1
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The area (in sq. units) of the portion lying above the $X$-axis and enclosed between the curves $y^2=2 a x-x^2$ and $y^2=a x$ is

A

$a^2\left(\frac{-\pi}{2}+\frac{2}{3}\right)$

B

$a^2\left(\frac{2}{3}-\frac{\pi}{4}\right)$

C

$a^2\left(\frac{\pi}{4}-\frac{2}{3}\right)$

D

$a^2\left(\frac{\pi}{4}+\frac{2}{3}\right)$

2
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The differential equation for which $l x^2+m y^2=x+y$ is the general solution is

A

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y^{\prime} y & y^{\prime}+1 \\ 2 & 2 y y^{\prime \prime} & y^{\prime \prime}\end{array}\right|=0$

B

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y y^{\prime} & 1+y^{\prime} \\ 2 & 2\left(y^{\prime 2}+y y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0$

C

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y y^{\prime} & y+1 \\ 2 & 2\left(y^{\prime 2}+y^{\prime} y^{\prime \prime}\right) & y^{\prime \prime}\end{array}\right|=0$

D

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2 x & 2 y & 1+y^{\prime} \\ 2 & 2 y y^{\prime} y^{\prime \prime} & y^{\prime \prime}\end{array}\right|=0$

3
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $(x-2 y+1) d y-(3 x-6 y+2) d x=0$ is

A

$\left|x+2 y+\frac{3}{5}\right|^{2 / 25} \cdot e^{115(x+2 y)}=C$

B

$\left|x-2 y+\frac{3}{5}\right|^{2 / 25} \cdot e^{1 / 5(x-2 y)}=C$

C

$\left|x-2 y+\frac{3}{5}\right|^{\frac{2}{25}} \cdot e^{1 / 5(6 x-2 y)}=C$

D

$\left|x-2 y+\frac{1}{5}\right|^{\frac{2}{25}} \cdot e^{1 / 5(x-2 y)}=C$

4
TS EAMCET 2020 (Online) 11th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$ is

A

$x=\left(\tan ^{-1} y\right)-1+C e^{-\tan ^{-1} y}$

B

$x=\left(\tan ^{-1} y\right)-1+C e^{\tan ^{-1} y}$

C

$x=\left(\tan ^{-1} y\right)-1+C$

D

$x=\left(\tan ^{-1} y\right)+C e^{-\tan ^{-1} y}$

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