1
TS EAMCET 2023 (Online) 13th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int_0^2 x^{\frac{5}{2}} \sqrt{2-x} d x= $$

A

$\frac{5 \pi}{16}$

B

$\frac{5}{4}$

C

$\frac{5 \pi}{8}$

D

$\frac{5}{8}$

2
TS EAMCET 2023 (Online) 13th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Area of the region bounded by the curve $y=2-x-3 x^2$, the $X$-axis, the $Y$-axis and the line $x=-2$ is

A

2

B

$\frac{44}{27}$

C

$\frac{9}{2}$

D

5

3
TS EAMCET 2023 (Online) 13th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $A$ and $B$ are arbitrary constants, then the differential equation having $y=A e^x+B \sin 2 x$ as its general solution is

A

$$ \begin{aligned} & (\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} -4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

B

$$ \begin{aligned} & (\cos 2 x+\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} -4(\sin 2 x-\cos 2 x) y=0 \end{aligned} $$

C

$$ \begin{aligned} & (\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} +4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

D

$$ \begin{aligned} & (\sin 2 x-\cos 2 x) \frac{d^2 y}{d x^2}-(4 \sin 2 x) \frac{d y}{d x} -4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

4
TS EAMCET 2023 (Online) 13th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $\frac{d y}{d x}=\sin (x-y)+\cos (x-y)$ is

A

$\log \left|\frac{\tan \frac{(x-y)}{2}+1}{\tan \frac{(x-y)}{2}}\right|=x+C$

B

$\log \left|\frac{\tan \frac{(x-y)}{2}-1}{\tan \frac{(x-y)}{2}}\right|=x+C$

C

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

D

$\log \left|\frac{\sin (x-y)+\cos (x-y)}{\cos (x-y)}\right|=x+C$

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