OCR C3 2013 June — Question 7 10 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2013
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeFind intersection points
DifficultyStandard +0.3 This is a standard C3 question covering routine inverse function techniques and a simple iterative method. Parts (i)-(ii) require straightforward manipulation of exponential functions and domain/range identification. Part (iii) is a basic fixed-point iteration with clear instructions. Part (iv) tests conceptual understanding that y=x is the line of symmetry between a function and its inverse. All parts are textbook-standard with no novel problem-solving required, making it slightly easier than average.
Spec1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

7 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-3_428_751_703_641} The diagram shows the curve \(y = \mathrm { f } ( x )\), where f is the function defined for all real values of \(x\) by $$\mathrm { f } ( x ) = 3 + 4 \mathrm { e } ^ { - x }$$
  1. State the range of f .
  2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\), and state the domain and range of \(\mathrm { f } ^ { - 1 }\).
  3. The straight line \(y = x\) meets the curve \(y = \mathrm { f } ( x )\) at the point \(P\). By using an iterative process based on the equation \(x = \mathrm { f } ( x )\), with a starting value of 3 , find the coordinates of the point \(P\). Show all your working and give each coordinate correct to 3 decimal places.
  4. How is the point \(P\) related to the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) ?

Question 7:
Part (i):
AnswerMarks Guidance
AnswerMark Guidance
State \(y>3\) or \(f(x)>3\) or \(f>3\) or 'greater than 3'B1 Must be \(>\) not \(\geq\); allow \(3
Part (ii):
AnswerMarks Guidance
AnswerMark Guidance
Obtain expression or eqn involving \(\ln\!\left(\frac{y-3}{4}\right)\) or \(\ln\!\left(\frac{x-3}{4}\right)\)M1 Or equivs such as \(\ln\!\left(\frac{4}{y-3}\right)\) or \(\ln\!\left(\frac{4}{x-3}\right)\)
Obtain \(\ln\!\left(\frac{4}{x-3}\right)\) or \(-\ln\!\left(\frac{x-3}{4}\right)\)A1 Or equiv
State domain is \(x>3\) or equivB1FT Following answer to part (i) (but with adjustment so that reference is to \(x\))
State range is all real numbers or equivB1
Part (iii):
AnswerMarks Guidance
AnswerMark Guidance
Obtain correct first iterateB1 Showing at least 3 dp; B0 if initial value not 3 but then M1A1A1 available
Show correct iteration processM1 Showing at least 3 iterates in all; may be implied by plausible converging values; M1 available if based on equation with just a slip in \(x=f(x)\) but M0 if based on clearly different equation
Obtain at least 3 correct iteratesA1 Allowing recovery after error; iterates to only 3 dp acceptable; values may be rounded or truncated
Obtain (3.168, 3.168)A1 Each coordinate required to exactly 3 dp; award A0 if fewer than 4 iterates shown; part (iii) consisting of answer only gets 0 out of 4. \([3 \to 3.199148 \to 3.163187 \to 3.169162 \to 3.168155 \to 3.168324]\)
Part (iv):
AnswerMarks Guidance
AnswerMark Guidance
State \(P\) is point where the curves meetB1 Or equiv
## Question 7:

### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $y>3$ or $f(x)>3$ or $f>3$ or 'greater than 3' | B1 | Must be $>$ not $\geq$; allow $3<y<\infty$ |

### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain expression or eqn involving $\ln\!\left(\frac{y-3}{4}\right)$ or $\ln\!\left(\frac{x-3}{4}\right)$ | M1 | Or equivs such as $\ln\!\left(\frac{4}{y-3}\right)$ or $\ln\!\left(\frac{4}{x-3}\right)$ |
| Obtain $\ln\!\left(\frac{4}{x-3}\right)$ or $-\ln\!\left(\frac{x-3}{4}\right)$ | A1 | Or equiv |
| State domain is $x>3$ or equiv | B1FT | Following answer to part (i) (but with adjustment so that reference is to $x$) |
| State range is all real numbers or equiv | B1 | |

### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain correct first iterate | B1 | Showing at least 3 dp; B0 if initial value not 3 but then M1A1A1 available |
| Show correct iteration process | M1 | Showing at least 3 iterates in all; may be implied by plausible converging values; M1 available if based on equation with just a slip in $x=f(x)$ but M0 if based on clearly different equation |
| Obtain at least 3 correct iterates | A1 | Allowing recovery after error; iterates to only 3 dp acceptable; values may be rounded or truncated |
| Obtain (3.168, 3.168) | A1 | Each coordinate required to exactly 3 dp; award A0 if fewer than 4 iterates shown; part (iii) consisting of answer only gets 0 out of 4. $[3 \to 3.199148 \to 3.163187 \to 3.169162 \to 3.168155 \to 3.168324]$ |

### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| State $P$ is point where the curves meet | B1 | Or equiv |

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7\\
\includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-3_428_751_703_641}

The diagram shows the curve $y = \mathrm { f } ( x )$, where f is the function defined for all real values of $x$ by

$$\mathrm { f } ( x ) = 3 + 4 \mathrm { e } ^ { - x }$$

(i) State the range of f .\\
(ii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$, and state the domain and range of $\mathrm { f } ^ { - 1 }$.\\
(iii) The straight line $y = x$ meets the curve $y = \mathrm { f } ( x )$ at the point $P$. By using an iterative process based on the equation $x = \mathrm { f } ( x )$, with a starting value of 3 , find the coordinates of the point $P$. Show all your working and give each coordinate correct to 3 decimal places.\\
(iv) How is the point $P$ related to the curves $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$ ?

\hfill \mbox{\textit{OCR C3 2013 Q7 [10]}}