# Question 9(i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| The interval being integrated over is unbounded / $\cosh x$ is undefined at the limit of $\infty$ / the upper limit is infinite | B1 | Must refer to the interval being unbounded. Accept "upper limit is infinity" but NOT "because it is infinity" without reference to what "it" is |

# Question 9(i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\infty} \cosh x \, dx = \lim_{t \to \infty} \int_0^t \cosh x \, dx$ or $\lim_{t \to \infty} \int_0^t \frac{1}{2}(e^x + e^{-x})dx$ | B1 | Writes integral in terms of a limit as $t \to \infty$ with limits 0 and $t$ |
| $\int_0^t \cosh x \, dx = [\sinh x]_0^t = \sinh t \, (-0)$ or $\frac{1}{2}\int_0^t e^x + e^{-x} \, dx = \frac{1}{2}[e^x - e^{-x}]_0^t$ | M1 | Integrates $\cosh x$ correctly and applies limits of 0 and $t$ |
| When $t \to \infty$, $e^t \to \infty$ and $e^{-t} \to 0$ therefore the integral is divergent | A1 | States $\sinh t \to \infty$ or $e^t \to \infty$ and $e^{-t} \to 0$, therefore divergent |

# Question 9(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\sinh x = p\cosh x \Rightarrow \tanh x = \frac{p}{4}$ or $4\tanh x = p$; Alternative: $\frac{4}{2}(e^x - e^{-x}) = \frac{p}{2}(e^x + e^{-x}) \Rightarrow e^{2x}(4-p) = p+4 \Rightarrow e^{2x} = \frac{p+4}{4-p}$ | M1 | Divides through by $\cosh x$ to find expression involving $\tanh x$, OR uses exponential definitions |
| $\left\{-1 < \frac{p}{4} < 1 \Rightarrow\right\} -4 < p < 4$ | A1 | Deduces correct inequality for $p$. Note $|p| < 4$ is acceptable |