Standard +0.8 This is a standard induction proof on derivatives requiring the product rule and chain rule, but involves factorial notation and alternating signs which students often find tricky. The base case and algebraic manipulation in the inductive step are moderately demanding but follow a well-established template for this type of question.
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]}}