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08 May 2020, 03:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
54% (01:39) correct
46%
(01:25)
wrong
based on 87
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(a < \sqrt{2}\)?
(1) Point (a,0) is inside the circle \(x^2 + y^2 =3\)
(2) Point (a,1) is inside the circle \(x^2 + y^2 =3\)
Are You Up For the Challenge: 700 Level Questions
(1) Point (a,0) is inside the circle \(x^2 + y^2 =3\)
(2) Point (a,1) is inside the circle \(x^2 + y^2 =3\)
Are You Up For the Challenge: 700 Level Questions
08 May 2020, 03:43
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Statement 1-
If Point (a,0) is inside the circle, it must satisfies the inequality \(x^2+y^2 < 3\)
\(a^2+0 <3\)
\(-√3 < a < √3\)
Insufficient
Statement 2-
If Point (a,1) is inside the circle, it must satisfies the inequality \(x^2+y^2 < 3\)
\(a^2+1 <3\)
\(-√2 < a < √2\)
Sufficient
If Point (a,0) is inside the circle, it must satisfies the inequality \(x^2+y^2 < 3\)
\(a^2+0 <3\)
\(-√3 < a < √3\)
Insufficient
Statement 2-
If Point (a,1) is inside the circle, it must satisfies the inequality \(x^2+y^2 < 3\)
\(a^2+1 <3\)
\(-√2 < a < √2\)
Sufficient
Bunuel
08 May 2020, 05:53
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Bunuel
Solution
Step 1: Analyse Question Stem
We need to find if \(a < \sqrt{2}\)
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: Point (a,0) is inside the circle \(x^2+y^2=3\)
- • The given circle has centre at (0,0) and has a radius of \(\sqrt {3}\)
• If point (a, 0) lies inside the given circle then the distance between the center of the circle and the point must be less than the radius of the circle.
- o So, \(\sqrt{(a -0)^2 + (0 – 0)^2} < \sqrt {3}\)
\(⟹ a^2 < 3\)
\(⟹ a^2 - 3 < 0\)
\(⟹ - \sqrt{3} < a < \sqrt{3}\)
- o Therefore, we cannot be sure if \(a < \sqrt{2}\)
Statement 2: Point (a,1) is inside the circle \(x^2+y^2=3\)
- • Since, point (a,1) lies inside the circle \(x^2+y^2=3\)
- This means, \(\sqrt{(a- 0)^2 + (1-0)^2} < \sqrt {3}\)
\(⟹ a^2 + 1 < 3\)
\(⟹ a^2 < 2\)
\(⟹ a^2 - 2 < 0\)
\(⟹ - \sqrt {2}< a < \sqrt {2}\)
Thus, the correct answer is Option B.
