The number of solutions of ^-14x+ ^-16x=( )/(6), where -12√6

Question

The number of solutions of ${	an }^{-1}4x+{	an }^{-1}6x=\frac{\pi }{6}$, where $-\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}$, is equal to :

Accepted Answer

$$\tan^{-1}4x+\tan^{-1}6x=\frac{\pi}{6}\ \Rightarrow\ \tan^{-1}\left(\frac{10x}{1-24x^2}\right)=\frac{\pi}{6}\ \Rightarrow\ \frac{10x}{1-24x^2}=\frac{1}{\sqrt3}.$$

$$\Rightarrow 24x^2+10\sqrt3x-1=0\ \Rightarrow\ x=\frac{-5\sqrt3\pm3\sqrt{14}}{24}.$$

Checking the two roots against the given interval $\left(-\dfrac{1}{2\sqrt6},\dfrac{1}{2\sqrt6}\right)$ and against the sign condition needed for the arctangent addition formula to hold (i.e. $1-24x^2>0$), only one of the two roots is admissible. Hence the number of solutions is 1.

Solutions · 1