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

The slope of the normal drawn at a point $P$ to the curve $y=x^3-10 x^2+31 x-30$ is $-\frac{1}{14}$. If the co-ordinates of $P$ are integers, then the $X$-intercept of the tangent drawn at $P$ to the given curve is

A

$\frac{-11}{7}$

B

22

C

$\frac{11}{7}$

D

-22

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

$x$ and $y$ are two positive integers such that $2 x+3 y=50$. If $x^2 y^3$ is maximum for $x=\alpha$ and $y=\beta$, then $\frac{\alpha}{2}+\frac{\beta}{5}=$

A

10

B

$10 / 3$

C

5

D

7

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

$$ \int \frac{1}{16-7 \sin ^2 x} d x= $$

A

$\frac{1}{12} \tan ^{-1}\left(\frac{3 \tan x}{4}\right)+C$

B

$\frac{1}{3} \sin ^{-1}\left(\frac{3 \sin x}{4}\right)+C$

C

$\frac{1}{12} \log \left(\frac{4-\sqrt{7} \sin x}{4+\sqrt{7} \sin x}\right)+C$

D

$\frac{1}{12} \log \left(\frac{4+\sqrt{7} \sin x}{4-\sqrt{7} \sin x}\right)+C$

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

$$ \int \frac{\sec ^2 x}{(\sec x+\tan x)^2} d x= $$

A

$\frac{3+(\sec x+\tan x)^2}{2(\sec x+\tan x)^3}+C$

B

$-\frac{1+3(\sec x+\tan x)^2}{6(\sec x+\tan x)^3}+C$

C

$-\frac{3+(\sec x+\tan x)^2}{2(\sec x+\tan x)^3}+C$

D

$-\frac{1+(\sec x+\tan x)}{3(\sec x+\tan x)^2}+C$

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