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Original question & working

The decreasing sine branch · source reference

[2016/NYJC/Prelim/I/10a] · Question pp. 3 · Working pp. 15

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The working and teaching notes are included in the interactive lesson.

Source discrepancy: solution PDF p.15 prints arcsin(x/2)−π/3. This maps 2 to π/6 but maps −2 to −5π/6, outside the required domain. Porisma shows the corrected branch and retains the original source.

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Full original question, working & notes

Complete question

(i) Express sin⁡x+3cos⁡x\sin x+\sqrt3\cos x as Rsin⁡(x+α)R\sin(x+\alpha) with exact positive R and alpha. [1] (ii) Define f by that expression on π/6≤x≤k\pi/6\le x\le k. Find the largest exact k for which f has an inverse, define the inverse including its domain, and state the inputs where ff−1(x)=f−1f(x)ff^{-1}(x)=f^{-1}f(x). [5]

[2016/NYJC/Prelim/I/10a]

Printed working

Part (i)

sin⁡x+3cos⁡x=Rsin⁡(x+α)=Rcos⁡αsin⁡x+Rsin⁡αcos⁡x.\sin x+\sqrt3\cos x=R\sin(x+\alpha)=R\cos\alpha\sin x+R\sin\alpha\cos x.

tan⁡α=3/1⇒α=π/3\tan\alpha=\sqrt3/1\Rightarrow\alpha=\pi/3 and R=12+(3)2=2R=\sqrt{1^2+(\sqrt3)^2}=2.

Thus sin⁡x+3cos⁡x=2sin⁡(x+π/3)\sin x+\sqrt3\cos x=2\sin(x+\pi/3).

Part (ii): printed working

f(x)=2sin⁡(x+π/3)f(x)=2\sin(x+\pi/3), π/6≤x≤k\pi/6\le x\le k. For an inverse, ff must be one-one. The largest k=7π/6k=7\pi/6.

The source plot shows the closed endpoints (π/6,2)(\pi/6,2), (7π/6,−2)(7\pi/6,-2), the x-intercept (2π/3,0)(2\pi/3,0) and dotted continuations of the unrestricted sine curve.

Let y=2sin⁡(x+π/3)y=2\sin(x+\pi/3), π/6≤x≤7π/6\pi/6\le x\le7\pi/6. The source prints

x=sin⁡−1(y/2)−π/3,f−1:x↦sin⁡−1(x/2)−π/3,−2≤x≤2.x=\sin^{-1}(y/2)-\pi/3,\qquad f^{-1}:x\mapsto\sin^{-1}(x/2)-\pi/3,\quad-2\le x\le2.

The printed inverse domain Df−1=Rf=[−2,2]D_{f^{-1}}=R_f=[-2,2] is correct, but the printed inverse rule is the wrong sine branch.

Corrected inverse branch

On the decreasing half-wave, x+π/3=π−sin⁡−1(y/2)x+\pi/3=\pi-\sin^{-1}(y/2). Thus

f−1(x)=2π3−sin⁡−1(x/2),−2≤x≤2.f^{-1}(x)=\frac{2\pi}3-\sin^{-1}(x/2),\qquad-2\le x\le2.

Composite domain comparison

Both identity composites equal xx where defined. The printed working gives

x∈Df∩Df−1=[π/6,7π/6]∩[−2,2]=[π/6,2].x\in D_f\cap D_{f^{-1}}=[\pi/6,7\pi/6]\cap[-2,2]=[\pi/6,2].

Alternatively {x∈R:π/6≤x≤2}\{x\in\mathbb R:\pi/6\le x\le2\}. This composite-domain answer is unchanged after correcting the branch.

Teacher margin notes

Dff−1=Df−1D_{ff^{-1}}=D_{f^{-1}} and Df−1f=DfD_{f^{-1}f}=D_f. To compare them, intersect their domains: x∈Dff−1∩Df−1f⇒x∈Df−1∩Df⇒x∈[−2,2]∩[π/6,7π/6]=[π/6,2]x\in D_{ff^{-1}}\cap D_{f^{-1}f}\Rightarrow x\in D_{f^{-1}}\cap D_f\Rightarrow x\in[-2,2]\cap[\pi/6,7\pi/6]=[\pi/6,2].

Source diagrams

Original restricted descending sine half-wave with closed endpoints(π/6,2),(7π/6,−2), zero2π/3 and dotted unrestricted continuations.

Review qualification

Source discrepancy: solution PDF p.15 prints arcsin(x/2)−π/3. This maps 2 to π/6 but maps −2 to −5π/6, outside the required domain. Porisma shows the corrected branch and retains the original source.

The corrected or qualified authored derivation is retained in the question article.

Provenance

Question page references: 3. Solution page references: 15. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.