1
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

At an election a voter may vote for any number of candidates not exceeding the number to be elected. If 4 candidates are to be elected out of the 12 contested in the election and voter votes for at least one candidate, then the number of ways in which a voter can vote is

A

793

B

298

C

781

D

1585

2
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $x \in \mathbf{R}$ be so small that the powers of $x$ beyond two are insignificant and negligibly small. For such $x$, if $(1-x)^3(2+x)^6$ is approximated by $a+b x+c x^2$, then $a+b+c=$

A

-80

B

144

C

80

D

127

3
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

For $0 < x < 1$, the expansion of $\left(1+\frac{1}{x}\right)^{\frac{1}{2}}$ is

A

$1+\frac{1}{2 x}-\frac{1}{2!}\left(\frac{1}{2 x}\right)^2+\frac{1 \cdot 3}{3!}\left(\frac{1}{2 x}\right)^3-\frac{1 \cdot 3 \cdot 5}{4!}\left(\frac{1}{2 x}\right)^4+\ldots \infty$

B

$\frac{1}{\sqrt{x}}+\frac{1}{2} \sqrt{x}-\frac{1}{2!} \frac{x \sqrt{x}}{2^2}+\frac{1 \cdot 3}{3!} \frac{x^2 \sqrt{x}}{2^3}-\ldots . \infty$

C

$1+\frac{1}{\sqrt{x}}+\frac{1}{2} x \sqrt{x}+\frac{1}{2!} \frac{x^2 \sqrt{x}}{2^3}+\frac{1 \cdot 3}{3!} \frac{x^3 \sqrt{x}}{2^4}+\ldots . \infty$

D

$\frac{1}{\sqrt{x}}+\frac{1}{2 x \sqrt{x}}-\frac{1}{2!}\left(\frac{1}{2 x}\right)^2 \frac{1}{\sqrt{x}}+\frac{1 \cdot 3}{3!}\left(\frac{1}{2 x}\right)^3 \frac{1}{\sqrt{x}}-\ldots \ldots \infty$

4
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1} +\frac{C x^3+D x^2+E x+F}{x^4+x^2+1}$, then $B+2(D+F+E)-C \cdot A=$

A

0

B

3

C

1

D

-3

TS EAMCET Papers

All year-wise previous year question papers