If f(x)=2≤ft(tan^-1≤ft(e^x)-(π)/4) , then f(x) is:

The correct answer is Option (3) → odd and is strictly increasing in \((-\infty, \infty)\)

Odd or even: since \(e^{-x} = \frac{1}{e^x}\) and \(e^x \gt 0\), the identity \(\tan^{-1}\frac{1}{u} = \frac{\pi}{2} - \tan^{-1}u\) for u > 0 gives

    \( \displaystyle f(-x) = 2\left(\tan^{-1}\frac{1}{e^x} - \frac{\pi}{4}\right) \)

    \( \displaystyle = 2\left(\frac{\pi}{2} - \tan^{-1}e^x - \frac{\pi}{4}\right) \)

    \( \displaystyle = -2\left(\tan^{-1}e^x - \frac{\pi}{4}\right) = -f(x) \)

So f is an odd function.

Increasing or decreasing: by the chain rule,

    \( \displaystyle f'(x) = 2\times\frac{1}{1+e^{2x}}\times e^x = \frac{2e^x}{1+e^{2x}} \)

Both \(e^x\) and \(1+e^{2x}\) are positive for every x, so \(f'(x) \gt 0\) and f is strictly increasing on \((-\infty, \infty)\).

Note: This topic was not in the CUET 2026 syllabus.