If \(\tan^{-1}\left(\displaystyle\frac{2}{3^{-x}+1}\right)=\cot^{-1}\left(\displaystyle\frac{3}{3^{x}+1}\right)\), then which one of the following is true?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → There is one positive and one negative real value of x satisfying the above equation.
Since \(\frac{3}{3^x+1} \gt 0\), we can use \(\cot^{-1} y = \tan^{-1}\frac{1}{y}\) for y > 0:
\( \displaystyle \tan^{-1}\left(\frac{2}{3^{-x}+1}\right) = \tan^{-1}\left(\frac{3^{x}+1}{3}\right) \)
The function \(\tan^{-1}\) is one-one, so
\( \displaystyle \frac{2}{3^{-x}+1} = \frac{3^{x}+1}{3} \)
\(6 = \left(3^{x}+1\right)\left(3^{-x}+1\right)\)
\(6 = 1 + 3^{x} + 3^{-x} + 1\)
\(3^{x} + 3^{-x} = 4\)
Put \(t = 3^{x}\), which is positive:
\( \displaystyle t + \frac{1}{t} = 4 \)
\(t^2 - 4t + 1 = 0\)
\(t = 2 \pm \sqrt{3}\)
Both roots are positive, so each gives a real x:
\(3^{x} = 2 + \sqrt{3} \gt 1\), so \(x \gt 0\)
\(3^{x} = 2 - \sqrt{3} \lt 1\), so \(x \lt 0\)
Since \((2 + \sqrt{3})(2 - \sqrt{3}) = 1\), the two values are \(x = \pm\log_3\left(2 + \sqrt{3}\right)\): one positive and one negative.
Note: This topic was not in the CUET 2026 syllabus.