A-Level Equations and inequalities: worked solution

Question

(i) In terms of a, solve 1/(x-a)²=|x-a|. (ii) On one set of axes, sketch y=1/(x-a)² and y=|x-a| for a>1, and hence solve 1/(x-a)²>|x-a|.

(i) Solve 1/(x - a)² = |x - a|. Step 1: simplify with a substitution, since a appears only in the combination x - a. Let u = x - a. The left-hand side requires u ≠ 0, because we cannot divide by zero, so x ≠ a. Keep that restriction in mind. The equation becomes 1/u² = |u|. Step 2: clear the fraction. Since u ≠ 0, u² > 0, so we may multiply both sides by u² without reversing anything: 1 = |u| × u² Step 3: write everything in terms of |u|. Note that u² = |u|², because squaring removes the sign either way. So 1 = |u| × |u|² = |u|³ Step 4: solve. |u|³ = 1, and |u| is a real number, so taking the cube root gives |u| = 1. (The cube root is unique for real numbers, so there is no second case here.) |u| = 1 means u = 1 or u = -1. Step 5: return to x. u = x - a, so x - a = 1 or x - a = -1. x = a + 1 or x = a - 1. Both satisfy x ≠ a, so both are valid. Answer to (i): x = a - 1 or x = a + 1. (ii) Sketch and solve 1/(x - a)² > |x - a|. Step 6: describe each curve before drawing it. y = 1/(x - a)² is the graph of y = 1/x² translated a units in the positive x direction. - It has a vertical asymptote at x = a, since the denominator tends to 0 there. - y is always positive, because (x - a)² > 0 for x ≠ a, so both branches lie entirely above the x-axis. - y → +∞ as x → a from either side, and y → 0+ as x → +∞ and as x → -∞, so y = 0 (the x-axis) is a horizontal asymptote. - The graph is symmetric about the line x = a. y = |x - a| is the graph of y = |x| translated a units in the positive x direction. - It is a V shape with its vertex at (a, 0). - Its right arm is y = x - a with gradient 1, its left arm is y = -(x - a) with gradient -1. - It is also symmetric about x = a. Since a > 1, both graphs sit to the right of the y-axis; this does not change their shape. Step 7: mark the intersections, which part (i) has already found. They meet where x = a - 1 and x = a + 1. At x = a + 1: y = 1/(1)² = 1 and y = |1| = 1, so the point is (a + 1, 1). By symmetry the other is (a - 1, 1). Label both points on the sketch, together with the asymptote x = a and the vertex (a, 0). Step 8: read off where the curve lies above the V. The inequality 1/(x - a)² > |x - a| asks for the x values where the curve y = 1/(x - a)² is strictly higher than the V. Test a point between the intersections, say just to the right of x = a: the curve is heading to +∞ while the V is heading to 0, so the curve is above there. Test a point outside, say x = a + 2: the curve gives 1/4 while the V gives 2, so the V is above there. So the curve is above the V exactly between the two intersection points, on both sides of the asymptote. Step 9: state the solution, remembering the excluded value. The inequality holds for a - 1 < x < a + 1, but x = a must be excluded because 1/(x - a)² is undefined there. Answer to (ii): a - 1 < x < a + 1, x ≠ a.