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BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

$$\left\{ {x \in R:{{2x - 1} \over {{x^3} + 4{x^2} + 3x}} \in R} \right\}$$ is equal to

A
$$R - \{ 0\} $$
B
$$R - \{ 0,1,3\} $$
C
$$R - \{ 0, - 1, - 3\} $$
D
$$R - \left\{ {0, - 1, - 3,{1 \over 2}} \right\}$$
2
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

The solution set of $${{|x - 2|\, - 1} \over {|x - 2|\, - 2}} \le 0$$ is

A
[0, 1] $$\cup$$ (3, 4)
B
[0, 1] $$\cup$$ [3, 4]
C
[$$-$$1, 1] $$\cup$$ (3, 4]
D
None of these
3
BITSAT 2020
MCQ (Single Correct Answer)
+3
-1

Let $$f(x) = {x \over {\sqrt {1 + {x^2}} }}$$, $$\underbrace {fofofo.....of(x)}_{x\,times}$$ is

A
$${x \over {\sqrt {1 + \left( {\sum\limits_{r = 1}^n r } \right){x^2}} }}$$
B
$${x \over {\sqrt {1 + \left( {\sum\limits_{r = 1}^n 1 } \right){x^2}} }}$$
C
$${\left( {{x \over {\sqrt {1 + {x^2}} }}} \right)^x}$$
D
$${x \over {\sqrt {1 + n{x^2}} }}$$
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