9231 P11 - Nov 2025 - Q3 - 7 marks
5883
Prove by mathematical induction that, for every positive integer \(n\),
\(\frac{d^{2n-1}}{dx^{2n-1}}(x\cos x)=(-1)^n\left(x\sin x-(2n-1)\cos x\right).\)
Solution
Answer: The identity is true for all positive integers \(n\).
Base case. For \(n=1\),
\(\frac d{dx}(x\cos x)=\cos x-x\sin x=(-1)^1(x\sin x-(2\cdot1-1)\cos x),\)
so the result is true for \(n=1\).
Inductive step. Assume the result is true for \(n=k\):
\(\frac{d^{2k-1}}{dx^{2k-1}}(x\cos x)=(-1)^k(x\sin x-(2k-1)\cos x).\)
Differentiate once:
\(\frac{d^{2k}}{dx^{2k}}(x\cos x)=(-1)^k(x\cos x+2k\sin x).\)
Differentiate again:
\(\frac{d^{2k+1}}{dx^{2k+1}}(x\cos x)=(-1)^k(-x\sin x+\cos x+2k\cos x).\)
Thus
\(\frac{d^{2k+1}}{dx^{2k+1}}(x\cos x)=(-1)^{k+1}(x\sin x-(2k+1)\cos x).\)
This is the required result for \(n=k+1\). Therefore, by induction, the identity is true for all positive integers \(n\).
Mark scheme
| Part | Marks | Expected result | Notes |
|---|
| 3 | B1 | \(\frac d{dx}(x\cos x)=\cos x-x\sin x=(-1)^1(x\sin x-(2(1)-1)\cos x)\) | Checks base case using product rule. |
| 3 | B1 | Assume true for \(n=k\), so \(\frac{d^{2k-1}}{dx^{2k-1}}(x\cos x)=(-1)^k(x\sin x-(2k-1)\cos x)\). | States inductive hypothesis. |
| 3 | M1 A1 | \(\frac{d^{2k}}{dx^{2k}}(x\cos x)=(-1)^k(x\cos x+2k\sin x)\) | Differentiates once. |
| 3 | M1 A1 | \(\frac{d^{2k+1}}{dx^{2k+1}}(x\cos x)=(-1)^k(-x\sin x+\cos x+2k\cos x)=(-1)^{k+1}(x\sin x-(2k+1)\cos x)\) | Differentiates again. |
| 3 | A1 | Conclusion: true for \(n=k+1\), hence true for all positive integers. | States conclusion. |