OCR C3 — Question 8 12 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind inverse function
DifficultyStandard +0.2 This is a standard C3 functions question covering routine techniques: finding range from a restricted quadratic, sketching a function and its inverse as reflections in y=x, finding inverse by swapping and rearranging, and evaluating composite functions. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than the average A-level question.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02z Models in context: use functions in modelling

The function f is defined by $$\text{f}(x) \equiv 3 - x^2, \quad x \in \mathbb{R}, \quad x \geq 0.$$
  1. State the range of f. [1]
  2. Sketch the graphs of \(y = \text{f}(x)\) and \(y = \text{f}^{-1}(x)\) on the same diagram. [3]
  3. Find an expression for f\(^{-1}(x)\) and state its domain. [3]
The function g is defined by $$\text{g}(x) \equiv \frac{8}{3-x}, \quad x \in \mathbb{R}, \quad x \neq 3.$$
  1. Evaluate fg\((-3)\). [2]
  2. Solve the equation $$\text{f}^{-1}(x) = \text{g}(x).$$ [3]

(i)
AnswerMarks
\(f(x) \le 3\)B1
(ii)
AnswerMarks
Graph showing \(y = f(x)\) and \(y = f^{-1}(x)\) with correct reflection about \(y = x\)B3
(iii)
\(y = 3 - x^2\)
AnswerMarks
\(x^2 = 3 - y\), \(x = \pm\sqrt{3-y}\)M1
\(f^{-1}(x) = \sqrt{3-x}\), \(x \in \mathbb{R}\), \(x \le 3\)A2
(iv)
AnswerMarks
\(=f\left(\frac{4}{3}\right) = \frac{11}{9}\)M1 A1
(v)
\(\sqrt{3-x} = \frac{8}{3-x}\)
\(3 - x = \frac{64}{(3-x)^2}\)
\((3-x)^3 = 64\)
\(3 - x = 4\)
AnswerMarks Guidance
\(x = -1\)M1, M1, M1, A1 (12)
## (i)
$f(x) \le 3$ | B1 |

## (ii)
Graph showing $y = f(x)$ and $y = f^{-1}(x)$ with correct reflection about $y = x$ | B3 |

## (iii)
$y = 3 - x^2$

$x^2 = 3 - y$, $x = \pm\sqrt{3-y}$ | M1 |

$f^{-1}(x) = \sqrt{3-x}$, $x \in \mathbb{R}$, $x \le 3$ | A2 |

## (iv)
$=f\left(\frac{4}{3}\right) = \frac{11}{9}$ | M1 A1 |

## (v)
$\sqrt{3-x} = \frac{8}{3-x}$

$3 - x = \frac{64}{(3-x)^2}$

$(3-x)^3 = 64$

$3 - x = 4$

$x = -1$ | M1, M1, M1, A1 | (12)

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The function f is defined by
$$\text{f}(x) \equiv 3 - x^2, \quad x \in \mathbb{R}, \quad x \geq 0.$$

\begin{enumerate}[label=(\roman*)]
\item State the range of f. [1]
\item Sketch the graphs of $y = \text{f}(x)$ and $y = \text{f}^{-1}(x)$ on the same diagram. [3]
\item Find an expression for f$^{-1}(x)$ and state its domain. [3]
\end{enumerate}

The function g is defined by
$$\text{g}(x) \equiv \frac{8}{3-x}, \quad x \in \mathbb{R}, \quad x \neq 3.$$

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Evaluate fg$(-3)$. [2]
\item Solve the equation
$$\text{f}^{-1}(x) = \text{g}(x).$$ [3]
\end{enumerate}

\hfill \mbox{\textit{OCR C3  Q8 [12]}}