October 2021 Paper 1 Q8
8 Functions f and g are defined for \(0 \leqslant x \leqslant 2\pi\) by \(\mathrm{f}(x) = 2\tan x\) and \(\mathrm{g}(x) = \sec x\).
Solve the equation \((\mathrm{f}(x))^2 + 6\mathrm{g}(x) = 0\). [5]
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{f}(x) \in \mathbb{R}\) | B1 | 2.5 |
| [1] | ||
| (ii) \(\mathrm{g}(x) \in (-\infty, -1] \cup [1, \infty)\) | B1 | 2.5 |
| [1] |
Notes
(a)(i) B1: Allow alternative notation, or worded equivalent
Allow \(y\), or just f, but not \(x\)
Accept just \(\mathbb{R}\)
Allow \((-\infty, \infty)\)
(a)(ii) B1: Allow alternative notation, or worded equivalent
Allow \(y\), or just g, but not \(x\)
Or \((-\infty, \infty)\) with \((-1, 1)\) clearly excluded
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\cos(0.6) = 0.8253\), so \(\sec(0.6) = \frac{1}{0.8253} = 1.2116\) \(2\tan(1.2116) = 2 \times 2.6634 = 5.3269\) | M1 | 2.1 |
| hence \(\mathrm{fg}(0.6) = 5.33\) A.G. | A1 | 2.1 |
| [2] | ||
| (ii) \(\mathrm{f}(x)\) is a many to one function so has no inverse | B1 | 2.4 |
| [1] |
Notes
(b)(i)
M1: Attempt correct composition of functions
At least one interim value required
A1: Conclude with 5.33
SC B1 for stating \(2\tan(1 \div \cos 0.6) = 5.33\)
(b)(ii)
B1: Must refer to inverse of f not existing, with reason
Must be clear that referring to the function f
| Scheme | Marks | AO |
|---|---|---|
| DR \(4\tan^2 x + 6\sec x = 0\) \(4(\sec^2 x - 1) + 6\sec x = 0\) | M1 | 3.1a |
| \(4\sec^2 x + 6\sec x - 4 = 0\) | A1 | 1.1 |
| \(\sec x = -2\), \(\sec x = \frac{1}{2}\) | M1 | 1.1 |
| \(x = \frac{2}{3}\pi,\ \frac{4}{3}\pi\) | A1 | 1.1 |
| \(\sec x = \frac{1}{2}\) has no solutions as \(|\sec x| \geqslant 1\) | A1 | 2.3 |
| [5] |
Notes
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
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 \((4\cos^2 x - 6\cos x - 4 = 0)\)
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 \(x\)
A1: Obtain at least one correct value
Allow decimals or in degrees
Must be from correct solution method of correct quadratic (condone second root not being seen – but must be correct if seen)
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 \(\cos x\)