2010 A-level H2 Mathematics (9740) Paper 2 Question 4 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.




For f^{-1} to exist f must be one-one, thus least value of k=0


= f(\frac{1}{x-3})

= \frac{1}{\frac{1}{(x-3)^2}-1}

= \frac{(x-3)^2}{(4-x)(x-2)}

fg(x) > 0

\frac{(x-3)^2}{(4-x)(x-2)} > 0

Since (x-3)^2 \ge 0, ((4-x)(x-2) > 0

\therefore, 2 \textless x \textless 4, x \neq 3

Range of fg(x) = (-\infty, -1 ) \cup (0, \infty)

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