The diagram shows part of the curve of \(y = \frac{1}{4}x^2 - \frac{1}{2}\ln\left(\frac{x}{3}\right)

Question

The diagram shows part of the curve of $y = \frac{1}{4}x^2 - \frac{1}{2}\ln\left(\frac{x}{3}\right)$ and $P(3, 2.25)$ is a point on the curve. By using calculus, determine, with full working, whether $y = \frac{4}{3}x - 1$ is the equation of the tangent to the curve at $P$.[5] Differentiate $x\ln \left(\frac{x}{3}\right)-x$ with respect to $x$[2] Using the result from part (b), find the exact area of the region enclosed by the curve $y=\frac{1}{4}x^2-\frac{1}{2}\ln \left(\frac{x}{3}\right)$, the x-axis, the line x=1 and the line x=3.[3]

Accepted Answer

(b)$\ln (\frac{x}{3})$ (c)$\frac{19}{6}+\frac{1}{2}\ln \frac{1}{3}$

2025 SCGS PRELIMS P1 Q12

2025 SCGS PRELIMS P1 Q12