1
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$\int_{\frac{-\pi}{8}}^{\frac{\pi}{8}} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x=$
A
$\frac{3 \pi}{128}$
B
$\frac{3 \pi}{256}$
C
$\frac{3 \pi}{64}$
D
$\frac{3 \pi}{32}$
2
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The area of the region enclosed by the curves $y^2=4(x+1)$ and $y^2=5(x-4)$ is
A
$\frac{280}{3}$
B
150
C
140
D
$\frac{200}{3}$
3
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $A$ and $B$ are arbitrary constants, then the differential equation having $y=A e^{-x}+B \cos x$ as its general solution is
A
$(\sin x-\cos x) \frac{d^2 y}{d x^2}+2 \cos x \frac{d y}{d x}-(\sin x+\cos x) y=0$
B
$(\cos x-\sin x) \frac{d^2 y}{d x^2}+2 \cos x \frac{d y}{d x}+(\sin x+\cos x) y=0$
C
$(\cos x+\sin x) \frac{d^2 y}{d x^2}+2 \sin x \frac{d y}{d x}-(\sin x-\cos x) y=0$
D
$(\cos x-\sin x) \frac{d^2 y}{d x^2}-2 \sin x \frac{d y}{d x}+(\cos x+\sin x) y=0$
4
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The general solution of the differential equation $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ is
A
$(\sec x+\tan x)[\operatorname{cosec}(2 x+y)-\cot (2 x+y)]=c$
B
$\sin (2 x+y) \cos x=c$
C
$\cos (2 x+y) \sin x=c$
D
$(\operatorname{cosec} x-\cot x)(\sec (2 x+y)-\tan (2 x+y))=c$
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