Question bank/JEE Main/Mathematics

Let ABC be a triangle with $\angle BAC = \frac{2\pi}{3}$ and AB
= x such that (AB) (AC) = 1. If x varies then the longest possible
length of the internal angle bisector AD equals

A$\frac{1}{3}$B$\frac{1}{2}$C$\frac{2}{3}$D$\frac{\sqrt{2}}{3}$

Show answer and solution▾

[IMAGE] bx = 1
$AD = \frac{2bx}{b + x}\cos\frac{\pi}{3}$
$\frac{1}{b + x} = \frac{1}{x + \frac{1}{x}}$
$x + \frac{1}{x} \geq 2,x > 0$

Source: Grass-G / M-24 Mathematics Paper 31 Dec.docx