# Question 15:

## Part (b) - Finding values of k where circles touch:

**Main Scheme (Distance/Geometric approach):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct limits found | A1 | Allow decimals to 3sf e.g. awrt 4.06 and awrt 14.1 |
| $\{k : k \in \mathbb{R}, \sqrt{82}-5 < k < \sqrt{82}+5\}$ | A1 | Correct answer with exact values using set notation. Allow equivalent forms e.g. $\{k : \sqrt{82}-5 < k < \sqrt{82}+5\}$, $k \in (\sqrt{82}-5, \sqrt{82}+5)$. Allow "\|" for ":" and allow the "$k$:" or "$k \in$" to be missing. **Do not** allow $\sqrt{82}-5 < k < \sqrt{82}+5$ (not in set notation) |

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## Part (b) - Algebraic Approach:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Solves 2 circle equations simultaneously to find $y$ in terms of $x$ and $k$, or $x$ in terms of $y$ and $k$; substitutes; attempts $b^2-4ac=0$ | M1 | Valid strategy required across all three bullet points |
| $k^4 - 214k^2 + 3249 = 0$ (correct simplified 3TQ in $k^2$) | A1 | |
| Solves their 3TQ in $k^2$ by any correct method including calculator | dM1 | |
| $k = 5+\sqrt{82},\ -5+\sqrt{82}$ (both correct, exact or decimals) | A1 | May have extra values which can be ignored |
| $\{k : \sqrt{82}-5 < k < \sqrt{82}+5\}$ | A1 | Exact values in set notation as main scheme |

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## Part (b) - Implicit Differentiation Approach:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiates both circles implicitly, equates derivatives, substitutes into $C_1$ to get equation in one variable | M1 | Valid strategy across all bullet points |
| $82y^2 + 1148y + 3993 = 0$ or $82x^2 - 492x - 1287 = 0$ (correct 3TQ in $y$ or $x$) | A1 | |
| Solves 3TQ, finds at least one intersection point, substitutes into $C_2$ for $k$ | dM1 | |
| $k = 5+\sqrt{82},\ -5+\sqrt{82}$ | A1 | Exact or decimals as main scheme |
| $\{k : \sqrt{82}-5 < k < \sqrt{82}+5\}$ | A1 | Exact values in set notation |

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## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3(x+y)^2\left(1+\frac{dy}{dx}\right) = 6x - 3\frac{dy}{dx}$ | M1, A1, A1 | AOs 3.1a, 1.1b, 1.1b. Award M1 for $(x+y)^3 \to k(x+y)^2\left(\lambda + \frac{dy}{dx}\right)$ where $\lambda$ is 1, $x$ or 0. A1 for either $3(x+y)^2\left(1+\frac{dy}{dx}\right)$ **or** $6x-3\frac{dy}{dx}$. Second A1 for both seen/equated |
| $\left(3(x+y)^2+3\right)\frac{dy}{dx} = 6x - 3(x+y)^2 \Rightarrow \frac{dy}{dx} = \ldots$ | M1 | AO 2.1. Valid attempt to make $\frac{dy}{dx}$ subject with exactly 2 **different** $\frac{dy}{dx}$ terms |
| $\frac{dy}{dx} = \frac{6x-3(x+y)^2}{3(x+y)^2+3}$ | A1 | AO 1.1b. Fully correct expression. Allow equivalent correct forms |

**Alternative (expanding $(x+y)^3$ first):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 + 3x^2\frac{dy}{dx} + 6xy + 6xy\frac{dy}{dx} + 3y^2 + 3y^2\frac{dy}{dx} = 6x - 3\frac{dy}{dx}$ | M1, A1, A1 | AOs 3.1a, 1.1b, 1.1b |
| $\left(3x^2+6xy+3y^2+3\right)\frac{dy}{dx} = 6x-3x^2-6xy-3y^2$ | M1 | AO 2.1. Valid attempt with exactly 4 **different** $\frac{dy}{dx}$ terms |
| $\frac{dy}{dx} = \frac{6x-3x^2-6xy-3y^2}{3x^2+6xy+3y^2+3}$ | A1 | AO 1.1b |

# Mark Scheme Extraction

## Part (a) - Implicit Differentiation

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\frac{dy}{dx}$ and $-8\frac{dy}{dx}$ terms present | M1 | Look for $(\ldots \pm \ldots \pm \ldots \pm \ldots)\frac{dy}{dx} = \ldots \Rightarrow \frac{dy}{dx} = \ldots$ implied by working |
| Fully correct expression for $\frac{dy}{dx}$ | A1 | Allow equivalent correct forms; condone e.g. $3x2y$ for $6xy$ |

**Alternative - making y the subject:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(3x^2-3y-2)^{\frac{1}{3}} \rightarrow \ldots(3x^2-3y-2)^{-\frac{2}{3}}$ or $3x^2-3y-2 \rightarrow \ldots x + \ldots\frac{dy}{dx}$ | M1 | Cube roots both sides, makes $x+y$ or $y$ the subject |
| $\frac{1}{3}(3x^2-3y-2)^{-\frac{2}{3}}$ **or** $6x-3\frac{dy}{dx}$ | A1 | |
| Fully correct $\frac{dy}{dx} = \frac{2x(3x^2-3y-2)^{-\frac{2}{3}}-1}{1+(3x^2-3y-2)^{-\frac{2}{3}}}$ | A1 | |
| Valid attempt to make $\frac{dy}{dx}$ subject with exactly 2 **different** terms in $\frac{dy}{dx}$ | M1 | |
| Correct expression | A1 | |

**Alternative - using partial derivatives:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct structure for either partial derivative; doesn't expand: $\frac{\partial f}{\partial x} = \ldots(x+y)^2 + \ldots x$ or $\frac{\partial f}{\partial y} = \ldots(x+y)^2 + \ldots$; or expands: $\frac{\partial f}{\partial x} = \ldots x^2 + \ldots xy + \ldots y^2 + \ldots x$ | M1 | Where "$\ldots$" are non-zero constants |
| Correct $\frac{\partial f}{\partial x}$ **or** correct $\frac{\partial f}{\partial y}$ | A1 | |
| Correct $\frac{\partial f}{\partial x}$ **and** correct $\frac{\partial f}{\partial y}$ | A1 | |
| Attempts $\frac{dy}{dx} = -\frac{\partial f}{\partial x} \div \frac{\partial f}{\partial y}$ | M1 | |
| Correct expression e.g. $\frac{dy}{dx} = \frac{6x-3(x+y)^2}{3(x+y)^2+3}$ | A1 | |

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## Part (b) - Normal at Point

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{6(1)-3(0+1)^2}{3(0+1)^2+3} = \frac{1}{2}$, then uses negative reciprocal and point $(1,0)$ | M1 | Substitutes $x=1$, $y=0$ into $\frac{dy}{dx}$, obtains tangent gradient, uses negative reciprocal with correct straight line method |
| $y = -2x + 2$ | A1* | Correct equation following a correct $\frac{dy}{dx}$ from part (a) |
| **(2 marks)** | | Note: gradient $\frac{1}{2}$ could be deduced from given equation — check solutions carefully |

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## Part (c) - Normal Meets Curve Again

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $y = -2x+2$ into curve equation to get equation in one variable | M1 | Uses equation from part (a), substitutes $y = \pm 2x \pm 2$ or $x = \frac{\pm 2 \pm y}{2}$ |
| $x^3 - 3x^2 + 18x - 16 = 0$ **or** $y^3 + 60y = 0$ | A1 | Correct cubic with terms collected and "$=0$" seen or implied |
| Uses $(x-1)$ or $y$ as factor to establish quadratic: $(x-1)(x^2-2x+16)=0$ or $y(y^2+60)=0$ | dM1 | Must attempt to use factor $(x-1)$ with cubic in $x$, or factor of $y$ with cubic in $y$ |
| For $x^2-2x+16=0$: $b^2-4ac = 4-4\times1\times16$ **or** for $y^2+60=0$: $y^2 \neq -60$ | ddM1 | Requires: correct cubic, correct quadratic factor, attempt to show no real roots |
| As $b^2-4ac < 0$ or $y^2 \neq -60$, there are no other real roots so the normal does not meet $C$ again | A1 | Fully correct argument with justification and conclusion |
| **(5 marks)** | | Do **not** accept "gives a math error so they do not meet again" without justification |

# Alternative to (c) by showing the cubic is strictly increasing (or decreasing):

**M1A1:** | As in the main scheme | M1A1 | As in the main scheme then

$$f(x) = x^3 - 3x^2 + 18x - 16 \Rightarrow f'(x) = 3x^2 - 6x + 18$$ | M1A1 |

$$3x^2 - 6x + 18 = 3(x^2 - 2x + 6) = 3(x-1)^2 + 15$$ | dM1 | Differentiates their cubic of the form $ax^3 + bx^2 + cx + d = 0$, $a,b,c,d \neq 0$ to obtain a 3 term quadratic expression with only coefficient errors on the non-constant terms.

$$3(x-1)^2 + 15 > 0$$ so $f(x)$ is an increasing function | ddM1 | This mark requires:
- a correct cubic equation in $x$
- the correct derivative or a multiple of it
- an attempt to show that the quadratic expression is always positive (or negative)

Hence there can only be one intersection (at $x = 1$) so the normal and curve do not intersect again. | A1 | Fully correct concluding argument e.g. that as the derivative is always positive (or always negative) the function is strictly increasing (or decreasing) and therefore there can only be one intersection (at $x = 1$) so the normal and curve do not meet again.

**(12 marks)**