A-Level Differentiation: worked solution

Question

The curve y = ax + b + c/(x^2 - 1) has real constants a, b, c. It crosses the x-axis at x = 2, and the normal to the curve at (0, 3) reaches the x-axis at x = 7.5. Determine a, b and c.

The curve is y = ax + b + c/(x² - 1), so three facts are needed to pin down the three constants. Step 1: use the point (0, 3), which lies on the curve. Put x = 0: y = a(0) + b + c/(0 - 1) = b - c. Since y = 3 there, b - c = 3. ...(1) Step 2: use the x-intercept at x = 2, where y = 0. Put x = 2, y = 0: 2a + b + c/(4 - 1) = 0, so 2a + b + c/3 = 0. ...(2) Step 3: turn the normal into the gradient of the curve at (0, 3). The normal passes through (0, 3) and (7.5, 0), so its gradient is (0 - 3)/(7.5 - 0) = -3/7.5 = -2/5. A tangent and its normal are perpendicular, so their gradients multiply to -1: gradient of the tangent at (0, 3) = -1 / (-2/5) = 5/2. Step 4: differentiate, so that this gradient can be read off. Write the last term as c(x² - 1)^(-1) and use the chain rule: d/dx [ c(x² - 1)^(-1) ] = c * (-1)(x² - 1)^(-2) * 2x = -2cx/(x² - 1)². So dy/dx = a - 2cx/(x² - 1)². At x = 0 that second term has a factor x in its numerator, so it is 0, leaving dy/dx = a. Comparing with Step 3: a = 5/2. Step 5: solve (1) and (2) for b and c. Put a = 5/2 in (2): 2(5/2) + b + c/3 = 0, i.e. 5 + b + c/3 = 0. From (1), b = 3 + c. Substitute: 5 + (3 + c) + c/3 = 0 8 + c + c/3 = 0 8 + 4c/3 = 0 4c/3 = -8 c = -6. Then b = 3 + c = 3 + (-6) = -3. So a = 5/2, b = -3, c = -6.