Question

The number of values of \(x\) satisfying \[ \tan^{-1}(4x)+\tan^{-1}(6x)=\frac{\pi}{6}, \quad x\in\left[-\frac{1}{2\sqrt{6}},\,\frac{1}{2\sqrt{6}}\right] \] is

• A. \(1\)
• B. \(0\)
• C. \(2\)
• D. \(3\)

Hint:

While solving inverse trigonometric equations:
Always check the validity condition of identities
After solving algebraically, \textbf{verify solutions with the given interval}

Answer 1

The Correct Option is A

Concept: For inverse tangent expressions, the identity \[ \tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\left(\frac{a+b}{1-ab}\right) \] holds when \(ab<1\), with the principal value restricted to the interval \((-\pi/2,\pi/2)\).
Step 1: Use the identity. Let \[ a=4x,\quad b=6x \] Then, \[ \tan^{-1}(4x)+\tan^{-1}(6x) = \tan^{-1}\!\left(\frac{10x}{1-24x^2}\right) \] Given that: \[ \tan^{-1}\!\left(\frac{10x}{1-24x^2}\right)=\frac{\pi}{6} \] 
Step 2: Take tangent on both sides. \[ \frac{10x}{1-24x^2}=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} \] 
Step 3: Solve the resulting equation for \(x\). \[ 10\sqrt{3}\,x=1-24x^2 \] \[ 24x^2+10\sqrt{3}\,x-1=0 \] Solving the quadratic: \[ x=\frac{-10\sqrt{3}\pm\sqrt{300+96}}{48} =\frac{-10\sqrt{3}\pm\sqrt{396}}{48} =\frac{-10\sqrt{3}\pm6\sqrt{11}}{48} \] 
Step 4: Apply the interval restriction. Given: \[ x\in\left[-\frac{1}{2\sqrt{6}},\,\frac{1}{2\sqrt{6}}\right] \] On numerical checking: 
One root lies within the specified interval 
The other root lies outside the interval Therefore, only one value of \(x\) satisfies both the equation and the given interval condition.