A curve is such that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9}{(2x-1)^2}$, where \(x \

Question

A curve is such that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9}{(2x-1)^2}$, where $x \neq \frac{1}{2}$. The point $(2, 12)$ lies on the curve. Find the equation of the curve.[3] Find the value of $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ when $x = 0$.[2] A student concludes from (ii) that since the value of $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$ at $x = 0$, hence there is a minimum point at $x = 0$. State with reason and mathematical working whether the above conclusion is correct.[2]

Accepted Answer

(i) $y=-9/[2(2x-1)]+27/2$ for $x>1/2$; an independent constant for $x<1/2$. (ii) 36. (iii) No: $y'(0)=9\ne0$.

AMKSS 4AM 2022 Prelims P1 Q12

AMKSS 4AM 2022 Prelims P1 Q12

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