## Question 5:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| When $x=0$, $f(x) = a = 2$ | B1 | NB **AG** '$a$ is the $y$-intercept' not enough; but allow verification ($2 + \sin 0 = 2$); or equiv transformation arguments e.g. 'curve is shifted up 2 so $a=2$' |
| When $x = \pi$, $f(\pi) = 2 + \sin b\pi = 3$ | M1 | or when $x = -\pi$, $f(-\pi) = 2 + \sin(-b\pi) = 1$ |
| $\sin b\pi = 1$ | | |
| $b\pi = \frac{1}{2}\pi$, so $b = \frac{1}{2}$ | A1 | $\Rightarrow \sin(-b\pi) = -1$, condone using degrees; $\Rightarrow -b\pi = -\frac{1}{2}\pi$, $b = \frac{1}{2}$ **NB AG**; e.g. period of sine curve is $4\pi$, or stretched by sf. 2 in $x$-direction |
| or $1 = a + \sin(-\pi b) = a - \sin\pi b$ | | M1 for both points substituted |
| $3 = a + \sin(\pi b)$ | | A1 solving for $b$ or $a$ |
| $\Rightarrow 2 = 2\sin\pi b$, $\sin\pi b = 1$, $\pi b = \frac{\pi}{2}$, $b = \frac{1}{2}$ | | A1 substituting to get $a$ (or $b$) |
| $\Rightarrow 3 = a+1$ or $1 = a-1 \Rightarrow a = 2$ | | If $y = 2 + \sin\frac{1}{2}x$ verified at two points, SC2 |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = \frac{1}{2}\cos\frac{1}{2}x$ | M1 | $\pm k\cos\frac{1}{2}x$ |
| | A1 | cao |
| $f'(0) = \frac{1}{2}$ | A1 | www |
| Maximum value of $\cos\frac{1}{2}x$ is 1 | M1 | or $f''(x) = -\frac{1}{4}\sin\frac{1}{2}x$ |
| max value of gradient is $\frac{1}{2}$ | A1 | $f''(x) = 0 \Rightarrow x = 0$, so max val of $f'(x)$ is $\frac{1}{2}$ |

### Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2 + \sin\frac{1}{2}x$, $x \leftrightarrow y$ | M1 | Attempt to invert formula; viz solve for $x$ in terms of $y$ or vice-versa – one step enough; condone use of $a$ and $b$ in inverse function |
| $x - 2 = \sin\frac{1}{2}y$ | | |
| $\arcsin(x-2) = \frac{1}{2}y$ | A1 | or $\arcsin(y-2) = \frac{1}{2}x$ |
| $y = f^{-1}(x) = 2\arcsin(x-2)$ | A1 | must be $y = \ldots$ or $f^{-1}(x) = \ldots$; or $\sin^{-1}(y-2)$, condone no bracket for 1st A1 only; or $2\sin^{-1}(x-2)$, condone $f'(x)$, must have bracket in final ans |
| Domain $1 \leq x \leq 3$ | B1 | or $[1,3]$ |
| Range $-\pi \leq y \leq \pi$ | B1 | or $[-\pi, \pi]$ or $-\pi \leq f^{-1}(x) \leq \pi$; but not $-\pi \leq x \leq \pi$ |
| Gradient at $(2,0)$ is 2 | B1ft | ft their answer in (ii) (except $\pm 1$) $1/\text{their } \frac{1}{2}$; or by differentiating $\arcsin(x-2)$ or implicitly |

### Part (iv):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \int_0^{\pi}\left(2 + \sin\frac{1}{2}x\right)dx$ | M1 | correct integral and limits; soi from subsequent work, condone no $dx$ but not 180 |
| $= \left[2x - 2\cos\frac{1}{2}x\right]_0^{\pi}$ | M1 | $\left[2x - k\cos\frac{1}{2}x\right]$ where $k$ is positive |
| | A1 | $k = 2$ |
| $= 2\pi - (-2)$ | | |
| $= 2\pi + 2$ $(= 8.2831\ldots)$ | A1cao | answers rounding to 8.3; unsupported correct answers score 1st M1 only |