# Question 5:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dx}{dy} = 2\times 3\sec 3y \cdot \sec 3y\tan 3y = 6\sec^2 3y\tan 3y$ $\left(\text{oe } \dfrac{6\sin 3y}{\cos^3 3y}\right)$ | M1A1 | |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}}$ to obtain $\dfrac{dy}{dx} = \dfrac{1}{6\sec^2 3y\tan 3y}$ | M1 | |
| $\tan^2 3y = \sec^2 3y - 1 = x - 1$ | B1 | |
| Uses $\sec^2 3y = x$ and $\tan^2 3y = x-1$ to get $\dfrac{dy}{dx}$ or $\dfrac{dx}{dy}$ in just $x$ | M1 | |
| $\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{6x(x-1)^{\frac{1}{2}}}$ | A1* (CSO) | Given answer — all steps must be shown |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{d^2y}{dx^2} = \dfrac{0 - [6(x-1)^{\frac{1}{2}} + 3x(x-1)^{-\frac{1}{2}}]}{36x^2(x-1)}$ | M1A1 | |
| $\dfrac{d^2y}{dx^2} = \dfrac{6-9x}{36x^2(x-1)^{\frac{3}{2}}} = \dfrac{2-3x}{12x^2(x-1)^{\frac{3}{2}}}$ | dM1A1 | |

## Question 5:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses chain rule to get $A\sec 3y \sec 3y \tan 3y = (A\sec^2 3y \tan 3y)$ | M1 | Alternatively chain rule on $(\cos 3y)^{-2} \Rightarrow A(\cos 3y)^{-3}\sin 3y$, or quotient rule on $\frac{1}{(\cos 3y)^2} \Rightarrow \frac{\pm A\cos 3y \sin 3y}{(\cos 3y)^4}$ |
| $\frac{dx}{dy} = 2 \times 3\sec 3y \sec 3y \tan 3y$ or equivalent | A1 | Both sides must be correct. No need to simplify rhs |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ to get expression for $\frac{dy}{dx}$ | M1 | Follow through on their $\frac{dx}{dy}$. Allow slips on coefficient but not trig expression |
| Writes $\tan^2 3y = \sec^2 3y - 1$ or equivalent such as $\tan 3y = \sqrt{\sec^2 3y - 1}$, uses $x = \sec^2 3y$ to obtain $\tan^2 3y = x-1$ or $\tan 3y = (x-1)^{\frac{1}{2}}$ | B1 | All elements must be present. Accept $\frac{\sqrt{x}}{3y}$ with $\sqrt{x-1}$, $\cos 3y = \frac{1}{\sqrt{x}} \Rightarrow \tan 3y = \sqrt{x-1}$. If differential in terms of $\sin 3y, \cos 3y$, awarded for $\sin 3y = \frac{\sqrt{x-1}}{\sqrt{x}}$ |
| Uses $\sec^2 3y = x$ and $\tan^2 3y = \sec^2 3y - 1 = x-1$ or equivalent to get $\frac{dy}{dx}$ in just $x$ | M1 | Allow slips on signs in $\tan^2 3y = \sec^2 3y - 1$. May be implied |
| Correct simplified final answer (given/CSO) | A1* | All steps must be shown |

**Note:** Example scoring M1, B0, M1, A0:
$$\frac{dx}{dy} = 6\sec^2 3y \tan 3y = 6x(x-1)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{6x(x-1)^{\frac{1}{2}}}$$

### Part (c) — Using Quotient and Product Rules:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses quotient rule $\frac{vu'-uv'}{v^2}$ with $u=1$, $v=6x(x-1)^{\frac{1}{2}}$, achieving $u'=0$ and $v' = A(x-1)^{\frac{1}{2}} + Bx(x-1)^{-\frac{1}{2}}$ | M1 | If formulae quoted, both must be correct. If not quoted, allow forms: $\frac{0-[A(x-1)^{\frac{1}{2}}+Bx(x-1)^{-\frac{1}{2}}]}{(6x(x-1)^{\frac{1}{2}})^2}$ or $\frac{0-A(x-1)^{\frac{1}{2}} \pm Bx(x-1)^{-\frac{1}{2}}}{Cx^2(x-1)}$ |
| $\frac{d^2y}{dx^2} = \frac{0-[6(x-1)^{\frac{1}{2}}+3x(x-1)^{-\frac{1}{2}}]}{36x^2(x-1)}$ or equivalent | A1 | Correct unsimplified expression |
| Multiply numerator and denominator by $(x-1)^{\frac{1}{2}}$ producing a linear numerator, simplified by collecting like terms. Alternatively factor out $(x-1)^{-\frac{1}{2}}$ from numerator | dM1 | Dependent on 1st M1 |
| $\frac{d^2y}{dx^2} = \frac{2-3x}{12x^2(x-1)^{\frac{3}{2}}}$ or equivalent | A1 | — |

### Part (c) — Using Product and Chain Rules:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\frac{dy}{dx} = \frac{1}{6x(x-1)^{\frac{1}{2}}} = Ax^{-1}(x-1)^{-\frac{1}{2}}$ and uses product rule with $u$ or $v = Ax^{-1}$ and $v$ or $u = (x-1)^{-\frac{1}{2}}$ | M1 | If rule quoted it must be correct. If not quoted, award for expression of form $(x-1)^{-\frac{3}{2}} \times Bx^{-1} \pm C(x-1)^{-\frac{1}{2}} \times x^{-2}$ |
| $\frac{d^2y}{dx^2} = \frac{1}{6}[x^{-1}(-\frac{1}{2})(x-1)^{-\frac{3}{2}} + (-1)x^{-2}(x-1)^{-\frac{1}{2}}]$ | A1 | — |
| Factorise/use common denominator of $x^{-2}(x-1)^{-\frac{3}{2}}$ producing linear factor/numerator, collecting like terms, single fraction | dM1 | — |
| $\frac{d^2y}{dx^2} = \frac{1}{12}x^{-2}(x-1)^{-\frac{3}{2}}[2-3x]$ or equivalent | A1 | — |

### Part (c) — Using Quotient and Chain Rules:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses quotient rule with $u=(x-1)^{-\frac{1}{2}}$ and $v=6x$, achieving $u'=A(x-1)^{-\frac{3}{2}}$ and $v'=B$ | M1 | If formula quoted must be correct. If not quoted allow form $\frac{Cx(x-1)^{-\frac{3}{2}}-D(x-1)^{-\frac{1}{2}}}{Ex^2}$ |
| $\frac{d^2y}{dx^2} = \frac{6x \times (-\frac{1}{2})(x-1)^{-\frac{3}{2}} - (x-1)^{-\frac{1}{2}} \times 6}{(6x)^2}$ | A1 | Correct unsimplified |
| Multiply numerator and denominator by $(x-1)^{\frac{3}{2}}$ producing linear numerator, collecting like terms. Alternatively factor $(x-1)^{-\frac{3}{2}}$ from numerator | dM1 | Dependent on 1st M1 |
| $\frac{d^2y}{dx^2} = \frac{2-3x}{12x^2(x-1)^{\frac{3}{2}}}$ or $\frac{(2-3x)x^{-2}(x-1)^{-\frac{3}{2}}}{12}$ | A1 | — |

### Part (c) — Using just the Chain Rule:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\frac{dy}{dx} = \frac{1}{6x(x-1)^{\frac{1}{2}}} = \frac{1}{(36x^3-36x^2)^{\frac{1}{2}}} = (36x^3-36x^2)^{-\frac{1}{2}}$ and proceeds by chain rule to $A(36x^3-36x^2)^{-\frac{3}{2}}(Bx^2-Cx)$ | M1 | — |
| Would automatically follow under this method if first M scored | M1 | — |

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