Standard +0.3 This is a straightforward modulus equation requiring case-by-case analysis (typically 2-3 cases based on sign changes at x=0 and x=3/2), then solving linear equations and checking validity. It's slightly above average difficulty as it involves systematic reasoning about absolute values, but the algebraic manipulation is routine and the question type is standard for C3.
If 3 or more final answers offered, \(-1\) for each incorrect additional answer
or \(\quad 3 - 2x = -4x, \quad x = -1\frac{1}{2}\)
M1 A1
or \((3 - 2x)^2 = 16x^2\)
M1
squaring both sides
\(\Rightarrow 12x^2 + 12x - 9 = [0]\)
A1
correct quadratic o.e. but with single \(x^2\) term
\(\Rightarrow x = 1/2, -1\frac{1}{2}\)
A1 A1 [4]
\(-1\) for final ans written as an inequality; \((3-2x)^2 = 4x^2\) is M0
Question 4(i)
Answer
Marks
Guidance
\(a = 2, \quad b = 1/2\)
B1 B1 [2]
(may be seen later)
Question 4(ii)
Answer
Marks
Guidance
\(y = 2 + \cos 1/2 x \Leftrightarrow y\)
M1
subtracting [their] \(a\) from both sides (first); need not substitute for \(a, b\)
\(x = 2 + \cos 1/2 y\)
\(\Rightarrow x - 2 = \cos 1/2 y\)
M1
or with \(x \Leftrightarrow y\), need not subst for \(a, b\)
\(\Rightarrow \arccos(x - 2) = 1/2 y\)
M1
\(\arccos(x - [their] a) = \) [their] \(b \times y\)
\(\Rightarrow y = f^{-1}(x) = 2\arccos(x - 2)\)
A1
cao or \(2\cos^{-1}(x - 2)\); may be implied by flow diagram
Domain \(1 \le x \le 3\)
M1
domain 1 to 3, range 0 to \(2\pi\)
Range \(0 \le y \le 2\pi\)
A1
correctly specified: must be \(\le, x\) for domain, \(y\) or \(f^{-1}\) or \(f^{-1}(x)\) for range
[5]
if not stated, assume first is domain; allow \([1, 3], [0, 2\pi]\) not \(360°\) (not f)
$|3 - 2x| = 4|x|$ | | |
$\Rightarrow 3 - 2x = 4x, \quad x = 1/2$ | M1 A1 | If 3 or more final answers offered, $-1$ for each incorrect additional answer |
or $\quad 3 - 2x = -4x, \quad x = -1\frac{1}{2}$ | M1 A1 | |
or $(3 - 2x)^2 = 16x^2$ | M1 | squaring both sides |
$\Rightarrow 12x^2 + 12x - 9 = [0]$ | A1 | correct quadratic o.e. but with single $x^2$ term |
$\Rightarrow x = 1/2, -1\frac{1}{2}$ | A1 A1 [4] | $-1$ for final ans written as an inequality; $(3-2x)^2 = 4x^2$ is M0 |
# Question 4(i)
$a = 2, \quad b = 1/2$ | B1 B1 [2] | (may be seen later) |
# Question 4(ii)
$y = 2 + \cos 1/2 x \Leftrightarrow y$ | M1 | subtracting [their] $a$ from both sides (first); need not substitute for $a, b$ |
$x = 2 + \cos 1/2 y$ | | |
$\Rightarrow x - 2 = \cos 1/2 y$ | M1 | or with $x \Leftrightarrow y$, need not subst for $a, b$ |
$\Rightarrow \arccos(x - 2) = 1/2 y$ | M1 | $\arccos(x - [their] a) = $ [their] $b \times y$ |
$\Rightarrow y = f^{-1}(x) = 2\arccos(x - 2)$ | A1 | cao or $2\cos^{-1}(x - 2)$; may be implied by flow diagram |
Domain $1 \le x \le 3$ | M1 | domain 1 to 3, range 0 to $2\pi$ |
Range $0 \le y \le 2\pi$ | A1 | correctly specified: must be $\le, x$ for domain, $y$ or $f^{-1}$ or $f^{-1}(x)$ for range |
| [5] | if not stated, assume first is domain; allow $[1, 3], [0, 2\pi]$ not $360°$ (not f) |