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03 Jan 2025, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
48% (02:41) correct
52%
(02:23)
wrong
based on 128
sessions
History
Date
Time
Result
Not Attempted Yet
For how many integral values of x, the expression x^2 – 7x -2 is always less than 6 and greater than -8?
A) 0
B) 1
C) 2
D) 3
E) 4
A) 0
B) 1
C) 2
D) 3
E) 4
We can rewrite to:
\(x^2 - 7x -2 < 6\)
\(x^2 - 7x -2 > -8\)
OR working out further:
\(x(x-7) < 8 \)
\(x(x-7) > - 6\)
The first equation will be 8 if x = - 1 or x = 8 and the second equation will be -6 if x = 1 or x = 6
So for the 2 equations you get:
x element of \(]-1, 8[\)
x element of \(]inf , 1[\) and \(]6 , inf[\)
Combined you get that x= 0 or x= 7 are the only two options: 2
IMO: C
\(x^2 - 7x -2 < 6\)
\(x^2 - 7x -2 > -8\)
OR working out further:
\(x(x-7) < 8 \)
\(x(x-7) > - 6\)
The first equation will be 8 if x = - 1 or x = 8 and the second equation will be -6 if x = 1 or x = 6
So for the 2 equations you get:
x element of \(]-1, 8[\)
x element of \(]inf , 1[\) and \(]6 , inf[\)
Combined you get that x= 0 or x= 7 are the only two options: 2
IMO: C
03 Jan 2025, 04:21
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Bunuel
One way could be to plot rough graph and identify overlapping areas.
Minima of this quadratic equation would be at -b/2a = 7/2, for which value of this equation would be 7/2*7/2 -7*7/2 - 2 = -57/4 = ~(-14)
So minima is (x, y) = (7/2, -57/4)
As the given range (-8 , 6) would be above the minimal there would be 2 overlapping areas both on the left and right side of the minima.
During test it would be easier to identify both these overlaps by trying some obvious values like 0 and 7, as that cancels both the x^2 and x terms in the quadratic equation giving us -2 which falls in our required range.
If we try integer values around 0 ie. -1 and 1 we will get values 6 and -8 respectively which falls outside our required range
Similarly trying values around 7 ie. 6 and 8 will also result in -8 and 6 respectively which is again outside the range.
If we closely look at the attached diagram, we can see that the quadratic equation has one integer solution within the given range on both the sides of the minima.
Hence, no. of integer values of x withing the given range should be 2.
IMO: C
Attachments
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