1
TG EAPCET 2024 (Online) 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$\int \frac{\left(1-4 \sin ^2 x\right) \cos x}{\cos (3 x+2)} d x=$
A
$(\cos 2) x-\frac{1}{3}(\sin 2) \log |\sec (3 x+2)|+c$
B
$(\sin 2) x-\frac{1}{3}(\cos 2) \log |\cos (3 x+2)|+c$
C
$(\sin 2) x+\frac{1}{3}(\cos 2) \log |\cos (3 x+2)|+c$
D
$(\cos 2) x+\frac{1}{3}(\sin 2) \log |\sec (3 x+2)|+c$
2
TG EAPCET 2024 (Online) 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$\int \frac{\left(1-4 \sin ^2 x\right) \cos x}{\cos (3 x+2)} d x=$
A
$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^4-1}{\sqrt{2} x^2}\right)+c$
B
$\log \left(x^5+x^2\right)-\log \left(x^3+x\right)+\log (x+1)+c$
C
$\frac{2}{9} x^8-\frac{4}{9} x^6+\frac{1}{9} x^4-\frac{1}{3} x^2+c$
D
$\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^5-1}{\sqrt{2} x^3}\right)+c$
3
TS EAMCET 2023 (Online) 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $\frac{x+1}{\left(x^2+1\right)(x-1)^2}=\frac{A x+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$, then $A+B+C+D=$

A
$-\frac{1}{2}$
B
$\frac{1}{2}$
C
1
D
$\frac{3}{2}$
4
TS EAMCET 2023 (Online) 12th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
$$ \text { Match the following items from List I into List II } $$
List-I List-II
1. sin 2 x cos 4 x d x sin 2 x cos 4 x d x int(sin^(2)x)/(cos^(4)x)dx A. tan 2 x 2 + ln | cos x | + C tan 2 x 2 + ln | cos x | + C quad(tan^(2)x)/(2)+ln |cos x|+C
2. sin 4 x cos 2 x d x sin 4 x cos 2 x d x int(sin^(4)x)/(cos^(2)x)dx B. cos x + sec x + C cos x + sec x + C cos x+sec x+C
3. sin 3 x cos 2 x d x sin 3 x cos 2 x d x int(sin^(3)x)/(cos^(2)x)dx C. tan 3 x 3 + C tan 3 x 3 + C (tan^(3)x)/(3)+C
4. sin 3 x cos 3 x d x sin 3 x cos 3 x d x int(sin^(3)x)/(cos^(3)x)dx D. tan x + sin 2 x 4 3 x 2 + C tan x + sin 2 x 4 3 x 2 + C tan x+(sin 2x)/(4)-(3x)/(2)+C
E. cos x sec x + C cos x sec x + C cos x-sec x+C
Select the correct choice
A
1-C, 2-E, 3-B, 4-A
B
1,-C, 2-D, 3-B, 4-A
C
1-D, 2-C, 3-A, 4-B
D
1-C, 2-E, 3-A, 4-D
TS EAMCET Subjects
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