Question

The number of real solutions of $\tan^{-1}x + \tan^{-1}(2x) = \frac{\pi}{4}$ is

• A. 2
• B. 1
• C. 0
• D. infinitely many

Hint:

When solving equations involving inverse trigonometric functions, always check the validity of the solutions obtained. Extraneous roots can be introduced by the formulas used, so substitute the final answers back into the original equation or check against the domain/range constraints of the functions.

Answer 1

The Correct Option is B

Step 1: Apply the formula for the sum of inverse tangent functions.
\[ \tan^{-1}x + \tan^{-1}(2x) = \tan^{-1}\left(\frac{x+2x}{1-x\cdot 2x}\right) = \tan^{-1}\left(\frac{3x}{1-2x^2}\right) \] The formula is valid for $2x^2<1$.

Step 2: Solve the resulting equation.
\[ \frac{3x}{1-2x^2} = \tan\frac{\pi}{4} = 1 \implies 3x = 1 - 2x^2 \implies 2x^2 + 3x - 1 = 0 \]
Step 3: Find the roots using the quadratic formula.
\[ x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2\cdot 2} = \frac{-3 \pm \sqrt{17}}{4} \]
Step 4: Validate the solutions.
$x_2 = \frac{-3-\sqrt{17}}{4}<0$ is invalid since sum of two negative arctangents cannot be $\pi/4$.
$x_1 = \frac{-3+\sqrt{17}}{4}>0$ and satisfies $2x_1^2<1$.

Step 5: Conclusion.
Only one real solution exists.