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Original question & working

Keep the parameter branch when eliminating t · source reference

2025 TJC H2 Maths (9758) JC2 MSM · original exercise attribution retained on source pages · Question pp. 3 · Working pp. 10, 11

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Confirmed branch qualification: the printed Cartesian expression is valid only for 0≤t≤π/2. Counterexample t=π gives (x,y)=(π,0), while the printed expression gives x=0. Question PDF p.3 (printed 53); solution PDF p.11 explicitly adds the rising-arc restriction.

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Printed question

C: x=t−sin⁡tx=t-\sin t, y=sin⁡2ty=\sin^2t, 0≤t≤π0\le t\le\pi. A has parameter π/2\pi/2 and is a maximum point. Line l passes through the origin and A. R is bounded by C and l as shown.

(i) Find A. [1]

(ii) Using an algebraic method, find the exact area of R. [5]

(iii) Show the Cartesian equation of C is x=sin⁡−1y−yx=\sin^{-1}\sqrt y-\sqrt y. [1]

(iv) Find the volume when R is rotated 2π2\pi radians about the yy-axis. [3]

[2023/ACJC/Prelim/I/8]

Printed working

(i)

At t=π/2t=\pi/2, x=π/2−sin⁡(π/2)=π/2−1x=\pi/2-\sin(\pi/2)=\pi/2-1, y=sin⁡2(π/2)=1y=\sin^2(\pi/2)=1: A=(π/2−1,1)A=(\pi/2-1,1).

(ii)

Area =∫0π/2−1yCdx−12(π/2−1)(1)=\int_0^{\pi/2-1}y_Cdx-\frac12(\pi/2-1)(1).

∫0π/2−1yCdx=∫0π/2sin⁡2t(dx/dt)dt=∫0π/2sin⁡2t(1−cos⁡t)dt=∫0π/2(sin⁡2t−sin⁡2tcos⁡t)dt=∫0π/21−cos⁡2t2dt−[sin⁡3t/3]0π/2=12[t−sin⁡2t/2]0π/2−[sin⁡3t/3]0π/2=12(π/2)−1/3=π/4−1/3.\begin{aligned}\int_0^{\pi/2-1}y_Cdx&=\int_0^{\pi/2}\sin^2t(dx/dt)dt=\int_0^{\pi/2}\sin^2t(1-\cos t)dt\\&=\int_0^{\pi/2}(\sin^2t-\sin^2t\cos t)dt\\&=\int_0^{\pi/2}\frac{1-\cos2t}2dt-[\sin^3t/3]_0^{\pi/2}\\&=\frac12[t-\sin2t/2]_0^{\pi/2}-[\sin^3t/3]_0^{\pi/2}=\frac12(\pi/2)-1/3=\pi/4-1/3.\end{aligned}

Therefore area =(π/4−1/3)−(π/4−1/2)=1/6=(\pi/4-1/3)-(\pi/4-1/2)=1/6.

(iii)

x=t−sin⁡tx=t-\sin t, y=sin⁡2ty=\sin^2t; sin⁡t=y\sin t=\sqrt y since sin⁡t≥0\sin t\ge0. t=sin⁡−1yt=\sin^{-1}\sqrt y for 0≤t≤π/20\le t\le\pi/2, hence x=sin⁡−1y−yx=\sin^{-1}\sqrt y-\sqrt y.

(iv)

V=13π(π/2−1)2(1)−π∫01(xC)2dy=13π(π/2−1)2−π∫01(sin⁡−1y−y)2dy=0.25261≈0.253 (3 s.f.).V=\frac13\pi(\pi/2-1)^2(1)-\pi\int_0^1(x_C)^2dy=\frac13\pi(\pi/2-1)^2-\pi\int_0^1(\sin^{-1}\sqrt y-\sqrt y)^2dy=0.25261\approx0.253\ (3\text{ s.f.}).

Teacher notes

Area of blue region minus area of red triangle. Triangle rotated about yy-axis gives a cone of radius (π/2−1)(\pi/2-1) units and height 11 unit. Native panels retain A,R,O, level 1, x=π/2−1x=\pi/2-1, line l and rotation arrows.

Printed answer key

(i) A=(π/2−1,1)A=(\pi/2-1,1); (ii) 1/61/6; the short key labels the volume “(iii) 0.2530.253 (3 s.f.)”, whereas the question places volume in (iv).

Source diagrams in native form

Faithful native source geometry and annotations for question 8.

Mathematical review

The stated Cartesian equation is only the left branch 0≤t≤π/20\le t\le\pi/2 used for R. The full C has a second branch x=π−sin⁡−1y−yx=\pi-\sin^{-1}\sqrt y-\sqrt y for π/2≤t≤π\pi/2\le t\le\pi. The solution correctly states its restricted branch; the question’s global claim is qualified here.