Given: \(\tan^{-1} x + \tan^{-1} 2x = \frac{\pi}{4}\), where \(x > 0\).
\(\implies \tan^{-1} 2x = \frac{\pi}{4} - \tan^{-1} x\)
Taking tangent on both sides:
\(\implies 2x = \frac{1 - x}{1 + x}\)
\(\implies 2x(1 + x) = 1 - x\)
\(\implies 2x^2 + 3x - 1 = 0\)
Solving the quadratic equation:
\(x = \frac{-3 \pm \sqrt{9 + 8}}{4}\)
\(x = \frac{-3 \pm \sqrt{17}}{4}\)
Since \(x > 0\), the only possible solution is:
\(x = \frac{-3 + \sqrt{17}}{4}\)
Thus, the number of positive real values of \(x\) is \(1\).
The Correct answer is: 1
Prove that:
\( \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right), \quad x \in [0, 1] \)
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: