The curve $y = ax + \frac{b}{(2x - 1)^3}$, where $a$ and $b$ are constants, has a station

Question

The curve $y = ax + \frac{b}{(2x - 1)^3}$, where $a$ and $b$ are constants, has a stationary point at $ \left( \frac{3}{2}, \frac{11}{2} \right)$. Find the value of $a$ and of $b$.[4] Find the $x$-coordinate of the other stationary point.[3] Find $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$.[1] Hence, determine the nature of the stationary point at $ \left(\frac{3}{2}, \frac{11}{2} \right)$.[2]

Accepted Answer

(a) $a=3,\ b=8$ (b) $x=-\dfrac12$ (c) $\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{384}{(2x-1)^5}$ (d) Minimum point.

2024 CGSS P1 Q8

2024 CGSS P1 Q8

Solution:

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