If the 9th and 10th terms are the numerically greatest terms in the expansion of $(5 x-6 y)^n$ when $x=2 / 5$ and $y=1 / 2$, then the absolute value of the middle terms of that expansion is
$$ 1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^2-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^3+\ldots $$
If $\frac{2 x+1}{(x-1)^2\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+1}$, then $A+B+C+D=$
Let $a$ be maximum value of $(3 \cos \theta-4 \sin \theta)$ and $\theta \neq \frac{n \pi}{2}$. If $\alpha=a \sin ^2 \theta \cdot \cos ^3 \theta$ and $\beta=a \sin ^3 \theta \cdot \cos ^2 \theta$, then $\sqrt{\frac{\left(\alpha^2+\beta^2\right)^5}{(\alpha \beta)^4}}=$
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