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).