Question
For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$.
If $f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}$, then
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$f'(x)$ does not exist
To determine whether the derivative of the function \(f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}\) exists for all \(x\), we need to analyze the behavior of the function. Here's a detailed explanation:
- The function involves the floor function \([x]\), which denotes the greatest integer less than or equal to \(x\). This can lead to discontinuities at integer points, as the value of \([x]\) changes abruptly.
- The expression \(\tan(\pi [x-\pi])\) needs careful consideration. Since tangent is discontinuous at odd multiples of \(\frac{\pi}{2}\), the function may exhibit discontinuities.
- For any non-integer \(x\), \([x]\) remains constant across a small neighborhood around that number, so in these cases, the derivative can be considered.
- At integer values of \(x\), \([x]\) changes suddenly, causing a discontinuity in the function. This affects the tangent term and consequently the entire expression, which can result in undefined or non-differentiable points due to jumps in function values.
As a result of the discontinuities at integer values of \(x\) caused by the floor function, \(f'(x)\) does not exist for these points. Therefore, the correct answer is:
\(f'(x)\) does not exist
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