If $\sum\limits_{n=1}^{2026}\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right)$, where $x \neq 0$, then $x = $

Let $f(x) = ax + b$ and $g(x) = cx + d$. The condition $f(g(x)) = g(f(x))$ holds for all $x$ if and only if ...

The number of point / points where the function $f(x) = \dfrac{1}{x^2-5|x|+6}$ is discontinuous is......

If $x\,e^{xy} = y + \sin^2 x$, then the value of $\dfrac{dy}{dx}$ at $x = 0$ is equal to