Hint
An equation in \(\displaystyle k^{2}\) does not by itself choose a sign. Use the additional information only when it is introduced.
Worked solution
(a)
The coefficient is \(\binom82k^2=28k^2\). Hence \(\displaystyle 28k^{2}=112\), giving \(\displaystyle k^{2}=4\) and both real values \(\displaystyle k=\pm 2\).
Answer: \(k=-2\) or \(k=2\).
(b)
The coefficient is \(\binom83 k^3=56k^3\). Its sign is the sign of k, so the additional condition selects \(\displaystyle k=-2\). The coefficient is \(\displaystyle 56(-8)=-448\).
Answer: \(k=-2\); coefficient −448.
(c)
\(\binom84(-2)^4=70(16)=1120\). An even power makes this coefficient positive.
Answer: 1120.
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) · 3 marks
- 1 mark: Set \(\displaystyle 28k^{2}=112\).
- 1 mark: Obtain \(\displaystyle k^{2}=4\).
- 1 mark: Retain both \(\displaystyle k=-2\) and \(\displaystyle k=2\).
Alternative methods: A correct factorisation of \(\displaystyle k^{2}-4\) is accepted.
Common error: Selecting the positive root without any stated sign condition.
(b) · 2 marks
- 1 mark: Select \(\displaystyle k=-2\) using the sign of the odd-power coefficient.
- 1 mark: Evaluate the \(\displaystyle x^{3}\) coefficient as −448.
Alternative methods: Comparing the two candidate cubic coefficients directly is accepted.
Common error: Using the negative sign of a squared quantity to select a branch.
(c) · 2 marks
- 1 mark: Use \(\displaystyle C(8,4)k^{4}\) with the selected parameter.
- 1 mark: Give 1120.
Alternative methods: A full expansion using \(\displaystyle k=-2\) is accepted.
Common error: Making every coefficient negative because k is negative.
Total: 7 marks.