IBDP Maths AHL 5.19 Maclaurin series AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)(i)
Let \(f(x)=(1-x)^{-4}\).
The Maclaurin series of \(f\) begins with:
\(f(x)=f(0)+f'(0)x+\dfrac{f”(0)}{2!}x^2+\dfrac{f”'(0)}{3!}x^3+\cdots\)
Differentiate \(f(x)\).
\(f'(x)=4(1-x)^{-5}\)
Substitute \(x=0\).
\(f'(0)=4(1-0)^{-5}=4\)
The coefficient of \(x\) is \(f'(0)\), so:
\(a=4\)
✅ Hence, \(a=4\).
(a)(ii)
Differentiate again.
\(f”(x)=20(1-x)^{-6}\)
Therefore:
\(f”(0)=20\)
The coefficient of \(x^2\) in a Maclaurin series is \(\dfrac{f”(0)}{2!}\).
\(b=\dfrac{20}{2!}=\dfrac{20}{2}=10\)
Thus, the first four terms are:
\((1-x)^{-4}\approx1+4x+10x^2+20x^3\)
✅ Answer: \(b=10\)
(b)
A depreciation of \(10\%\) each year means that the car retains \(90\%\), or \(0.9\), of its value each year.
If \(V\) is the value four years ago, then:
\(1000=V(0.9)^4\)
Therefore:
\(V=1000(0.9)^{-4}=1000(1-0.1)^{-4}\)
Using the Maclaurin approximation from part (a) with \(x=0.1\):
\((1-0.1)^{-4}\approx1+4(0.1)+10(0.1)^2+20(0.1)^3\)
\((1-0.1)^{-4}\approx1+0.4+0.1+0.02\)
\((1-0.1)^{-4}\approx1.52\)
Hence:
\(V\approx1000(1.52)=1520\)
The value is an estimate because only the first four terms of the infinite Maclaurin series have been used.
✅ Answer: \(\$1520\)