It is given that $y=\mathrm{f}(x)$ is such that \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\

Question

It is given that $y=\mathrm{f}(x)$ is such that $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}y}{\mathrm{d}x}\left(2\tan x+p\sec^2x\right),$$ where $p$ is a constant. The Maclaurin series for $\mathrm{f}(x)$ is given by $$y=1+x+\frac12x^2+qx^3+\cdots,$$ where $q$ is a constant. State the values of $\mathrm{f}(0)$ and $\mathrm{f}'(0)$.[1] Find the values of $p$ and $q$.[5] Using the given series for $\mathrm{f}(x)$ and standard series from the List of Formulae (MF27), find the first three non-zero terms of the series expansion of $\sqrt{\mathrm{f}(x)}$.[2] The given Maclaurin series for $\mathrm{f}(x)$ is used to estimate $\displaystyle \int_0^{0.2}\mathrm{f}(x)\,\mathrm{d}x$ and $\displaystyle \int_{0.2}^{0.4}\mathrm{f}(x)\,\mathrm{d}x$. Without any further computations, determine which estimate has a smaller percentage error. Give an explanation for your choice.[1]

Accepted Answer

(a) $\mathrm{f}(0)=1$, $\mathrm{f}'(0)=1$ (b) $p=1$, $q=\frac12$ (c) $1+\frac12x+\frac18x^2$ (d) $\int_0^{0.2}\mathrm{f}(x)\,\mathrm{d}x$

2026 TJC P1 Q5

2026 TJC P1 Q5

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