1
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$\alpha, \beta$ are the roots of the equation $x^{2}+2 x+4=0$. If the point representing $\alpha$ in the argand diagram lies in the 2nd quadrant and $\alpha^{2024}-\beta^{2024}=i k,(i=\sqrt{-1})$, then $k=$
A
$-2^{2025} \sqrt{3}$
B
$2^{2025} \sqrt{3}$
C
$-2^{2024} \sqrt{3}$
D
$2^{2004} \sqrt{3}$
2
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If $\alpha$ is a root of the equation $x^{2}-x+1=0$, then $\left(\alpha+\frac{1}{\alpha}\right)^{3}+\left(\alpha^{2}+\frac{1}{\alpha^{2}}\right)^{3}+\left(\alpha^{3}+\frac{1}{\alpha^{3}}\right)^{3}+\left(\alpha^{4}+\frac{1}{\alpha^{4}}\right)^{3}=$
A
0
B
1
C
-3
D
-9
3
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
$\alpha, \beta$ are the real roots of the equation $x^{2}+a x+b=0$. If $\alpha+\beta=\frac{1}{2}$ and $\alpha^{3}+\beta^{3}=\frac{37}{8}$, then $a-\frac{1}{b}=$
A
$\frac{-1}{6}$
B
$\frac{3}{2}$
C
$\frac{-3}{2}$
D
$\frac{1}{6}$
4
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
The solution set of the inequation $\sqrt{x^{2}+x-2} > (1-x)$ is
A
$(-\infty, 2)$
B
$(-\infty,-2)$
C
$(1, \infty)$
D
$(0, \infty)$
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