1
TG EAPCET 2025 (Online) 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{\left(x^2+1\right)}$, then $3 A+2 B-C=$

A

$\frac{8}{5}$

B

$\frac{16}{5}$

C

$\frac{3}{5}$

D

$\frac{19}{5}$

2
TG EAPCET 2025 (Online) 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $5 \sin \theta+3 \cos \left(\theta+\frac{\pi}{3}\right)+3$ lies between $\alpha$ and $\beta$ (including $\alpha, \beta$ also), then $(\alpha-\beta)(\alpha+\beta-6)=$

A

$28-5 \sqrt{3}$

B

0

C

3

D

$28+5 \sqrt{3}$

3
TG EAPCET 2025 (Online) 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots \sin 89^{\circ}}{2\left(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ}\right)+1}= $$

A

2

B

$\frac{1}{\sqrt{2}}$

C

$\frac{1}{2}$

D

$\sqrt{2}$

4
TG EAPCET 2025 (Online) 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $3 \sin (\alpha-\beta)=5 \cos (\alpha+\beta)$ and $\alpha+\beta \neq \frac{\pi}{2}$, then $\frac{\tan \left(\frac{\pi}{4}-\alpha\right)}{\tan \left(\frac{\pi}{4}-\beta\right)}=$

A

0

B

-4

C

$-\frac{1}{4}$

D

$\frac{1}{2}$

TS EAMCET Papers

All year-wise previous year question papers