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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Inverse Trigonometric Functions

The number of solutions of ^ -1 4x + ^ -1 6x = (π)/(6) , where - 1...

The number of solutions of tan⁡−14x+tan⁡−16x=π6\tan^{-1} 4x + \tan^{-1} 6x = \frac{\pi}{6}, where −126<x<126-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}, is equal to

Options

  1. A3
  2. B0
  3. C2
  4. D1

Correct answer

D. 1

Step-by-step solution

For x∈(−126,126)x \in \left(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}}\right), 24x2<124x^2 < 1
So tan⁡−1(4x)+tan⁡−1(6x)=tan⁡−1(10x1−24x2)=π6\tan^{-1}(4x) + \tan^{-1}(6x) = \tan^{-1}\left(\frac{10x}{1-24x^2}\right) = \frac{\pi}{6}.
10x1−24x2=13⇒24x2+103x−1=0\frac{10x}{1-24x^2} = \frac{1}{\sqrt{3}} \Rightarrow 24x^2 + 10\sqrt{3}x - 1 = 0.
x=−103±300+9648=−53±31124x = \frac{-10\sqrt{3} \pm \sqrt{300+96}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}.
x1=−53+31124≈0.054x_1 = \frac{-5\sqrt{3}+3\sqrt{11}}{24} \approx 0.054 (valid, lies in range).
x2=−53−31124≈−0.776x_2 = \frac{-5\sqrt{3}-3\sqrt{11}}{24} \approx -0.776 (rejected, outside range).
Number of solutions =1= 1.
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