Question ID: #740
The number of solutions of $\tan^{-1}4x+\tan^{-1}6x=\frac{\pi}{6}$ where $-\frac{1}{2\sqrt{6}}< x < \frac{1}{2\sqrt{6}}$ is equal to
- (1) 3
- (2) 0
- (3) 1
- (4) 2
Solution:
Using the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ for $AB < 1$. Here $A = 4x, B = 6x$. The condition $AB < 1 \Rightarrow 24x^2 < 1$ is satisfied by the given interval $-\frac{1}{\sqrt{24}} < x < \frac{1}{\sqrt{24}}$.
$$ \tan^{-1}\left(\frac{4x+6x}{1-24x^2}\right) = \frac{\pi}{6} $$
$$ \frac{10x}{1-24x^2} = \tan\frac{\pi}{6} $$
$$ \frac{10x}{1-24x^2} = \frac{1}{\sqrt{3}} $$
Cross-multiplying:
$$ 10\sqrt{3}x = 1 – 24x^2 $$
$$ 24x^2 + 10\sqrt{3}x – 1 = 0 $$
Solving the quadratic equation for $x$:
$$ x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 – 4(24)(-1)}}{2(24)} $$
$$ x = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} $$
$$ x = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} $$
$$ x = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24} $$
We check which solution lies in the interval $(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}) \approx (-0.204, 0.204)$.
Approximating values: $\sqrt{3} \approx 1.732, \sqrt{11} \approx 3.316$.
Case 1: Positive sign
$$ x_1 = \frac{-5(1.732) + 3(3.316)}{24} = \frac{-8.66 + 9.948}{24} = \frac{1.288}{24} \approx 0.053 $$
This value is within the interval $(-0.204, 0.204)$.
Case 2: Negative sign
$$ x_2 = \frac{-5(1.732) – 3(3.316)}{24} = \frac{-8.66 – 9.948}{24} = \frac{-18.608}{24} \approx -0.775 $$
This value is outside the interval.
Thus, there is only 1 solution.
Ans. (3)
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