O-Level A-Math 2022 Paper 1, Question 13: Quotient rule

(a) Differentiating by the quotient rule, \(\dfrac{dy}{dx} = \dfrac{4(x + 1) - (4x + 2)(1)}{(x + 1)^2} = \dfrac{4x + 4 - 4x - 2}{(x + 1)^2} = \dfrac{2}{(x + 1)^2}\). Since the numerator \(2\) is a non-zero constant, \(\dfrac{dy}{dx} \neq 0\) for every \(x \neq -1\). A stationary point requires \(\dfrac{dy}{dx} = 0\), which never happens, so the curve has no stationary point. (shown)

(b) Set the gradient equal to \(\tfrac12\): \(\dfrac{2}{(x + 1)^2} = \dfrac12 \Rightarrow (x + 1)^2 = 4 \Rightarrow x + 1 = \pm 2 \Rightarrow x = 1\) or \(x = -3\). Since the curve is for \(x > -1\), \(x = 1\), and then \(y = \dfrac{4(1) + 2}{1 + 1} = 3\), so \(P(1, 3)\). The normal at \(P\) has gradient \(-\dfrac{1}{1/2} = -2\), so its equation is \(y - 3 = -2(x - 1) \Rightarrow y = -2x + 5\). At \(B\) (on the \(y\)-axis, \(x = 0\)): \(y = 5\), so \(B(0, 5)\). At \(A\) (on the \(x\)-axis, \(y = 0\)): \(0 = -2x + 5 \Rightarrow x = \tfrac52\), so \(A\left(\tfrac52, 0\right)\). Triangle \(AOB\) is right-angled at \(O\), with legs \(OA = \tfrac52\) and \(OB = 5\), so area \(= \dfrac12 \times \dfrac52 \times 5 = \dfrac{25}{4} = 6\dfrac{1}{4}\) units\(^2\).

(c) Rewrite \(\dfrac{4x + 2}{x + 1} = \dfrac{4(x + 1) - 2}{x + 1} = 4 - \dfrac{2}{x + 1}\), so \(a = 4\), \(b = -2\). For \(x > -1\), \(x + 1 > 0\), so \(-\dfrac{2}{x + 1} < 0\) and hence \(4 - \dfrac{2}{x + 1} < 4\). The curve always lies below \(y = 4\), so a horizontal line \(y = c\) fails to meet it precisely when \(c \geqslant 4\). (shown)