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Updated on: 23 May 2014, 01:09
Originally posted by peacewarriors on 21 May 2014, 03:20.
Last edited by WoundedTiger on 23 May 2014, 01:09, edited 2 times in total.
Last edited by WoundedTiger on 23 May 2014, 01:09, edited 2 times in total.
Attachment removed.Fig in png format added
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
53% (02:22) correct
47%
(02:53)
wrong
based on 74
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A coordinate plane is shown in the figure. A curve can be drawn using the equation "a*f(x) = bx – 2b" (ab ≠ 0). If (1, t) is a point on the curve, then in which quadrant does the point lie?
(1) The curve crosses the y-axis on its positive side.
(2) The curve crosses the x-axis on its positive side.
21 May 2014, 03:53
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For Statement I:
Since (1,t) point lies on the curve, so we have t = b-2b i.e b= -t.
Now since, curve crosses the y-axis on its positive side, so we have b*0-2b = +ve i.e b= -ve.
From above information, as b= -t, so t is +ve. Thus (1,t) lies in quarter 1.
Statement 1 is sufficient
For Statement II:
+ve = b*x - 2b -> Sign of b can't be determined. Hence statement 2 is insufficient.
Since (1,t) point lies on the curve, so we have t = b-2b i.e b= -t.
Now since, curve crosses the y-axis on its positive side, so we have b*0-2b = +ve i.e b= -ve.
From above information, as b= -t, so t is +ve. Thus (1,t) lies in quarter 1.
Statement 1 is sufficient
For Statement II:
+ve = b*x - 2b -> Sign of b can't be determined. Hence statement 2 is insufficient.
23 May 2014, 00:41
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Hi Anujpadia,
Hope it helps
Rgds,
Rajat Jain
Hope it helps
Rgds,
Rajat Jain
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