Q38 / 75 · JEE Main 2026 (06 April Shift 1) · Mathematics

Let $0 < \alpha < 1$, $\beta = \dfrac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}$. Then $6(\alpha + \beta)$ is equal to:

  • A
    $6$
  • Correct answer
    B
    $7$
  • C
    $8$
  • D
    $9$

Solution

Given $\tan^{-1}(1-\alpha) + \tan^{-1}(1-\beta) = \dfrac{\pi}{4}$

Applying the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$, we get:

$\dfrac{(1-\alpha) + (1-\beta)}{1 - (1-\alpha)(1-\beta)} = \tan\left(\dfrac{\pi}{4}\right) = 1$

$2 - \alpha - \beta = 1 - (1 - \alpha - \beta + \alpha\beta)$

$2 - \alpha - \beta = \alpha + \beta - \alpha\beta$

$2(\alpha + \beta) = 2 + \alpha\beta$

Given $\beta = \dfrac{1}{3\alpha}$, we have $\alpha\beta = \dfrac{1}{3}$. Substituting this into the equation:

$2(\alpha + \beta) = 2 + \dfrac{1}{3}$

$2(\alpha + \beta) = \dfrac{7}{3}$

$\alpha + \beta = \dfrac{7}{6}$

Therefore, $6(\alpha + \beta) = 6 \times \dfrac{7}{6} = 7$

Answer: $7$

Hence, the correct answer is option: 'B'.

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