 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/11/d/ii/solution

The graph is the upper branch of the hyperbola

d^{2} - (t + 1)^{2} = 1/2

. It is symmetric about

t = - 1

, positive everywhere, and has its lowest point at

(- 1, 1/ 2)

d^{'} (t) = \frac{t + 1}{d ( t )}, d^{''} (t) = \frac{1}{2 d ( t ) ^{3}} > 0,

so it is strictly convex. Since

d (t) - ∣ t + 1∣ = \frac{1/2}{( t + 1 ) ^{2} + 1/2 + ∣ t + 1∣} ⟶ 0,

its two oblique asymptotes are

d = t + 1 (t \to + \infty), d = - t - 1 (t \to - \infty) .

 New to topics? Read the docs here!