Compose sine and logarithm carefully · source reference

Printed question

It is given that $y=\ln(2+\sin2x)$.

(i) Show that $e^y\frac{d^2y}{dx^2}+e^y(\frac{dy}{dx})^2=-4\sin2x$. [2]

(ii) By further differentiation of the above results, find the Maclaurin series for $y$, up to and including the term in $x^3$. [3]

(iii) Verify that the series for $\ln(2+\sin2x)$ is the same as the result obtained in part (ii), if the standard series from the List of Formulae (MF26) are used. [3]

(iv) Hence or otherwise, find the series expansion for $\frac{\ln(2+\sin2x)}{\sqrt{1-x}}$, up to and including the term in $x^2$. [2]

[2022/ACJC/Prelim/I/7]

Printed working

$e^y=2+\sin2x$, $e^yy'=2\cos2x$, $e^yy''+e^y(y')^2=-4\sin2x$.

Differentiate: $e^yy'''+(e^yy')y''+e^y(2y'y'')+e^yy'(y')^2=-8\cos2x$, so $e^yy'''+3e^yy'y''+e^y(y')^3=-8\cos2x$.

Printed source discrepancy: p.17 shows $-2\cos2x$ on its two third-derivative lines; differentiation requires $-8\cos2x$. The subsequent values $y' =1,y''=-1,y'''=-2$ and final series agree with $-8$.

At 0: $y=\ln2$, $y'=1$, $y''=-1$, $y'''=-2$. Hence $y=\ln2+x-x^2/2-x^3/3+\cdots$.

Standard-series verification: $\ln[2+2x-(2x)^3/6]=\ln[2(1+x-2x^3/3)]=\ln2+\ln(1+x-2x^3/3)$.

$=\ln2+(x-2x^3/3)-\frac12(x-2x^3/3)^2+\frac13(x-2x^3/3)^3+\cdots=\ln2+x-x^2/2-x^3/3+\cdots$.

$(\ln2+x-x^2/2)(1+x/2+3x^2/8)=\ln2+(1+\ln2/2)x+(3\ln2/8)x^2+\cdots$.

Teacher note 1

Let $u=y'$. Using chain rule: $\frac d{dx}(y')^2=\frac d{du}u^2\frac{du}{dx}=2u\frac{du}{dx}=2y'y''$.

Teacher note 2

From MF27 replace $x$ by $2x$ in $\sin x=x-x^3/3!+\cdots$ to obtain $\sin2x$. Factorise out 2 to make the constant term 1. Apply $\ln(ab)=\ln a+\ln b$. Use $\ln(1+x)=x-x^2/2+x^3/3-\cdots$, replacing $x$ by $x-2x^3/3$.

Teacher note 3

Use the result from (iii). Use binomial expansion for $(1-x)^{-1/2}$.

Mathematical review

(i) $e^y=2+\sin2x$; two differentiations yield $e^y[(y')^2+y'']=-4\sin2x$.

(ii) At zero, $y=\ln2,y'=1,y''=-1$. Differentiate (i): $e^y[y^{(3)}+3y'y''+(y')^3]=-8\cos2x$; at zero this gives $2[y^{(3)}-3+1]=-8$, so $y^{(3)}=-2$. Thus $y=\ln2+x-x^2/2-x^3/3+O(x^4)$.

(iii) $\ln(2+\sin2x)=\ln2+\ln(1+x-2x^3/3+\cdots)$ gives the same terms. (iv) Multiply by $(1-x)^{-1/2}=1+x/2+3x^2/8+\cdots$, obtaining $\ln2+(1+\tfrac12\ln2)x+\tfrac38\ln2\,x^2+O(x^3)$.

The printed third-derivative right side on page 17 is $-2\cos2x$; it must be $-8\cos2x$. The source’s evaluated derivatives and final series remain correct.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 16. Working page(s): 16, 17, 18. 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.