## Question 8:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{1}{4}x^3 - 8x^{-\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{4} \times 3x^2 - 8 \times -\frac{1}{2}x^{-\frac{3}{2}}$ | M1 | Correct method of differentiation, at least one power reduced by 1. Award for $\frac{1}{4}x^3 \to ...x^2$ or $-8x^{-\frac{1}{2}} \to ...x^{-\frac{3}{2}}$ |
| | A1 | Correct differentiation, need not be simplified |
| $\left.\frac{dy}{dx}\right|_{x=4} = ...\left(\frac{25}{2}\right)$ | M1 | Substitutes $x = 4$ into a "changed" function to find a value. Note: using "12" to find gradient scores M0 |
| $y - 12 = \frac{25}{2}(x - 4)$ | dM1 | Depends on previous M. Correct method for tangent (not normal). Uses their $\frac{dy}{dx}$ at $x = 4$ with $y = 12$ and $x = 4$ correctly placed. If using $y = mx + c$ must proceed to finding $c$ |
| $25x - 2y - 76 = 0$ or e.g. $-25x + 2y + 76 = 0$ | A1 | Correct equation in required form including "$= 0$" or any non-zero integer multiple |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{4}x^3 - 8x^{-\frac{1}{2}}\, dx = \frac{1}{4}\cdot\frac{x^4}{4} - \frac{8x^{\frac{1}{2}}}{\frac{1}{2}} (+c)$ | M1 | Attempts to integrate, look for power increased by 1 on at least one term. Award for $\frac{1}{4}x^3 \to ...x^4$ or $-\frac{1}{8}x^{-\frac{1}{2}} \to ...x^{\frac{1}{2}}$ |
| | A1 | Correct integration, need not be simplified (no need for constant for this mark) |
| $f(4) = 12 \Rightarrow 16 - 32 + c = 12 \Rightarrow c = ...(28)$ | dM1 | Depends on first M. Sets $f(4) = 12$ and proceeds to find value of $c$ — must have a constant of integration to score this mark |
| | A1 | Correct value for $c$ found |
| $f(x) = \frac{x^4}{16} - 16\sqrt{x} + 28$ | A1ft | Correct answer following through their $c$ only (algebraic part must be correct). Accept fractional index rather than square root. Accept $0.0625x^4$ for $\frac{x^4}{16}$. The "$f(x) =$" is not required. Condone poor notation e.g. leaving final answer as $f(x) = \int \frac{x^4}{16} - 16x^{\frac{1}{2}} + 28$ |

# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{1}{4}x^3 - 8x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{4} \times 3x^2 - 8 \times \frac{1}{2}x^{-\frac{1}{2}}$ | M1 | Correct method of differentiation, at least one power reduced by 1. Award for $\frac{1}{4}x^3 \to ...x^2$ or $-8x^{\frac{1}{2}} \to ...x^{-\frac{1}{2}}$ |
| Correct differentiation (need not be simplified) | A1 | Ignore any spurious "$= 0$" |
| $\frac{dy}{dx}\bigg|_{x=4} = ...(10)$ | M1 | Finds $\frac{dy}{dx}$ at $x=4$. Requires substitution of $x=4$ into a "changed" function. Note: using "12" to find gradient e.g. $12 = \frac{1}{4}\times3(4)^2 - 8\times\frac{1}{2}(4)^{-\frac{1}{2}} + c \Rightarrow c=2 \Rightarrow m=2$ scores M0 |
| $y - 12 = \text{"10"}(x-4)$ | dM1 | Depends on previous M. Correct method for tangent **not** normal. Uses their $\frac{dy}{dx}$ at $x=4$ with $y=12$ and $x=4$ correctly placed. If using $y=mx+c$ must proceed to finding $c$ |
| $10x - y - 28 = 0$ oe e.g. $-10x + y + 28 = 0$ | A1 | Correct equation in required form including "$= 0$" or any non-zero integer multiple |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = \int \frac{1}{4}x^3 - 8x^{\frac{1}{2}}\, dx = \frac{1}{4}\cdot\frac{x^4}{4} - \frac{8x^{\frac{3}{2}}}{\frac{3}{2}} (+c)$ | M1 | Attempts to integrate $f(x)$, look for power increased by 1 on at least one term. Award for $\frac{1}{4}x^3 \to ...x^4$ or $-\frac{1}{8}x^{\frac{1}{2}} \to ...x^{\frac{3}{2}}$ |
| Correct integration (need not be simplified, no need for constant for this mark) | A1 | |
| $f(4) = 12 \Rightarrow 16 - \frac{128}{3} + c = 12 \Rightarrow c = ...\left(\frac{116}{3}\right)$ | dM1 | Depends on first M. Sets $f(4)=12$ and proceeds to find value of $c$ — must have a constant of integration |
| Correct value of $c$ found | A1 | |
| $f(x) = \frac{x^4}{16} - \frac{16}{3}x^{\frac{3}{2}} + \text{"}\frac{116}{3}\text{"}$ | A1ft | Correct answer following through their $c$ only; algebraic part must be correct. Accept $0.0625x^4$ for $\frac{x^4}{16}$. Allow equivalent mixed fractions for $\frac{16}{3}$ and $\frac{116}{3}$. Depends on previous method mark. Condone poor notation e.g. leaving final answer as $f(x) = \int \frac{x^4}{16} - \frac{16}{3}x^{\frac{3}{2}} + \frac{116}{3}$ |

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