Challenging +1.2 This is a standard induction proof on a derivative formula with moderate algebraic complexity. The base case is straightforward (differentiate x sin x once), and the inductive step requires applying the product rule and using standard trigonometric derivatives. While it involves careful bookkeeping of the (-1)^(n-1) term and coefficient (2n-1), it follows a predictable template for differentiation induction proofs without requiring novel insight or particularly challenging manipulation.
5 Prove by mathematical induction that, for every positive integer \(n\),
$$\frac { d ^ { 2 n - 1 } } { d x ^ { 2 n - 1 } } ( x \sin x ) = ( - 1 ) ^ { n - 1 } ( x \cos x + ( 2 n - 1 ) \sin x )$$
\(\frac{d}{dx}(x\sin x) = x\cos x + \sin x = (-1)^0(x\cos x + (2(1)-1)\sin x)\)
B1
Checks base case using product rule
Assume true for \(n=k\), so \(\frac{d^{2k-1}}{dx^{2k-1}}(x\sin x) = (-1)^{k-1}(x\cos x + (2k-1)\sin x)\)
B1
States inductive hypothesis
Then \(\frac{d^{2k}}{dx^{2k}}(x\sin x) = (-1)^{k-1}(-x\sin x + 2k\cos x)\)
M1 A1
Differentiates once. Must have correct LHS for A1
\(\frac{d^{2k+1}}{dx^{2k+1}}(x\sin x) = (-1)^{k-1}(-x\cos x - \sin x - 2k\sin x)\) \(= (-1)^k(x\cos x + (2k+1)\sin x)\)
M1 A1
Differentiates again
So it is also true for \(n = k+1\). Hence, by induction, true for all positive integers.
A1
States conclusion
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(x\sin x) = x\cos x + \sin x = (-1)^0(x\cos x + (2(1)-1)\sin x)$ | B1 | Checks base case using product rule |
| Assume true for $n=k$, so $\frac{d^{2k-1}}{dx^{2k-1}}(x\sin x) = (-1)^{k-1}(x\cos x + (2k-1)\sin x)$ | B1 | States inductive hypothesis |
| Then $\frac{d^{2k}}{dx^{2k}}(x\sin x) = (-1)^{k-1}(-x\sin x + 2k\cos x)$ | M1 A1 | Differentiates once. Must have correct LHS for A1 |
| $\frac{d^{2k+1}}{dx^{2k+1}}(x\sin x) = (-1)^{k-1}(-x\cos x - \sin x - 2k\sin x)$ $= (-1)^k(x\cos x + (2k+1)\sin x)$ | M1 A1 | Differentiates again |
| So it is also true for $n = k+1$. Hence, by induction, true for all positive integers. | A1 | States conclusion |
5 Prove by mathematical induction that, for every positive integer $n$,
$$\frac { d ^ { 2 n - 1 } } { d x ^ { 2 n - 1 } } ( x \sin x ) = ( - 1 ) ^ { n - 1 } ( x \cos x + ( 2 n - 1 ) \sin x )$$
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q5 [7]}}