Hint
Write the numerator as \(2(x-1)+5\). Keep every denominator restriction.
Worked solution
(a)
Since \(2x+3=2(x-1)+5\), the required form follows. The fraction cannot be zero, so the output cannot be 2. Conversely every real y other than 2 gives a valid input \(\displaystyle x= 1+ \frac{5}{y- 2}\).
Answer: \(\displaystyle f(x)= 2+ \frac{5}{x- 1}\); range \(\mathbb R\setminus\{2\}\).
(b)
\(y(x-1)=2x+3\), so \(x(y-2)=y+3\). Hence \(\displaystyle f^{- 1}(x)= \frac{x+ 3}{x- 2}\), with \(\displaystyle x\ne 2\).
Answer: \(\displaystyle \frac{x+ 3}{x- 2}\), \(x\ne2\).
(c)
\(x+3=4(x-2)\), giving \(\displaystyle x= \frac{11}{3}\). This is not the excluded value 2. Equivalently, \(\displaystyle x= f(4)= \frac{11}{3}\).
Answer: \(\displaystyle x= \frac{11}{3}\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Obtain \(\displaystyle 2+ \frac{5}{x- 1}\).
- 1 mark: Exclude exactly the output 2, with a nonzero-fraction justification.
Alternative methods: Rearranging \(\displaystyle y=f(x)\) and showing all \(\displaystyle y\ne 2\) are attainable is accepted.
Common error: Excluding the input value 1 from the range instead of the output value 2.
(b) · 3 marks
- 1 mark: Collect original x-terms correctly after cross-multiplication.
- 1 mark: Obtain \(\displaystyle \frac{x+ 3}{x- 2}\).
- 1 mark: State \(\displaystyle x\ne 2\) as the inverse domain.
Alternative methods: Equivalent algebraic forms are accepted.
Common error: Cancelling terms across a sum or losing the excluded value.
(c) · 2 marks
- 1 mark: Use a correct equation or the inverse relation \(\displaystyle x=f(4)\).
- 1 mark: Obtain the permitted solution \(\displaystyle \frac{11}{3}\).
Alternative methods: Evaluating f(4) directly is accepted.
Common error: Returning 4 as the input rather than reversing the mapping.
Total: 7 marks.