Rightarrow k\!\left(\frac{32}{3} - 4\right) + b = \frac{4}{15}$ or $\frac{20}{3}k + b = \frac{4}{15}$ | M1 | |
| $F(4) = 1 \Rightarrow k\!\left(\frac{256}{3} - 64\right) + b = 1$ or $\frac{64}{3}k + b = 1$ | M1 | |
| $\frac{44}{3}k = \frac{11}{15}$ | M1dd | Dep dep |
| $k = \frac{1}{20}$ or $b = -\frac{1}{15}$ | A1 | |
| $F(2.5) = \frac{1}{20}\!\left(\frac{4}{3}\times2.5^3 - \frac{2.5^4}{4}\right) + \left(-\frac{1}{15}\right)$ | M1 | |
| $= \frac{623}{1280}$ or $0.4867\ldots$ | A1cso | awrt 0.487 |

# Question (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to find $F'(x)$, $x^n \to x^{n-1}$ | M1 | Condone missing $k$. Implied by correct $F''(x)$ |
| Attempt to find $F''(x)$, $x^n \to x^{n-1}$ | M1 | Condone missing $k$ |
| Putting "their $2a - 3x'' = 0$" and substituting $x = 8/3$ | M1d | Dependent on 2nd M being awarded |
| Fully correct solution with no errors, including $k$, no incorrect notation | A1* | cso |

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# Question (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Form correct equation in terms of $k$ and $b$ using $F(2) = \frac{4}{15}$ | M1 | |
| Form correct equation in terms of $k$ and $b$ using $F(4) = 1$ | M1 | |
| Solve two equations simultaneously by eliminating either $k$ or $b$ | M1dd | Dependent on first two method marks |
| One of $k$ or $b$ correct | A1 | |

**Alternative:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use $\frac{4}{15}$ and differentiate third line $x^n \to x^{n-1}$, must have $k$ | M1 | |
| Use pdf equations with correct limits, adding and setting equal to 1 | M1 | |
| Implied by $\left[\frac{4}{15}x\right]_1^2 + k\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_2^4 = 1$ or $k\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_2^4 = \frac{11}{15}$ | NB | These first two marks implied by this expression |
| Correct integration and attempt to substitute limits | ddM1 | Dependent on previous method marks |
| $k$ correct | A1 | |
| Correct method for $F(2.5)$ using values of $k$ and $b$, or with letters $a$(or 4), $k$ and $b$ | M1 | May be implied by correct answer, otherwise working must be shown |
| $\dfrac{623}{1280}$ or awrt $0.487$ | A1cso | All previous method marks must be awarded |

**Alternative for F(2.5):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct method for $F(2.5)$ using value of $k$, or letters $a$(or 4) and $k$ | M1 | May be implied by correct answer, otherwise working must be shown |
| $\dfrac{623}{1280}$ or awrt $0.487$ | A1cso | All previous method marks must be awarded |