The no. of solution in \(x \in \left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)\) of equation \(\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}\) is :
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are asked to find the number of solutions to a trigonometric equation involving inverse tangent functions within a specific interval.
Step 2: Key Formula or Approach:
We use the formula for the sum of two inverse tangent functions: \[ \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right), \quad \text{provided } AB<1 \] Here, \(A = 4x\) and \(B = 6x\). The condition is \( (4x)(6x)<1 \Rightarrow 24x^2<1 \Rightarrow x^2<\frac{1}{24} \).
This means \(|x|<\frac{1}{\sqrt{24}} = \frac{1}{2\sqrt{6}}\). The given interval for x is \(\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)\), which satisfies the condition \(AB<1\).
Step 3: Detailed Explanation:
Applying the formula to the given equation: \[ \tan^{-1}(4x) + \tan^{-1}(6x) = \tan^{-1}\left(\frac{4x+6x}{1-(4x)(6x)}\right) = \tan^{-1}\left(\frac{10x}{1-24x^2}\right) \] We are given that this equals \(\frac{\pi}{6}\). \[ \tan^{-1}\left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6} \] Taking the tangent of both sides: \[ \frac{10x}{1-24x^2} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \] Now, we solve for x: \[ 10\sqrt{3}x = 1 - 24x^2 \] \[ 24x^2 + 10\sqrt{3}x - 1 = 0 \] This is a quadratic equation in x. Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\): \[ x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} \] \[ x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} \] \[ x = \frac{-10\sqrt{3} \pm \sqrt{36 \times 11}}{48} = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24} \] The two possible solutions are \(x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24}\) and \(x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24}\).
Step 4: Checking the Solutions against the Interval:
The interval is \(\left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right)\).
\(\frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12} \approx \frac{2.449}{12} \approx 0.204\). So the interval is approx \((-0.204, 0.204)\).
Approximate values of the solutions: \(\sqrt{3} \approx 1.732\), \(\sqrt{11} \approx 3.317\).
\(x_1 \approx \frac{-5(1.732) + 3(3.317)}{24} = \frac{-8.66 + 9.951}{24} = \frac{1.291}{24} \approx 0.0538\).
This value lies inside the interval \((-0.204, 0.204)\).
\(x_2 \approx \frac{-5(1.732) - 3(3.317)}{24} = \frac{-8.66 - 9.951}{24} = \frac{-18.611}{24} \approx -0.775\).
This value lies outside the interval \((-0.204, 0.204)\).
Therefore, there is only one solution in the given interval.
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