| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Solve with restricted domain |
| Difficulty | Standard +0.3 This question tests standard knowledge of reciprocal trig functions and identities. Part (c) requires using the identity tan²x + 1 = sec²x to form a quadratic in sec x, then solving within the restricted domain. While it requires multiple steps and careful consideration of the domain restrictions, the technique is straightforward and commonly practiced. The 'show detailed reasoning' instruction and domain considerations elevate it slightly above average, but it remains a standard A-level pure maths question without requiring novel insight. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((1.206, -0.102)\) | B1, B1 | Correct \(x\)-coordinate (3dp or better); correct \(y\)-coordinate. Could be given as single coordinate or \(x=1.206, y=-0.102\). Allow BOD if 1.206 given but not identified as \(x\)-value. |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) \in \mathbb{R}\) | B1 | Allow alternative notation or worded equivalent. Allow \(y\) or just f, but not \(x\). Accept just \(\mathbb{R}\); allow \((-\infty, \infty)\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(g(x) \in (-\infty, -1] \cup [1, \infty)\) | B1 | Allow alternative notation or worded equivalent. Allow \(y\) or just g, but not \(x\). Or \((-\infty, \infty)\) with \((-1,1)\) clearly excluded. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos(0.6) = 0.8253\), so \(\sec(0.6) = \frac{1}{0.8253} = 1.2116\); \(2\tan(1.2116) = 2 \times 2.6634 = 5.3269\); hence \(fg(0.6) = 5.33\) | M1, A1 | Attempt correct composition of functions (at least one interim value required); conclude with 5.33. SC B1 for stating \(2\tan(1 \div \cos 0.6) = 5.33\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)\) is a many to one function so has no inverse | B1 | Must refer to inverse of f not existing, with reason. Must be clear that referring to the function f. |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\tan^2 x + 6\sec x = 0\) | ||
| \(4(\sec^2 x - 1) + 6\sec x = 0\) | M1 | Attempt use of identity in their equation to obtain quadratic. Allow \(\tan^2 x = \pm\sec^2 x \pm 1\). Award M1 when reduced to single trig ratio. |
| \(4\sec^2 x + 6\sec x - 4 = 0\) | A1 | Obtain correct equation in \(\sec x\) — possibly still with brackets. Or correct quadratic in \(\cos x\) — possibly still with brackets but with no fractions \((4\cos^2 x - 6\cos x - 4 = 0)\). |
| \(\sec x = -2,\ \sec x = \frac{1}{2}\) | M1 | Solve 3 term quadratic and attempt to find at least one value for \(x\). Could solve quadratic BC. Must be using root that would give solution for \(x\). |
| \(x = \frac{2}{3}\pi,\ \frac{4}{3}\pi\) | A1 | Obtain at least one correct value. Allow decimals or degrees. Must be from correct solution method. |
| \(\sec x = \frac{1}{2}\) has no solutions as \( | \sec x | \geq 1\) |
# Question 8:
## Part (d)
$(1.206, -0.102)$ | B1, B1 | Correct $x$-coordinate (3dp or better); correct $y$-coordinate. Could be given as single coordinate or $x=1.206, y=-0.102$. Allow BOD if 1.206 given but not identified as $x$-value.
## Part (a)(i)
$f(x) \in \mathbb{R}$ | B1 | Allow alternative notation or worded equivalent. Allow $y$ or just f, but not $x$. Accept just $\mathbb{R}$; allow $(-\infty, \infty)$.
## Part (a)(ii)
$g(x) \in (-\infty, -1] \cup [1, \infty)$ | B1 | Allow alternative notation or worded equivalent. Allow $y$ or just g, but not $x$. Or $(-\infty, \infty)$ with $(-1,1)$ clearly excluded.
## Part (b)(i)
$\cos(0.6) = 0.8253$, so $\sec(0.6) = \frac{1}{0.8253} = 1.2116$; $2\tan(1.2116) = 2 \times 2.6634 = 5.3269$; hence $fg(0.6) = 5.33$ | M1, A1 | Attempt correct composition of functions (at least one interim value required); conclude with 5.33. SC B1 for stating $2\tan(1 \div \cos 0.6) = 5.33$.
## Part (b)(ii)
$f(x)$ is a many to one function so has no inverse | B1 | Must refer to inverse of f not existing, with reason. Must be clear that referring to the function f.
## Part (c)
$4\tan^2 x + 6\sec x = 0$ | | |
$4(\sec^2 x - 1) + 6\sec x = 0$ | M1 | Attempt use of identity in their equation to obtain quadratic. Allow $\tan^2 x = \pm\sec^2 x \pm 1$. Award M1 when reduced to single trig ratio.
$4\sec^2 x + 6\sec x - 4 = 0$ | A1 | Obtain correct equation in $\sec x$ — possibly still with brackets. Or correct quadratic in $\cos x$ — possibly still with brackets but with no fractions $(4\cos^2 x - 6\cos x - 4 = 0)$.
$\sec x = -2,\ \sec x = \frac{1}{2}$ | M1 | Solve 3 term quadratic and attempt to find at least one value for $x$. Could solve quadratic BC. Must be using root that would give solution for $x$.
$x = \frac{2}{3}\pi,\ \frac{4}{3}\pi$ | A1 | Obtain at least one correct value. Allow decimals or degrees. Must be from correct solution method.
$\sec x = \frac{1}{2}$ has no solutions as $|\sec x| \geq 1$ | A1 | Obtain both correct values and no others, and explain that $\sec x = 0.5$ has no solutions as outside range. Now exact and in radians. Or equiv explanation for $\cos x$.
---8 Functions f and g are defined for $0 \leqslant x \leqslant 2 \pi$ by $\mathrm { f } ( x ) = 2 \tan x$ and $\mathrm { g } ( x ) = \sec x$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State the range of f .
\item State the range of $g$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that $\operatorname { fg } ( 0.6 ) = 5.33$, correct to 3 significant figures.
\item Explain why $\mathrm { f } ^ { - 1 } \mathrm {~g} ( 0.6 )$ is not defined.
\end{enumerate}\item In this question you must show detailed reasoning.
Solve the equation $( \mathrm { f } ( x ) ) ^ { 2 } + 6 \mathrm {~g} ( x ) = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2021 Q8 [10]}}