It is given that $\mathrm{f}\left( x \right)={{\mathrm{e}}^{{{\tan }^{-1}}x}}$, where \(

Question

It is given that $\mathrm{f}\left( x \right)={{\mathrm{e}}^{{{\tan }^{-1}}x}}$, where ${{\tan }^{-1}}x$ denotes the principal values. Show that $\left( 1+{{x}^{2}} \right)\mathrm{f}\,\mathrm{ }\!\!'\!\!\mathrm{ }\left( x \right)=\mathrm{f}\left( x \right)$.[1] By further differentiation of the above result, find the Maclaurin series for $\mathrm{f}(x)$, up to and including the term in ${{x}^{3}}$.[5] Using the series in part (ii) find an approximate value of $\displaystyle\displaystyle \displaystyle\int_{0}^{0.5}{\mathrm{f}\left( x \right)\,}\mathrm{d}x$, giving your answer to $4$ significant figures.[1] Comment on the suitability of substituting $x=1$ into the series in part (ii) to estimate the value of ${{\mathrm{e}}^{\frac{\pi }{4}}}$.[1]

Accepted Answer

(i) $(1+x^2)f'(x)=f(x)$ (shown) (ii) $f(x)=1+x+\frac{x^2}{2}-\frac{x^3}{6}+O(x^4)$ (iii) $0.6432$ (4 s.f.) (iv) Not suitable because $x=1$ is not close to the Maclaurin centre $x=0$

2024 JPJC P2 Q3

2024 JPJC P2 Q3

Video Solution:

Solution:

Solution locked