1
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int e^{-3 x}\left(x^2+\sin 4 x\right) d x= $$

A

$-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C$

B

$-e^{-3 x}\left(\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x+\frac{4}{25} \cos 4 x\right)+C$

C

$-e^{-3 x}\left(\frac{x^2}{3}+\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x-\frac{4}{25} \cos 4 x\right)+C$

D

$-e^{-3 x}\left(\frac{x^2}{3}-\frac{2 x}{9}+\frac{2}{27}+\frac{3}{25} \sin 4 x-\frac{4}{25} \cos 4 x\right)+C$

2
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$, then $\frac{m-l}{r}=$

A

3

B

4

C

5

D

6

3
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \mathop {\lim }\limits_{x \to \infty }\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}= $$

A

e

B

$2 e$

C

$2 e^{\frac{\pi-2}{2}}$

D

$2 e^{\frac{\pi-4}{2}}$

4
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int_{\pi / 4}^{\pi / 2} \frac{3 d x}{1+e^{\sqrt{8} \sin \left(x-\frac{3 \pi}{8}\right)}}= $$

A

$\frac{3 \sqrt{2}}{4} \pi$

B

$\frac{3}{4} \pi$

C

$\frac{\pi}{8}$

D

$\frac{3}{8} \pi$

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