The point $P$ lies on the curve $y = \ln\left(\frac{x + 1}{x - 1}\right)$ for $x 1$.

Question

The point $P$ lies on the curve $y = \ln\left(\frac{x + 1}{x - 1}\right)$ for $x > 1$. The normal to the curve at $P$ is parallel to the line $2y = 3x + 2$. The equation of the tangent to the curve at $P$ cuts the $y$-axis at $Q$. Find the area of the triangle $POQ$, where $O$ is the origin.

Accepted Answer

$\ln3+\dfrac43\text{ units}^2\approx2.43\text{ units}^2$

2024 GESS PRELIMS P1 Q7

2024 GESS PRELIMS P1 Q7

Solution:

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