Question 9 - A-Level Maths

OCR MEI C3 — Question 9 18 marks

Exam Board	OCR MEI
Module	C3 (Core Mathematics 3)
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Composite & Inverse Functions
Type	Find inverse function
Difficulty	Standard +0.3 This is a standard C3 composite functions question covering routine topics: ranges, inverses, composition, and solving equations. Parts (i)-(iv) are textbook exercises requiring only direct application of learned techniques. Part (v) requires slightly more thought (completing the square and discriminant analysis) but follows a predictable pattern for this module. Overall slightly easier than average due to straightforward function forms and standard question structure.
Spec	1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence 1.07q Product and quotient rules: differentiation

The functions f(x) and g(x) are defined by $$f(x) = x^2, \quad g(x) = 2x - 1,$$ for all real values of $x$.

State the ranges of f(x) and g(x). Explain why f(x) has no inverse. [3]
Find an expression for the inverse function g$^{-1}$(x) in terms of $x$. Sketch the graphs of $y = g(x)$ and $y = g^{-1}(x)$ on the same axes. [4]
Find expressions for gf(x) and fg(x). [2]
Solve the equation gf(x) = fg(x). Sketch the graphs of $y = gf(x)$ and $y = fg(x)$ on the same axes to illustrate your answer. [4]
Show that the equation f(x + a) = g$^{-1}$(x) has no solution if $a > \frac{1}{4}$. [5]

Show mark scheme Show mark scheme source

(i)

Answer	Marks	Guidance
Range of $f$ is $[0, \infty)$, of $g$ is $(-\infty, \infty)$. $f$ has no inverse because (say) for any value of $f > 0$ there are 2 corresponding values of $x$	B1, B1, E1	3 marks

(ii)

Answer	Marks	Guidance
$y = 2x - 1 \Rightarrow x = \frac{1}{2}y + \frac{1}{2} \Rightarrow g^{-1}(x) = \frac{1}{2}x + \frac{1}{2}$	M1, A1	One mark for one line, and one mark for second correctly related.
Graph showing $y = g(x)$ and $y = x$ and $y = g^{-1}(x)$	B1, B1	4 marks

(iii)

Answer	Marks	Guidance
$gf(x) = 2x^2 - 1$	B1
$fg(x) = (2x - 1)^2$	B1	2 marks

(iv)

Answer	Marks	Guidance
$2x^2 - 1 = (2x - 1)^2 \Rightarrow 0 = 2x^2 - 4x + 2 \Rightarrow 0 = 2(x - 1)^2 \Rightarrow x = 1$	M1, A1
Graph showing both curves meeting at $(1, 1)$	B1, B1	$y = (2x-1)^2$; $y = (2x-1)^2$. 4 marks

(v)

Answer	Marks	Guidance
$f(x + a) = (x + a)^2$	B1
$g^2(x) = 2(2x - 1) - 1 = 4x - 3$	B1	Both fns correct; Equating
$f(x + a) = g^2(x) \Rightarrow (x + a)^2 = 4x - 3$	M1
$\Rightarrow x^2 + (2a - 4)x + a^2 + 3 = 0$	M1	Using $b^2 - 4ac$; Correct inequality
$\Rightarrow$ There are two roots to this equation if $(2a - 4)^2 > 4(a^2 + 3)$ i.e. $4a^2 - 16a + 16 > 4a^2 + 12$	A1
$\Rightarrow 16a < 4 \Rightarrow a < \frac{1}{4}$	A1	Result. 5 marks

### (i)
Range of $f$ is $[0, \infty)$, of $g$ is $(-\infty, \infty)$. $f$ has no inverse because (say) for any value of $f > 0$ there are 2 corresponding values of $x$ | B1, B1, E1 | 3 marks

### (ii)
$y = 2x - 1 \Rightarrow x = \frac{1}{2}y + \frac{1}{2} \Rightarrow g^{-1}(x) = \frac{1}{2}x + \frac{1}{2}$ | M1, A1 | One mark for one line, and one mark for second correctly related.

Graph showing $y = g(x)$ and $y = x$ and $y = g^{-1}(x)$ | B1, B1 | 4 marks

### (iii)
$gf(x) = 2x^2 - 1$ | B1

$fg(x) = (2x - 1)^2$ | B1 | 2 marks

### (iv)
$2x^2 - 1 = (2x - 1)^2 \Rightarrow 0 = 2x^2 - 4x + 2 \Rightarrow 0 = 2(x - 1)^2 \Rightarrow x = 1$ | M1, A1

Graph showing both curves meeting at $(1, 1)$ | B1, B1 | $y = (2x-1)^2$; $y = (2x-1)^2$. 4 marks

### (v)
$f(x + a) = (x + a)^2$ | B1

$g^2(x) = 2(2x - 1) - 1 = 4x - 3$ | B1 | Both fns correct; Equating

$f(x + a) = g^2(x) \Rightarrow (x + a)^2 = 4x - 3$ | M1

$\Rightarrow x^2 + (2a - 4)x + a^2 + 3 = 0$ | M1 | Using $b^2 - 4ac$; Correct inequality

$\Rightarrow$ There are two roots to this equation if $(2a - 4)^2 > 4(a^2 + 3)$ i.e. $4a^2 - 16a + 16 > 4a^2 + 12$ | A1

$\Rightarrow 16a < 4 \Rightarrow a < \frac{1}{4}$ | A1 | Result. 5 marks

Show LaTeX source

The functions f(x) and g(x) are defined by
$$f(x) = x^2, \quad g(x) = 2x - 1,$$
for all real values of $x$.
\begin{enumerate}[label=(\roman*)]
\item State the ranges of f(x) and g(x).
Explain why f(x) has no inverse. [3]
\item Find an expression for the inverse function g$^{-1}$(x) in terms of $x$.
Sketch the graphs of $y = g(x)$ and $y = g^{-1}(x)$ on the same axes. [4]
\item Find expressions for gf(x) and fg(x). [2]
\item Solve the equation gf(x) = fg(x).

Sketch the graphs of $y = gf(x)$ and $y = fg(x)$ on the same axes to illustrate your answer. [4]
\item Show that the equation f(x + a) = g$^{-1}$(x) has no solution if $a > \frac{1}{4}$. [5]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C3  Q9 [18]}}

This paper (9 questions)

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Q1 5 Q2 4 Q3 6 Q4 4 Q5 4 Q6 7 Q7 6 Q8 18 Q9 18

Answer	Marks	Guidance
\(y = 2x - 1 \Rightarrow x = \frac{1}{2}y + \frac{1}{2} \Rightarrow g^{-1}(x) = \frac{1}{2}x + \frac{1}{2}\)	M1, A1	One mark for one line, and one mark for second correctly related.
Graph showing \(y = g(x)\) and \(y = x\) and \(y = g^{-1}(x)\)	B1, B1	4 marks

Answer	Marks	Guidance
\(2x^2 - 1 = (2x - 1)^2 \Rightarrow 0 = 2x^2 - 4x + 2 \Rightarrow 0 = 2(x - 1)^2 \Rightarrow x = 1\)	M1, A1
Graph showing both curves meeting at \((1, 1)\)	B1, B1	\(y = (2x-1)^2\); \(y = (2x-1)^2\). 4 marks

Answer	Marks	Guidance
\(f(x + a) = (x + a)^2\)	B1
\(g^2(x) = 2(2x - 1) - 1 = 4x - 3\)	B1	Both fns correct; Equating
\(f(x + a) = g^2(x) \Rightarrow (x + a)^2 = 4x - 3\)	M1
\(\Rightarrow x^2 + (2a - 4)x + a^2 + 3 = 0\)	M1	Using \(b^2 - 4ac\); Correct inequality
\(\Rightarrow\) There are two roots to this equation if \((2a - 4)^2 > 4(a^2 + 3)\) i.e. \(4a^2 - 16a + 16 > 4a^2 + 12\)	A1
\(\Rightarrow 16a < 4 \Rightarrow a < \frac{1}{4}\)	A1	Result. 5 marks