Question

If \(tan^{-1}ax+tan^{-1}3x = \frac{π}{4}\), where \(3ax^2 < 1\), then value of \(a\) for \(x = \frac{1}{6}\) is \(\ldots\)

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(9\)

Hint:

Use tan^-1 A + tan^-1 B = tan^-1((A+B)/(1-AB)) and put x = 1/6.

Answer 1

The Correct Option is A

Step 1: Plug in the values first:
At $x = \tfrac16$ the equation reads $\tan^{-1}\left(\tfrac a6\right) + \tan^{-1}\left(\tfrac12\right) = \tfrac{\pi}{4}$.

Step 2: Isolate the unknown:
$\tan^{-1}\left(\tfrac a6\right) = \tfrac{\pi}{4} - \tan^{-1}\tfrac12$.
Take tangent: $\dfrac a6 = \dfrac{1 - \frac12}{1 + \frac12} = \dfrac{1/2}{3/2} = \dfrac13$.

Step 3: Solve:
$a = 6\cdot\tfrac13 = 2$.

Step 4: Check the condition:
$3ax^2 = 3\cdot2\cdot\tfrac1{36} = \tfrac16 < 1$, so the combination formula was valid.

Final Answer:
$a = 2$, option (A). \[ \boxed{2 \text{ (A)}} \]