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
If $x$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\frac{\left(1+\frac{2 x}{3}\right)^{-4}(4+5 x)^{1 / 2}}{(9+x)^{3 / 2}}$, when $x=\frac{6}{371}$, is
TS EAMCET Papers
All year-wise previous year question papers