In a triangle $ABC$, $AB = 2$, angle $CAB = x$ radians and angle $CBA = \frac{\pi}{6}$ radia

Question

In a triangle $ABC$, $AB = 2$, angle $CAB = x$ radians and angle $CBA = \frac{\pi}{6}$ radians. Show that $AC = \frac{2}{\cos x + \sqrt{3} \sin x}$.[2] Given that $x$ is a sufficiently small angle, show that $$AC \approx a + bx + cx^2,$$ where $a$, $b$ and $c$ are constants to be determined.[3] It is given that $2xy + \ln y = \ln 3$. Show that $$(2xy^2 + y)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4y^2\frac{\mathrm{d}y}{\mathrm{d}x} - \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 0.$$ Hence find the Maclaurin series for $y$, up to and including the term in $x^2$.[5]

Accepted Answer

(a)(ii) $a=2,\ b=-2\sqrt3,\ c=7$ (b) $y=3-18x+162x^2+\cdots$

2025 RI P2 Q2

2025 RI P2 Q2

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