A curve is such that $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{1}{(2x-1)^2}$. The gradient of t

Question

A curve is such that $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{1}{(2x-1)^2}$. The gradient of the curve at the point $\left(1, \frac{1}{4}\right)$ is $-\frac{1}{2}$. Express $y$ in terms of $x$. Hence, find the coordinates of the point where the curve cuts the $x$-axis, giving your answer in exact form.

Accepted Answer

(a) $y=-\frac{1}{4}\ln(2x-1)+\frac{1}{4}$ (b) $\left(\frac{\mathrm e+1}{2},0\right)$

Integration - Finding Equation of Curve from Second Derivatives

Integration - Finding Equation of Curve from Second Derivatives

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