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# Quant Questions

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Intern
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08 Apr 2011, 09:35
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25% (00:00) correct 75% (01:13) wrong based on 4 sessions

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For which of the following values of x is √(1-√(2- √x) ) not defined as real number?

A. 1
B. 2
C. 3
D. 4
E. 5
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08 Apr 2011, 10:02
Krupa2 wrote:
For which of the following values of x is √(1-√(2- √x) ) not defined as real number?

A. 1
B. 2
C. 3
D. 4
E. 5

$$\sqrt{2-\sqrt{x}} \ge 0$$

$$\sqrt{2-\sqrt{x}} \ge 0$$

Squaring both sides:
$$2-\sqrt{x} \ge 0$$

$$-\sqrt{x} \ge -2$$

Multiplying both sides by -1
$$\sqrt{x} \le 2$$

Squaring both sides:
$$x \le 4$$

x can't be 5 because x should be less than or equal to 4.

Ans: "E"
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08 Apr 2011, 18:23
The answer is clearly E. If you evaluate the expression, anything that makes 2 - root(x) inside root(2 - root(x)) -ve will make the expression as complex number, hence 5 is the only answer choice that does so, because 2 - root(5) is -ve. The remaining choices still keep the expression as real number.
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Manager
Joined: 17 Jan 2011
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08 Apr 2011, 21:26
√(1-√(2- √x) )
if 2- √x is -ive, the result is an imaginary number.
For x=5, 2- √x is -ive
Ans. E
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14 Apr 2011, 16:21
2-sqrt(x) >=0

-sqrt(x) >= -2

=> x<=4

so cannot be 5.

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17 Apr 2011, 14:39
fluke wrote:
Krupa2 wrote:

$$\sqrt{2-\sqrt{x}} \ge 0$$

why is this being taken as greater than 0 but not 1
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17 Apr 2011, 15:03
ravsg wrote:
fluke wrote:
Krupa2 wrote:

$$\sqrt{2-\sqrt{x}} \ge 0$$

why is this being taken as greater than 0 but not 1

Square root of a non-negative real number will always be greater than equal to 0. Square root of a negative real number is non-existential.
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Re: Quant Questions   [#permalink] 17 Apr 2011, 15:03
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