Easy -1.2 This is a straightforward application of the Remainder Theorem requiring only substitution of x=2 into f(x), setting equal to 3, and solving a simple linear equation for a. It's a single-step problem testing basic recall of the theorem with minimal algebraic manipulation, making it easier than average.
1.
$$f ( x ) = a x ^ { 3 } + 3 x ^ { 2 } - 8 x + 2 \quad \text { where } a \text { is a constant }$$
Given that when \(\mathrm { f } ( x )\) is divided by \(( x - 2 )\) the remainder is 3 , find the value of \(a\).
Attempts to set \(f(2) = 3\). Values embedded and expression set equal to 3 is sufficient. May be implied by further work. Condone slips in evaluation if intent to substitute 2 is clear. May also be seen as \((f(2)-3=)\ a(2)^3 + 3(2)^2 - 8(2) - 1 = 0\)
\(\Rightarrow 8a = 5 \Rightarrow a = \dfrac{5}{8}\)
M1A1
M1: Solves a linear equation in \(a\) arising from setting \(f(\pm 2) = \pm 3\). Condone slips in rearrangement proceeding to \(a = ...\). A1: \(a = \dfrac{5}{8}\) or exact equivalent
## Question 1:
$f(x) = ax^3 + 3x^2 - 8x + 2$
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sets $f(2) = 3 \Rightarrow 8a + 12 - 16 + 2 = 3$ | M1 | Attempts to set $f(2) = 3$. Values embedded and expression set equal to 3 is sufficient. May be implied by further work. Condone slips in evaluation if intent to substitute 2 is clear. May also be seen as $(f(2)-3=)\ a(2)^3 + 3(2)^2 - 8(2) - 1 = 0$ |
| $\Rightarrow 8a = 5 \Rightarrow a = \dfrac{5}{8}$ | M1A1 | M1: Solves a linear equation in $a$ arising from setting $f(\pm 2) = \pm 3$. Condone slips in rearrangement proceeding to $a = ...$. A1: $a = \dfrac{5}{8}$ or exact equivalent |
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1.
$$f ( x ) = a x ^ { 3 } + 3 x ^ { 2 } - 8 x + 2 \quad \text { where } a \text { is a constant }$$
Given that when $\mathrm { f } ( x )$ is divided by $( x - 2 )$ the remainder is 3 , find the value of $a$.
\hfill \mbox{\textit{Edexcel P2 2024 Q1 [3]}}