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MioMath 2024 Paper 1 · Q19

Paper 1Geometry

Find the area enclosed by the curves

y=p∣x∣−pandy=−1p∣x∣+py=p|x|-p\quad\text{and}\quad y=-\frac{1}{p}|x|+p

when pp is a positive constant.

A4p3p2+1\dfrac{4p^3}{p^2+1}
B2p3p2+1\dfrac{2p^3}{p^2+1}
C8p3p2+1\dfrac{8p^3}{p^2+1}
D4p3p−1\dfrac{4p^3}{p-1}

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