# Question 3:

## Part (a):

**Way 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1+\left(\sqrt{x^2-1}\right)^2}} \times \frac{1}{2}(x^2-1)^{-\frac{1}{2}}(2x)$ | M1 A1 | M1: Obtains $\frac{1}{\sqrt{1+\left(\sqrt{x^2-1}\right)^2}} \times f(x)$ or e.g., $\frac{1}{x}\times f(x)$, $f(x)\neq k$. A1: Fully correct unsimplified expression |
| $= \frac{1}{\sqrt{1+x^2-1}}\times\frac{x}{\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}$* or e.g., $= \frac{1}{x}\times\frac{x}{\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}$* | A1* | Correct completion with intermediate line of working and no errors |

**Way 2 (Takes sinh of both sides):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arsinh}\left(\sqrt{x^2-1}\right) \Rightarrow \sinh y = \sqrt{x^2-1} \Rightarrow \cosh y\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(x^2-1)^{-\frac{1}{2}}(2x)$ | M1 A1 | M1: Takes sinh of both sides and differentiates to obtain $\cosh y\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)$, $f(x)\neq k$. A1: Fully correct unsimplified equation |
| $\cosh y = \sqrt{1+\sinh^2 y}$ or $\sqrt{1+x^2-1} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{x^2-1}}$* | A1* | Correct completion with clear use of identity (must see more than just $\cosh y = x$) and no errors |

**Way 3 (Takes sinh & squares):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arsinh}\left(\sqrt{x^2-1}\right) \Rightarrow \sinh y = \sqrt{x^2-1} \Rightarrow \sinh^2 y = x^2-1 \Rightarrow 2\sinh y\cosh y\frac{\mathrm{d}y}{\mathrm{d}x} = 2x$ | M1 A1 | M1: Takes sinh of both sides, squares and differentiates to obtain $c\sinh y\cosh y\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)$, $f(x)\neq k$. A1: Fully correct unsimplified expression or equation |
| $\cosh y = \sqrt{1+\sinh^2 y}$ or $\sqrt{1+x^2-1} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{x^2-1}}$* | A1* | Correct completion with clear use of identity (must see more than just $\cosh y = x$) and no errors |

**Way 4 (Takes sinh & squares & uses identity):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Rightarrow \sinh y = \sqrt{x^2-1} \Rightarrow \sinh^2 y = x^2-1 \Rightarrow \cosh^2 y = 1+(x^2-1) \Rightarrow \cosh^2 y = x^2 \Rightarrow 2\sinh y\cosh y\frac{\mathrm{d}y}{\mathrm{d}x} = 2x$ | M1 A1 | M1: Takes sinh of both sides, squares, uses identity and differentiates to obtain $c\sinh y\cosh y\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)$, $f(x)\neq k$. Allow sign errors with identity for the M mark. A1: Fully correct unsimplified expression or equation |
| $\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{x^2-1}}$ | A1* | Correct completion with clear use of identity and no errors |

**Way 5 (Takes sinh & squares & uses identity & roots):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Rightarrow \sinh y = \sqrt{x^2-1} \Rightarrow \sinh^2 y = x^2-1 \Rightarrow \cosh^2 y = 1+(x^2-1) \Rightarrow \cosh y = x \Rightarrow \sinh y\frac{\mathrm{d}y}{\mathrm{d}x} = 1$ | M1 A1 | M1: Takes sinh of both sides, squares, uses identity, roots and differentiates to obtain $c\sinh y\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)$ or $k$. Allow sign errors with identity. A1: Fully correct unsimplified expression or equation |
| $\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{x^2-1}}$ | A1* | Correct completion with clear use of identity and no errors |

**Way 6 (Uses log form of arsinh first):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{arsinh}\left(\sqrt{x^2-1}\right) \Rightarrow y = \ln\left(\sqrt{x^2-1}+\sqrt{x^2-1+1}\right) = \ln\left(\sqrt{x^2-1}+x\right) \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{1}{2}(x^2-1)^{-\frac{1}{2}}(2x)+1}{\sqrt{x^2-1}+x}$ | M1 A1 | M1: Use log form of arsinh correctly and differentiates to obtain $\frac{f(x)\neq k}{\sqrt{x^2-1}+x}$. A1: Fully correct unsimplified expression |
| $= \frac{\frac{x}{\sqrt{x^2-1}}+1}{\sqrt{x^2-1}+x}$ or $\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}}\times\frac{1}{\sqrt{x^2-1}+x} = \frac{1}{\sqrt{x^2-1}}$* | A1* | Correct completion with intermediate line of working and no errors |

## Question 3(b):

$f(x) = \frac{1}{3}\text{arsinh}\left(\sqrt{x^2-1}\right) - \arctan x$

| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = \frac{1}{3\sqrt{x^2-1}} - \frac{1}{1+x^2}$ | M1, B1 on ePen | $f'(x) = \frac{A}{\sqrt{x^2-1}} \pm \frac{1}{1\pm x^2}$, $A = \frac{1}{3}$, 3 or 1 |
| Sets $\frac{A}{\sqrt{x^2-1}} \pm \frac{1}{1+x^2} = 0$; $1+x^2 = 3\sqrt{x^2-1}$; $1+2x^2+x^4 = 9x^2-9$ | M1 | Denominator of derivative of $\arctan x$ must now be $1+x^2$. Cross multiplies and squares to obtain correct form for both sides. Do not condone e.g. $(1+x^2)^2 = 1+x^4$ |
| $x^4 - 7x^2 + 10 = 0 \Rightarrow (x^2-2)(x^2-5) = 0 \Rightarrow x^2 = 2, 5$ | ddM1 | Solves a 3TQ in $x^2$. One correct root if no working. Requires previous M marks. |
| $x = \sqrt{2},\ \sqrt{5}$ | A1 | Both exact and no other solutions. Negatives rejected. Must not reject either correct solution. |

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