MHTCET 18 Apr 2026 Shift 1 Paper

Concept:Use the identity $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$ to rewrite each term as a difference of inverse tangents.Explanation:Express the given fraction as a difference:$\frac{1}{n^{2}+n+1} = \frac{(n+1)-n}{1+n(n+1)}$.Therefore,$\tan^{-1}\left(\frac{1}{n^{2}+n+1}\right) = \tan^{-1}(n+1) - \tan^{-1}(n)$.The sum becomes a telescoping series:$\sum_{n=1}^{2026}\left[\tan^{-1}(n+1) - \tan^{-1}(n)\right] = \tan^{-1}(2027) - \tan^{-1}(1)$.Apply the same identity to combine the two terms:$\tan^{-1}(2027) - \tan^{-1}(1) = \tan^{-1}\left(\frac{2027-1}{1+2027}\right)$.Simplify the fraction:$\frac{2026}{2028} = \frac{1013}{1014}$.Hence, the given sum equals $\tan^{-1}\left(\frac{1013}{1014}\right)$.This is equal to $\tan^{-1}\left(1 - \frac{1}{x}\right)$, so:$1 - \frac{1}{x} = \frac{1013}{1014}$.Thus, $\frac{1}{x} = 1 - \frac{1013}{1014} = \frac{1}{1014}$.Therefore, $x = 1014$.Answer:$x = 1014$, which is Option C.