| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a multi-part question on inverse functions requiring sketching by reflection, algebraic manipulation to find an inverse (involving rationalizing with surds), domain considerations for composite functions, and function composition. While it covers several techniques, each part follows standard procedures taught in P1 with no novel insights required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Reflection of curve in \(y = x\)] | B1 | A reflection of the given curve in \(y = x\) (the line \(y = x\) can be implied by position of curve). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{-x}{\sqrt{4-x^2}}\) leading to \(x^2 = y^2(4 - x^2)\) | \*M1 | Squaring and clearing the fraction. Condone one error in squaring \(-x\) or \(y\) |
| \(x^2(1 + y^2) = 4y^2\) | DM1 | OE. Factorisation of the new subject with order of operations correct. Condone sign errors. |
| \(x = (\pm)\dfrac{2y}{\sqrt{1+y^2}}\) | DM1 | \(x = (\pm)\sqrt{\left(\dfrac{4y^2}{1+y^2}\right)}\) OE acceptable. Isolating the new subject. Order of operations correct. Condone sign errors. |
| \(f^{-1}(x) = \dfrac{-2x}{\sqrt{1+x^2}}\) | A1 | Selecting the correct square root. Must not have fractions in numerator or denominator. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1\) or \(a = 1\) | B1 | Do not allow \(x = 1\) or \(-1 < x < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{fg}(x) = f(2x) =] \dfrac{-2x}{\sqrt{4-4x^2}}\) | B1 | Allow \(\dfrac{-2x}{\sqrt{4-(2x)^2}}\) or any correct unsimplified form. |
| \(\text{fg}(x) = \dfrac{-x}{\sqrt{1-x^2}}\) or \(\dfrac{-x}{1-x^2}\sqrt{1-x^2}\) or \(\dfrac{x}{x^2-1}\sqrt{1-x^2}\) | B1 | Result of cancelling 2 in numerator and denominator. |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Reflection of curve in $y = x$] | B1 | A reflection of the given curve in $y = x$ (the line $y = x$ can be implied by position of curve). |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{-x}{\sqrt{4-x^2}}$ leading to $x^2 = y^2(4 - x^2)$ | \*M1 | Squaring and clearing the fraction. Condone one error in squaring $-x$ or $y$ |
| $x^2(1 + y^2) = 4y^2$ | DM1 | OE. Factorisation of the new subject with order of operations correct. Condone sign errors. |
| $x = (\pm)\dfrac{2y}{\sqrt{1+y^2}}$ | DM1 | $x = (\pm)\sqrt{\left(\dfrac{4y^2}{1+y^2}\right)}$ OE acceptable. Isolating the new subject. Order of operations correct. Condone sign errors. |
| $f^{-1}(x) = \dfrac{-2x}{\sqrt{1+x^2}}$ | A1 | Selecting the correct square root. Must not have fractions in numerator or denominator. |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1$ or $a = 1$ | B1 | Do not allow $x = 1$ or $-1 < x < 1$ |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{fg}(x) = f(2x) =] \dfrac{-2x}{\sqrt{4-4x^2}}$ | B1 | Allow $\dfrac{-2x}{\sqrt{4-(2x)^2}}$ or any correct unsimplified form. |
| $\text{fg}(x) = \dfrac{-x}{\sqrt{1-x^2}}$ or $\dfrac{-x}{1-x^2}\sqrt{1-x^2}$ or $\dfrac{x}{x^2-1}\sqrt{1-x^2}$ | B1 | Result of cancelling 2 in numerator and denominator. |
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\includegraphics[max width=\textwidth, alt={}, center]{cba7391d-2cf7-483c-9a21-bc9fd49860ee-08_608_597_258_772}
The diagram shows the graph of $y = \mathrm { f } ( x )$.
\begin{enumerate}[label=(\alph*)]
\item On this diagram sketch the graph of $y = \mathrm { f } ^ { - 1 } ( x )$.
It is now given that $\mathrm { f } ( x ) = - \frac { x } { \sqrt { 4 - x ^ { 2 } } }$ where $- 2 < x < 2$.
\item Find an expression for $\mathrm { f } ^ { - 1 } ( x )$.\\
The function g is defined by $\mathrm { g } ( x ) = 2 x$ for $- a < x < a$, where $a$ is a constant.
\item State the maximum possible value of $a$ for which fg can be formed.
\item Assuming that fg can be formed, find and simplify an expression for $\mathrm { fg } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q6 [8]}}