Standard +0.8 This is a standard induction proof on derivatives requiring the product rule and chain rule, but involves factorial notation and careful algebraic manipulation with powers. It's more demanding than typical A-level questions due to being Further Maths content and requiring rigorous proof technique, but follows a predictable induction structure without novel insight.
It is given that \(y = \ln(ax + 1)\), where \(a\) is a positive constant. Prove by mathematical induction that, for every positive integer \(n\),
$$\frac{d^n y}{dx^n} = (-1)^{n-1} \frac{(n-1)!a^n}{(ax+1)^n}.$$ [6]
Question 2:
2 | dy = a =( −1 )0 0!a1
dx ax+1 ( ax+1 )1
so true for n=1. | M1 A1 | Proves base case.
dky (k −1)!ak
= (−1)k−1
dxk (ax+1)k
Assume that for some positive integer k. | B1 | States inductive hypothesis.
Then
dk+1y (k −1)!ak k!ak+1
= −ka(−1)k−1 =(−1)k
dxk+1 (ax+1)k+1 (ax+1)k+1
so true for n=k+1. | M1 A1 | Differentiates k th derivative.
By induction, true for every positive integer n. | A1 | States conclusion.
6
Question | Answer | Marks | Guidance
It is given that $y = \ln(ax + 1)$, where $a$ is a positive constant. Prove by mathematical induction that, for every positive integer $n$,
$$\frac{d^n y}{dx^n} = (-1)^{n-1} \frac{(n-1)!a^n}{(ax+1)^n}.$$ [6]
\hfill \mbox{\textit{CAIE FP1 2019 Q2 [6]}}