OCR MEI C3 2010 January — Question 6 4 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2010
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeVerify composite identity
DifficultyModerate -0.3 This question tests recall of standard definitions (odd/even functions) and requires a straightforward algebraic proof using those definitions. The proof involves simple substitution: gf(-x) = g(f(-x)) = g(-f(x)) = g(f(x)), which is a direct application of the definitions with no novel insight required. While it's a proof question, it's more routine than the average A-level question.
Spec1.02u Functions: definition and vocabulary (domain, range, mapping)

6 Write down the conditions for \(\mathrm { f } ( x )\) to be an odd function and for \(\mathrm { g } ( x )\) to be an even function.
Hence prove that, if \(\mathrm { f } ( x )\) is odd and \(\mathrm { g } ( x )\) is even, then the composite function \(\mathrm { gf } ( x )\) is even.

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(f(-x) = -f(x)\), \(g(-x) = g(x)\)B1B1 condone f and g interchanged
\(gf(-x) = g[{-f(x)}] = gf(x)\)M1 forming \(gf(-x)\) or \(gf(x)\) and using \(f(-x)=-f(x)\)
\(\Rightarrow gf\) is evenE1 [4] www
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(-x) = -f(x)$, $g(-x) = g(x)$ | B1B1 | condone f and g interchanged |
| $gf(-x) = g[{-f(x)}] = gf(x)$ | M1 | forming $gf(-x)$ or $gf(x)$ and using $f(-x)=-f(x)$ |
| $\Rightarrow gf$ is even | E1 [4] | www |

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6 Write down the conditions for $\mathrm { f } ( x )$ to be an odd function and for $\mathrm { g } ( x )$ to be an even function.\\
Hence prove that, if $\mathrm { f } ( x )$ is odd and $\mathrm { g } ( x )$ is even, then the composite function $\mathrm { gf } ( x )$ is even.

\hfill \mbox{\textit{OCR MEI C3 2010 Q6 [4]}}