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Original question & working

Read a tangent from the first two coefficients · source reference

[modified DHS/2008/Prelim/I/5] · Question pp. 5 · Working pp. 5

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Printed question

Find the first three terms in the expansion of 1+x23−2x\frac{1+x^2}{\sqrt{3-2x}} in ascending powers of xx.

State the range of values of xx for which the expansion is valid. [3]

Deduce that the equation of the tangent to the curve y=2+2x23−2xy=\frac{2+2x^2}{\sqrt{3-2x}} at the point where x=0x=0, is given by 9y=23x+639y=2\sqrt3x+6\sqrt3. [2]

[modified DHS/2008/Prelim/I/5]

Printed working

1+x23−2x=(1+x2)(3−2x)−1/2=13(1+x2)(1−2x/3)−1/2=13(1+x2)[1+(−1/2)(−2x/3)+(−1/2)(−3/2)2!(−2x/3)2+⋯ ]=13(1+x2)(1+x/3+x2/6+⋯ )=13+x33+7x263+⋯ .\begin{aligned}\frac{1+x^2}{\sqrt{3-2x}}&=(1+x^2)(3-2x)^{-1/2}=\frac1{\sqrt3}(1+x^2)(1-2x/3)^{-1/2}\\&=\frac1{\sqrt3}(1+x^2)[1+(-1/2)(-2x/3)+\tfrac{(-1/2)(-3/2)}{2!}(-2x/3)^2+\cdots]\\&=\frac1{\sqrt3}(1+x^2)(1+x/3+x^2/6+\cdots)\\&=\frac1{\sqrt3}+\frac{x}{3\sqrt3}+\frac{7x^2}{6\sqrt3}+\cdots.\end{aligned}

∣−2x/3∣<1⇒∣x∣<3/2⇒−3/2<x<3/2|-2x/3|<1\Rightarrow|x|<3/2\Rightarrow-3/2<x<3/2.

y=2/3+2x/(33)+14x2/(63)+⋯y=2/\sqrt3+2x/(3\sqrt3)+14x^2/(6\sqrt3)+\cdots; at x=0x=0, y=2/3y=2/\sqrt3, y′=2/(33)y'=2/(3\sqrt3). Tangent y=2/3+2x/(33)⇒33y=6+2x⇒9y=63+23xy=2/\sqrt3+2x/(3\sqrt3)\Rightarrow3\sqrt3y=6+2x\Rightarrow9y=6\sqrt3+2\sqrt3x.

Teacher note 1

Need to factorise out 33 to make the constant term 1. Use the law of indices (AB)n=AnBn(AB)^n=A^nB^n. Then use the binomial theorem

(1+x)n=1+nx+n(n−1)2!x2+⋯ ,(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots,

where n=−1/2n=-1/2 and replace xx by −2x/3-2x/3.

Teacher note 2

At x=0x=0, y=2/3y=2/\sqrt3 and y′=2/(33)y'=2/(3\sqrt3). Thus tangent at x=0x=0 is y−2/3=(2/(33))(x−0)y-2/\sqrt3=(2/(3\sqrt3))(x-0). The constant and term in xx are the first two terms in the expansion of yy.

Mathematical review

(3−2x)−1/2=3−1/2(1+x/3+x2/6+O(x3))(3-2x)^{-1/2}=3^{-1/2}(1+x/3+x^2/6+O(x^3)). Multiplication gives (1+x2)/3−2x=1/3+x/(33)+7x2/(63)+O(x3)(1+x^2)/\sqrt{3-2x}=1/\sqrt3+x/(3\sqrt3)+7x^2/(6\sqrt3)+O(x^3) for ∣x∣<3/2|x|<3/2.

The curve is twice this expression: y(0)=2/3y(0)=2/\sqrt3, y′(0)=2/(33)y'(0)=2/(3\sqrt3). Its tangent is y=2/3+2x/(33)y=2/\sqrt3+2x/(3\sqrt3), equivalent to 9y=63+23 x9y=6\sqrt3+2\sqrt3\,x.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 5. Working page(s): 5. The two Series source files are byte-identical; their shared visible content is expressed once and mapped to both page sets. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.