Additional Mathematics 0606 / Equations, inequalities and graphs / 3.953.95: Area between y=12−x2y = 12 - x^{2}y=12−x2 and y=x4−4x2+8y = x^{4} - 4x^{2} + 8y=x4−4x2+80606/13/O/N/25 — Question 8 · 7 marksMark as done · Save for later · Show all solutionsThe diagram shows part of each of the curves y=12−x2y = 12 - x^{2}y=12−x2 and y=x4−4x2+8y = x^{4} - 4x^{2} + 8y=x4−4x2+8.Find the area of the shaded region enclosed by y=12−x2y = 12 - x^{2}y=12−x2 and y=x4−4x2+8y = x^{4} - 4x^{2} + 8y=x4−4x2+8.[7]▸ Answer▸ Official mark scheme← 3.94: Solve ∣x2−10∣=5−2x|x^{2} - 10| = 5 - 2x∣x2−10∣=5−2x3.96: Graphical inequality with y=∣x∣+3y = |x| + 3y=∣x∣+3 →
