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← JEE Main 2026 · 6 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 13 of 75

JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 13

Let 0<α<10<\alpha<1, β=13α\beta=\dfrac1{3\alpha} and tan⁡−1(1−α)+tan⁡−1(1−β)=π4\tan^{-1}(1-\alpha)+\tan^{-1}(1-\beta)=\dfrac\pi4. Then 6(α+β)6(\alpha+\beta) is equal to:

Choose one option
Answer and worked solution

Official answer: B

Tangent addition.
2−α−β1−(1−α)(1−β)=1 ⇒ 2−2(α+β)+αβ=0.\frac{2-\alpha-\beta}{1-(1-\alpha)(1-\beta)}=1\ \Rightarrow\ 2-2(\alpha+\beta)+\alpha\beta=0.

With αβ=13\alpha\beta=\tfrac13: α+β=76\alpha+\beta=\tfrac76, so 6(α+β)=76(\alpha+\beta)=7. (The values are α,β=23,12\alpha,\beta=\tfrac23,\tfrac12; both lie in (0, 1), so the principal-value condition holds.)

(B)

Topics:Inverse trigonometric functions
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