If $\alpha$ and $\beta$ are two complex roots of the equation $6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0$, then $\alpha+\beta=$
If $\alpha$ is a root of multiplicity 3 of the equation $x^5-8 x^4+25 x^3-38 x^2+28 x-8=0$, then $\alpha^2-5 \alpha+6=$
Consider the following statements:
I. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is 286 .
II. If $25!=10^n \times k,(k \in \mathbf{N})$, then $n=6$
Which one of the following options is true?
A student is allowed to select at least $(n+1)$ books but not all books from a collection of ( $2 n+1$ ) books. If the total number of ways in which he can select these books is 255 , then the number of books in that collection is
TS EAMCET Papers
All year-wise previous year question papers