Moderate -0.3 This is a slightly below-average A-level question covering standard function operations. Part (a) requires solving a simple equation, (b) involves routine composition of rational functions, (c) needs completing the square to find a range, and (d) tests basic understanding that non-injective functions lack inverses. All parts are textbook exercises requiring no novel insight.
5. The function f is defined by
$$\mathrm { f } : x \rightarrow \frac { 2 x - 3 } { x - 1 } \quad x \in R , x \neq 1$$
a. Find \(f ^ { - 1 } ( 3 )\).
b. Show that
$$\mathrm { ff } ( x ) = \frac { x + p } { x - 2 } \quad x \in R , \quad x \neq 2$$
where \(p\) is an integer to be found.
The function g is defined by
$$g : x \rightarrow x ^ { 2 } - 5 x \quad x \in R , 0 \leq x \leq 6$$
c. Find the range of g .
d. Explain why the function g does not have an inverse.
For either attempting to solve \(\frac{2x - 3}{x - 1} = 3\) leading to a value of \(x\) or score for substituting in \(x = 3\) in \(f^{-1}(x)\) where \(f^{-1}(x) = \frac{x - 3}{x - 2}\)
A1
\(f^{-1}(3) = 0\)
Part b: Show that \(ff(x) = \frac{x + p}{x - 2}\) where \(p\) is an integer to be found
Attempts to multiply all terms on the numerator and denominator by \((x - 1)\) to obtain a fraction \(\frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are linear expressions
Either applies the completing the square method to establish the minimum of g. Or differentiating the quadratic, setting the result equal to zero, finding x and inserting this value of x back into g(x) in order to find the minimum
B1
For either finding the correct minimum or maximum value of g
A1
\(-\frac{25}{4} \le g(x) \le 6\) or \(-\frac{25}{4} \le g \le 6\) or \(-\frac{25}{4} \le y \le 6\)
Part d: Explain why the function g does not have an inverse
Answer
Marks
\(g(x)\) does not have an inverse as it is not one-to-one
B1
either the function g is many-one or the function g is not one-to-one
(2) marks + (3) marks + (3) marks + (1) mark = (9) marks total for Question 5
**Part a:** Find $f^{-1}(3)$
| $y = \frac{2x - 3}{x - 1}$ | | |
| --- | --- | --- |
| $x = \frac{2y - 3}{y - 1} \Rightarrow x(y - 1) = 2y - 3 \Rightarrow xy - x = 2y - 3 \Rightarrow xy - 2y = x - 3$ | | |
| $y(x - 2) = x - 3 \Rightarrow y = \frac{x - 3}{x - 2}$ | | |
| $f^{-1}(x) = \frac{x - 3}{x - 2} \Rightarrow f^{-1}(3) = \frac{3 - 3}{3 - 2} = 0$ | M1 | For either attempting to solve $\frac{2x - 3}{x - 1} = 3$ leading to a value of $x$ or score for substituting in $x = 3$ in $f^{-1}(x)$ where $f^{-1}(x) = \frac{x - 3}{x - 2}$ |
| | A1 | $f^{-1}(3) = 0$ |
**Part b:** Show that $ff(x) = \frac{x + p}{x - 2}$ where $p$ is an integer to be found
| $ff(x) = f\left(\frac{2x - 3}{x - 1}\right) \Rightarrow ff(x) = \frac{4x - 6 - 3x + 3}{2x - 3 - x + 1}$ | M1 | For an attempt substituting $\frac{2x - 3}{x - 1}$ in $f(x)$ |
| --- | --- | --- |
| $ff(x) = \frac{4x - 6 - 3x + 3}{2x - 3 - x + 1} = \frac{x - 3}{x - 2}$ | dM1 | Attempts to multiply all terms on the numerator and denominator by $(x - 1)$ to obtain a fraction $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are linear expressions |
| | A1 cso | $\frac{x - 3}{x - 2}$ with $p = -3$ |
**Part c:** Find the range of g
| $\frac{dg}{dx} = 2x - 5$; $2x - 5 = 0 \Rightarrow x = \frac{5}{2}$ | | |
| --- | --- | --- |
| $g(6) = 36 - 30 = 6$ | | |
| $-\frac{25}{4} \le g(x) \le 6$ | C: M1 | Either applies the completing the square method to establish the minimum of g. Or differentiating the quadratic, setting the result equal to zero, finding x and inserting this value of x back into g(x) in order to find the minimum |
| | B1 | For either finding the correct minimum or maximum value of g |
| | A1 | $-\frac{25}{4} \le g(x) \le 6$ or $-\frac{25}{4} \le g \le 6$ or $-\frac{25}{4} \le y \le 6$ |
**Part d:** Explain why the function g does not have an inverse
| $g(x)$ does not have an inverse as it is not one-to-one | | |
| --- | --- | --- |
| | B1 | either the function g is many-one or the function g is not one-to-one |
**(2) marks + (3) marks + (3) marks + (1) mark = (9) marks total for Question 5**
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5. The function f is defined by
$$\mathrm { f } : x \rightarrow \frac { 2 x - 3 } { x - 1 } \quad x \in R , x \neq 1$$
a. Find $f ^ { - 1 } ( 3 )$.\\
b. Show that
$$\mathrm { ff } ( x ) = \frac { x + p } { x - 2 } \quad x \in R , \quad x \neq 2$$
where $p$ is an integer to be found.
The function g is defined by
$$g : x \rightarrow x ^ { 2 } - 5 x \quad x \in R , 0 \leq x \leq 6$$
c. Find the range of g .\\
d. Explain why the function g does not have an inverse.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q5 [9]}}