Pre-U Pre-U 9794/2 Specimen — Question 3 5 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
SessionSpecimen
Marks5
TopicTangents, normals and gradients
TypeFind stationary points
DifficultyStandard +0.8 This requires product rule differentiation with a fractional power and logarithm, setting the derivative to zero, and solving a transcendental equation involving ln x. The algebraic manipulation to find the exact stationary point (factoring out x^{1/2} and solving 3ln x + 2 = 0) is more demanding than routine calculus exercises, though still within standard A-level techniques.
Spec1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives
The equation of a curve is \(y = x^{\frac{3}{2}} \ln x\). Find the exact coordinates of the stationary point on the curve. [5]
AnswerMarks Guidance
Use of product rule and one summand correctM1 Obtain \(\frac{dy}{dx} = 4x^{-\frac{2}{3}} \ln x + x^2 \times \frac{1}{x}\) 
Use of product rule and one summand correct | M1 | Obtain $\frac{dy}{dx} = 4x^{-\frac{2}{3}} \ln x + x^2 \times \frac{1}{x}$ | A1 | Equation $\frac{dy}{dx}$ to zero and simplify ⇒ $\frac{\ln x + 2}{2x^2} = 0$ | M1 | Obtain $x = e^{-2}$ | A1 | Obtain $y = -2e^{-1}$ | A1 | 5 marks
The equation of a curve is $y = x^{\frac{3}{2}} \ln x$. Find the exact coordinates of the stationary point on the curve. [5]

\hfill \mbox{\textit{Pre-U Pre-U 9794/2  Q3 [5]}}
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