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

The $x$-coordinate changes on the curve $y=3 x^5+15 x-8$ at the rate of $\frac{1}{5}$ units/sec. $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ are the points on the curve at which the $y$-coordinate changes at the rate of 6 units/sec, then the slope of $A B=$

A

10

B

$\tan ^{-1}\left(\frac{1}{2}\right)$

C

18

D

$\tan ^{-1} 2$

2
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

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

3
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

$(-2,2)$

4
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

$\frac{-8}{3}$

TS EAMCET Papers

All year-wise previous year question papers