Moderate -0.8 This is a straightforward application of the Remainder Theorem requiring only substitution of x=2 into f(x), setting equal to 18, and solving a simple linear equation for k. It's easier than average as it's a direct single-concept question with minimal steps, though not trivial since students must recall and correctly apply the theorem.
or long division oe as far as obtaining a remainder (ie not involving \(x\)) and equating that remainder to 18 (there may be errors along the way)
\(32 + 2k - 20 = 18\) oe
A1
after long division: \(2(k + 16) - 20 = 18\) oe; A0 for just \(2k\) instead of 32 unless 32 implied by further work
\([k =] 3\)
A1 [3]
$I(2) = 18$ seen or used | M1 | or long division oe as far as obtaining a remainder (ie not involving $x$) and equating that remainder to 18 (there may be errors along the way)
$32 + 2k - 20 = 18$ oe | A1 | after long division: $2(k + 16) - 20 = 18$ oe; A0 for just $2k$ instead of 32 unless 32 implied by further work
$[k =] 3$ | A1 [3]
You are given that $\text{f}(x) = x^5 + kx - 20$. When $\text{f}(x)$ is divided by $(x - 2)$, the remainder is 18. Find the value of $k$. [3]
\hfill \mbox{\textit{OCR MEI C1 2013 Q5 [3]}}