The curve \(C\) has equation \[y = \frac{x^{2} + 3}{x - 1}, \qquad x \neq 1\] Show that \(\dfrac{dy}{dx} = \dfrac{x^{2} - 2x - 3}{(x - 1)^{2}}\). (3) Hence ...

(a) Show that \(\dfrac{dy}{dx} = \dfrac{x^{2} - 2x - 3}{(x - 1)^{2}}\) [3]

\(y = \dfrac{x^{2} + 3}{x - 1}\) is a genuine quotient whose numerator does not divide exactly by the denominator, so the quotient rule is the efficient route. Take

\[u = x^{2} + 3, \quad v = x - 1, \quad u' = 2x, \quad v' = 1.\] \[\frac{dy}{dx} = \frac{u'v - uv'}{v^{2}} = \frac{2x(x - 1) - \left(x^{2} + 3\right)(1)}{(x - 1)^{2}} \quad \textbf{[M1][A1]}\]

The M mark is for a correct quotient-rule structure, so it survives an arithmetic slip inside the numerator; the A mark requires the substitution itself to be right. The order \(u'v - uv'\) matters: reversing it changes every sign and loses the accuracy marks.

Simplify the numerator only, leaving \((x - 1)^{2}\) factorised:

\[\frac{dy}{dx} = \frac{2x^{2} - 2x - x^{2} - 3}{(x - 1)^{2}} = \frac{x^{2} - 2x - 3}{(x - 1)^{2}} \quad \textbf{[A1]}\]

The result is given in the question, so the marks are for the derivation. Quoting the printed answer earns nothing.

(b) Coordinates of the two stationary points [4]

A fraction is zero exactly when its numerator is zero and its denominator is not. Since \((x - 1)^{2} \neq 0\) for \(x \neq 1\), and \(x = 1\) is excluded from the domain anyway,

\[x^{2} - 2x - 3 = 0 \quad \textbf{[M1]}\] \[(x - 3)(x + 1) = 0 \quad \Rightarrow \quad x = 3 \ \text{ or } \ x = -1 \quad \textbf{[A1]}\]

Substitute back into the original \(y\), not into the derivative:

• \(x = 3\): \(y = \dfrac{9 + 3}{3 - 1} = \dfrac{12}{2} = 6\), so \((3,\, 6)\). [A1]
• \(x = -1\): \(y = \dfrac{1 + 3}{-1 - 1} = \dfrac{4}{-2} = -2\), so \((-1,\, -2)\). [A1]

The two coordinate marks are independent, so one wrong substitution does not destroy the other.

(c) Nature of each stationary point [2]

Two standard tests exist. The second derivative here would need the quotient rule a second time on an already awkward fraction, so the sign-of-gradient test is far quicker and is what the structure of part (a) invites.

Because \((x - 1)^{2} \gt 0\) throughout the domain, the sign of \(\dfrac{dy}{dx}\) is exactly the sign of the numerator \(x^{2} - 2x - 3\). That single observation turns a fraction test into a quadratic sign test.

Near \(x = -1\): at \(x = -2\), \(x^{2} - 2x - 3 = 4 + 4 - 3 = 5 \gt 0\); at \(x = 0\), \(0 - 0 - 3 = -3 \lt 0\). The gradient goes from positive to negative, so \((-1,\, -2)\) is a maximum point. [B1]

Near \(x = 3\): at \(x = 2\), \(4 - 4 - 3 = -3 \lt 0\); at \(x = 4\), \(16 - 8 - 3 = 5 \gt 0\). The gradient goes from negative to positive, so \((3,\, 6)\) is a minimum point. [B1]

These are independent B marks: each is awarded for a correct conclusion supported by a reason, and the question's word "reasons" means an unjustified statement of "maximum" and "minimum" scores nothing even though it is the right pair of words.

Common wrong turns on this question. Choosing test points that straddle \(x = 1\), where the curve is not defined, so the sign change reflects the vertical asymptote rather than the stationary point: keep both test values on the same side of \(x = 1\) as the point being tested. Cancelling \((x-1)\) wrongly at the start. Assuming, as with a parabola, that the smaller \(x\) gives the minimum: here the local maximum value \(-2\) is below the local minimum value \(6\), which is entirely normal for a curve with a vertical asymptote between them.

Check. The maximum value being less than the minimum value is the sanity check that you have the branches right: the left branch rises to \(-2\) then falls to the asymptote, and the right branch falls from the asymptote to \(6\) then rises.