Find the value of \(a\) .

Show that for this value of \(a\) there is a unique real solution to the equation \(f (x) = 0\) .

\(f(a) = 4{a^3} + 2{a^2} - 7a = - 10\)     M1

\(4{a^3} + 2{a^2} - 7a + 10 = 0\)

\(\left( {a + 2} \right)\left( {4{a^2} - 6a + 5} \right) = 0\) or sketch or GDC     (M1)

\(a = - 2\)     A1

[3 marks]

substituting \(a = - 2\) into \(f (x)\)

\(f(x) = 4{x^3} - 4x + 14 = 0\)     A1

EITHER

graph showing unique solution which is indicated (must include max and min)     R1

OR

convincing argument that only one of the solutions is real     R1

(−1.74, 0.868 ±1.12i)

[5 marks]

Candidates found this question surprisingly challenging. The most straightforward approach was use of the Remainder Theorem but a significant number of candidates seemed unaware of this technique. This lack of knowledge led many candidates to attempt an algebraically laborious use of long division. In (b) a number of candidates did not seem to appreciate the significance of the word unique and hence found it difficult to provide sufficient detail to make a meaningful argument. However, most candidates did recognize that they needed a technological approach when attempting (b).

Candidates found this question surprisingly challenging. The most straightforward approach was use of the Remainder Theorem but a significant number of candidates seemed unaware of this technique. This lack of knowledge led many candidates to attempt an algebraically laborious use of long division. In (b) a number of candidates did not seem to appreciate the significance of the word unique and hence found it difficult to provide sufficient detail to make a meaningful argument. However, most candidates did recognize that they needed a technological approach when attempting (b).