| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation with inverses |
| Difficulty | Moderate -0.8 This is a straightforward P1 question requiring routine inverse function finding (linear and rational functions using standard algebraic manipulation) and solving a simple composite function equation. All techniques are standard textbook exercises with no novel insight required, making it easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.02y Partial fractions: decompose rational functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((f^{-1}(x)) = \frac{x+2}{3}\) oe | B1 | |
| \(y = \frac{2x+3}{x-1} \rightarrow (x-1)y = 2x+3 \rightarrow x(y-2) = y+3\) | M1 | Correct method to obtain \(x =\) (or \(y =\) if interchanged) but condone \(+/-\) sign errors |
| \((g^{-1}(x)\) or \(y) = \frac{x+3}{x-2}\) oe \(\left(eg\ \frac{5}{x-2}+1\right)\) | A1 | Must be in terms of \(x\) |
| \(x \neq 2\) only | B1 | FT for value of \(x\) from their denominator \(= 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((fg(x) =)\ \frac{3(2x+3)}{x-1} - 2\ \left(= \frac{7}{3}\right)\) | B1 | |
| \(18x + 27 = 13x - 13\) or \(3(4x+11) = 7(x-1)\) giving \((5x = -40)\) | M1 | Correct method from their \(fg = \frac{7}{3}\) leading to a linear equation and collect like terms. Condone omission of \(2(x-1)\) |
| Alternative: \((f^{-1}(\frac{7}{3})) = \frac{13}{9}\) | B1 | |
| \(\frac{2x+3}{x-1} = \frac{13}{9} \rightarrow 9(2x+3) = 13(x-1)\ (\rightarrow 5x = -40)\) | M1 | Correct method from \(g(x) =\) their \(\frac{13}{9}\) leading to a linear equation and collect like terms |
| \(x = -8\) | A1 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(f^{-1}(x)) = \frac{x+2}{3}$ oe | B1 | |
| $y = \frac{2x+3}{x-1} \rightarrow (x-1)y = 2x+3 \rightarrow x(y-2) = y+3$ | M1 | Correct method to obtain $x =$ (or $y =$ if interchanged) but condone $+/-$ sign errors |
| $(g^{-1}(x)$ or $y) = \frac{x+3}{x-2}$ oe $\left(eg\ \frac{5}{x-2}+1\right)$ | A1 | Must be in terms of $x$ |
| $x \neq 2$ only | B1 | FT for value of $x$ from their denominator $= 0$ |
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(fg(x) =)\ \frac{3(2x+3)}{x-1} - 2\ \left(= \frac{7}{3}\right)$ | B1 | |
| $18x + 27 = 13x - 13$ or $3(4x+11) = 7(x-1)$ giving $(5x = -40)$ | M1 | Correct method from their $fg = \frac{7}{3}$ leading to a linear equation and collect like terms. Condone omission of $2(x-1)$ |
| **Alternative:** $(f^{-1}(\frac{7}{3})) = \frac{13}{9}$ | B1 | |
| $\frac{2x+3}{x-1} = \frac{13}{9} \rightarrow 9(2x+3) = 13(x-1)\ (\rightarrow 5x = -40)$ | M1 | Correct method from $g(x) =$ their $\frac{13}{9}$ leading to a linear equation and collect like terms |
| $x = -8$ | A1 | |7 Functions f and g are defined by
$$\begin{aligned}
& \mathrm { f } : x \mapsto 3 x - 2 , \quad x \in \mathbb { R } , \\
& \mathrm {~g} : x \mapsto \frac { 2 x + 3 } { x - 1 } , \quad x \in \mathbb { R } , x \neq 1
\end{aligned}$$
(i) Obtain expressions for $\mathrm { f } ^ { - 1 } ( x )$ and $\mathrm { g } ^ { - 1 } ( x )$, stating the value of $x$ for which $\mathrm { g } ^ { - 1 } ( x )$ is not defined. [4]\\
(ii) Solve the equation $\operatorname { fg } ( x ) = \frac { 7 } { 3 }$.\\
\hfill \mbox{\textit{CAIE P1 2019 Q7 [7]}}