Expand ${{\left( b-\frac{x}{2} \right)}^{n}}$ in ascending powers of $x$, where $b$ is a

Question

Expand ${{\left( b-\frac{x}{2} \right)}^{n}}$ in ascending powers of $x$, where $b$ is a positive constant and $n$ is a negative constant, up to and including the term in ${{x}^{2}}$.[2] It is given that the coefficient of $x$ is four times the coefficient of ${{x}^{2}}$ and the constant term in the expansion is $\frac{1}{2}$. Find the integer values of $b$ and $n$.[3] Explain why it is not possible to obtain a Maclaurin series for $\ln \left( 2{{x}^{2}} \right)$.[1] A Taylor series is an expansion of a real function $\mathrm{f}\left( x \right)$ about a point $x=a$ and it is defined by $\mathrm{f}\left( x \right)=\mathrm{f}\left( a \right)+\mathrm{f}'\left( a \right)\left( x-a \right)+\frac{\mathrm{f}''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+...+\frac{{{\mathrm{f}}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}+...$ where ${{\mathrm{f}}^{\left( n \right)}}\left( a \right)$ is the value of the $n$th derivative of $\mathrm{f}\left( x \right)$ when $x=a$. Find the first three exact non-zero terms of the Taylor series for $\ln \left( 2{{x}^{2}} \right)$ about the point $x=2$. [You need not simplify your answer.][3]

Accepted Answer

(a) $b=2$, $n=-1$ (b)(i) No Maclaurin series exists because $\ln(2x^2)$ is undefined at $x=0$ (b)(ii) $\ln(2x^2)=\ln8+(x-2)-\dfrac14(x-2)^2+\cdots$

2021 TMJC P2 Q4

2021 TMJC P2 Q4

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