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

$$ \text { Match the functions of List I with the items of List II. } $$

List I List II
A. 3 x 4 2 x 3 6 x 2 + 6 x + 1 3 x 4 2 x 3 6 x 2 + 6 x + 1 3x^(4)-2x^(3)-6x^(2)+6x+1 (I) has minimum value at x = 4 x = 4 x=4
B. x + 1 x , x < 0 x + 1 x , x < 0 x+(1)/(x),AA x < 0 (II) has maximum value at x = 1 x = 1 x=-1
C. x 4 ( 7 x ) 3 x 4 ( 7 x ) 3 x^(4)(7-x)^(3) (III) has maximum value at x = 4 x = 4 x=4
D. x 4 + ( 8 x ) 4 x 4 + ( 8 x ) 4 x^(4)+(8-x)^(4) (IV) is decreasing in [ 2 , ) [ 2 , ) [2,oo)
(V) is increasing in [ 2 , ) [ 2 , ) [2,oo)
A
A B C D
IV I II III
B
A B C D
V IV I III
C
A B C D
V II III I
D
A B C D
VI II I V
2
TS EAMCET 2020 (Online) 10th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

For $x \in\left(\frac{3 \pi}{4}, \pi\right), \int(\sqrt{1+\sin 2 x}+\sqrt{1-\sin 2 x}) d x=$

A

$-2 \cos x+C$

B

$2 \sin x+C$

C

$-2 \sin x+C$

D

$2 \cos x+C$

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

$$ \begin{aligned} & \text { If } \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x=A \log (|x \sin x+\cos x|) \\ & +B \frac{f(x)}{(x \tan x+1)}+C \text {, then } f(A+B)= \end{aligned} $$

A

1

B

0

C

-1

D

2

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

$$ \text { If } \begin{aligned} & \int x^3(\log x)^2 d x=x^4\left[A(\log x)^2+B(\log x)\right. \\ &+C \log e]+K, \text { then } A+B+C \end{aligned} $$

A

$\frac{7}{24}$

B

$\frac{4}{25}$

C

$\frac{3}{14}$

D

$\frac{5}{32}$

TS EAMCET Papers

All year-wise previous year question papers