The graphs have the equations 1e y = (p2+1) – x2 and y = p2 where p > 1 x2 1) x2 = p2 Intersection when (p2+ – x2 (p2 + 1)x2 x4 = p2 – 0 = x4 (p2 + 1)x2 + p2 – (x2 1)(x2 p2) = 0 – – Correct intersection x = ±1 and x = ± p for u area = = Correct idea of – area = area under cubic area under – 2p3 2p2 + 2p 2 – – = parabola (and vice 3 3 versa) shown with equations for r. Correct solution showing correct integrations and correct mathematical logic and statements.
1e The graphs have the equations y = (p2+1) – x2 and y = p2 where p > 1 x2 Intersection when (p2+ 1) – x2 = p2 x2 (p2 + 1)x2 – x4 = p2 0 = x4 – (p2 + 1)x2 + p2 (x2 – 1)(x2 – p2) = 0 x = ±1 and x = ± p area = = – = 2p3 – 2p2 + 2p – 2 3 3 Correct intersection for u Correct idea of area = area under cubic – area under parabola (and vice versa) shown with equations for r. Correct solution showing correct integrations and correct mathematical logic and statements.
4b_practice_examination__number_4_schedule.docx