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

In $\triangle A B C, \angle B=90^{\circ}$ and $(b+a)$ is always a constant. In order that $\triangle A B C$ encloses the maximum area, $\angle C=$

A

$\frac{\pi}{4}$

B

$\frac{\pi}{6}$

C

$\frac{\pi}{3}$

D

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

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

If $\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}}=A \cdot \tan ^{-1} \sqrt{f(x)}+$ constant, then the ordered pair $(A, f(-1))=$

A

$(2,1)$

B

$(2,-1)$

C

$(1,2)$

D

$(-2,2)$

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

If $f\left(\frac{2 x+3}{3 x+5}\right)=x+4, x \neq \frac{-5}{3}, \frac{2}{3}$ and $\int f(x) d x=A x+B \ln |3 x-2|+C$, then $3 B-A=$

A

$\frac{64}{9}$

B

$\frac{-52}{21}$

C

$\frac{-10}{3}$

D

$\frac{-8}{3}$

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

If $\int e^x\left(\frac{x^2-8 x+19}{(x-1)^5}\right) d x=\frac{e^x(l x+m)}{(x-1)^4}+C$, then $4 l+m=$

A

-5

B

-2

C

1

D

0

TS EAMCET Papers

All year-wise previous year question papers