1
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \begin{array}{r}\mathop {\lim }\limits_{n \to \infty }\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}\right. \\ +\frac{n^{3 / 2}}{(n+4)^{5 / 2}}-\frac{n^{1 / 2}}{(n+6)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}} \\ \left.-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]= \end{array} $$

A

$\frac{-\sqrt{2}}{3}$

B

$\frac{-1}{9 \sqrt{3}}$

C

$\frac{\sqrt{2}}{3}$

D

$\frac{1}{9 \sqrt{3}}$

2
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]= $$

A

$\frac{\pi}{4}+\frac{3}{2} \ln 2$

B

$\frac{\pi}{2}+\frac{3}{4} \ln 2$

C

$\frac{\pi}{4}-\frac{3}{2} \ln 2$

D

$\frac{\pi}{4}+\frac{1}{2} \ln 2$

3
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4 x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6 A-9 \sqrt{3}=$

A

$\pi$

B

$2 \pi$

C

$3 \pi$

D

$4 \pi$

4
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the order and degree of the differential equation corresponding to the family of curves $(x-2)^2+(y-a)^2=b^2$, (where $a$ and $b$ are parameters) are $m$ and $n$ respectively, then $m^2+n=$

A

7

B

5

C

4

D

3

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