Question 1 - A-Level Maths

OCR MEI C3 — Question 1 6 marks

Exam Board	OCR MEI
Module	C3 (Core Mathematics 3)
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Composite & Inverse Functions
Type	Verify composite identity
Difficulty	Moderate -0.3 Part (i) requires straightforward substitution to show f(f(x)) = x and recognizing that this means f is self-inverse. Part (ii) involves routine verification that g(-x) = g(x) and stating the symmetry property. Both parts are standard textbook exercises testing basic function properties with minimal problem-solving required, making this 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

The function f(x) is defined by $$f(x) = \frac{1-x}{1+x}, x \neq -1.$$ Show that f(f(x)) = x. Hence write down $f^{-1}(x)$. [3]
The function g(x) is defined for all real x by $$g(x) = \frac{1-x^2}{1+x^2}.$$ Prove that g(x) is even. Interpret this result in terms of the graph of $y = g(x)$. [3]

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
1	(i)	1x

1

1x 1x

ff(x)f( ) 

1x 1x

1

1x

1x1x 2x

  x*

1x1x 2

Answer	Marks
f−1(x) = f(x) = (1−x)/(1+x)	M1

A1

Answer	Marks
B1	substituting (1−x)/(1+x) for x in f(x)

correctly simplified to x NB AG

or just f−1(x) = f(x)

[3]

Answer	Marks
(ii)	1(x)2

g(x)

1(x)2

1x2

 g(x)

1x2

Answer	Marks
Graph is symmetrical about the y-axis.	M1

A1

B1

Answer	Marks
[3]	substituting –x for x in g(x) condone use of ‘f’

for g

must indicate that g(−x) = g(x) somewhere

Answer	Marks
allow ‘reflected’, ‘reflection’ for symmetrical	if brackets are omitted or misplaced

allow M1A0

condone use of ‘f’ for g

must state axis (y-axis or x = 0)

Question 1:
1 | (i) | 1x
1
1x 1x
ff(x)f( ) 
1x 1x
1
1x
1x1x 2x
  x*
1x1x 2
f−1(x) = f(x) = (1−x)/(1+x) | M1
A1
B1 | substituting (1−x)/(1+x) for x in f(x)
correctly simplified to x NB AG
or just f−1(x) = f(x)
[3]
(ii) | 1(x)2
g(x)
1(x)2
1x2
 g(x)
1x2
Graph is symmetrical about the y-axis. | M1
A1
B1
[3] | substituting –x for x in g(x) condone use of ‘f’
for g
must indicate that g(−x) = g(x) somewhere
allow ‘reflected’, ‘reflection’ for symmetrical | if brackets are omitted or misplaced
allow M1A0
condone use of ‘f’ for g
must state axis (y-axis or x = 0)

Show LaTeX source

\begin{enumerate}[label=(\roman*)]
\item The function f(x) is defined by
$$f(x) = \frac{1-x}{1+x}, x \neq -1.$$

Show that f(f(x)) = x.

Hence write down $f^{-1}(x)$. [3]

\item The function g(x) is defined for all real x by
$$g(x) = \frac{1-x^2}{1+x^2}.$$

Prove that g(x) is even. Interpret this result in terms of the graph of $y = g(x)$. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C3  Q1 [6]}}

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Q1 6 Q2 18 Q3 8 Q4 3 Q5 4 Q6 6 Q7 3 Q8 3