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24 Mar 2021, 02:30
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D
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:
55% (02:01) correct
45%
(01:55)
wrong
based on 514
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If x is an integer, which of the following inequality has one finite range of values of x satisfying it?
A. \(x^2 + 5x + 6 > 0\)
B. \(|x + 2| > 4\)
C. \(9x - 7 < 3x + 14\)
D. \(x^2 - 4x + 3 < 0\)
E. \(|x^2 - 5x - 6| > 0\)
A. \(x^2 + 5x + 6 > 0\)
B. \(|x + 2| > 4\)
C. \(9x - 7 < 3x + 14\)
D. \(x^2 - 4x + 3 < 0\)
E. \(|x^2 - 5x - 6| > 0\)
24 Mar 2021, 03:00
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Solution:
Let us use the options to reach at the correct answer choice.
A. x^2 + 5x + 6 > 0
=>(x+2)(x+3) > 0
=> x>-2 or x<-3
We have a range of values for x and hence infinite values of x.(Eliminate)
B . |x + 2| > 4
=> x+2>4 or x+2<-4
=> x>2 or x<-6
We have a range of values for x and hence infinite values of x.(Eliminate)
C. 9x - 7 < 3x + 14
=>x< 3.5
We have a range of values for x and hence infinite values of x.(Eliminate)
D:x^2 - 4x + 3 < 0
=> (x-3)(x-1)<0.
=> x is between 1<x<3
=> x has finite values.
Hence (option d)
Devmitra Sen (GMAT Quant Expert)
Let us use the options to reach at the correct answer choice.
A. x^2 + 5x + 6 > 0
=>(x+2)(x+3) > 0
=> x>-2 or x<-3
We have a range of values for x and hence infinite values of x.(Eliminate)
B . |x + 2| > 4
=> x+2>4 or x+2<-4
=> x>2 or x<-6
We have a range of values for x and hence infinite values of x.(Eliminate)
C. 9x - 7 < 3x + 14
=>x< 3.5
We have a range of values for x and hence infinite values of x.(Eliminate)
D:x^2 - 4x + 3 < 0
=> (x-3)(x-1)<0.
=> x is between 1<x<3
=> x has finite values.
Hence (option d)
Devmitra Sen (GMAT Quant Expert)
17 Jul 2021, 10:55
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We need to solve all the equations to see which one gives one finite range of values of x
A. \(x^2 + 5x + 6 > 0\)
=> \(x^2 + 2x + 3x + 6 > 0\)
=> \(x(x + 2) + 3(x + 2) > 0\)
=> \((x + 2) * (x + 3) > 0\)
Using, Sine Wave/Wavy Method [ Watch this video to learn about Sine Wave/Wavy method ]
image 5.jpg [ 9.79 KiB | Viewed 6901 times ]
As question is asking us about > 0 so we will take the values which are in the "+" zone
Solution will be x < 2 or x > 3 => NOT a finite range of values of x (because x can go from 2 to negative infinity and from 3 to positive infinity)
B. \(|x + 2| > 4\)
Theory: |x| > a => x > a or x < -a [ Watch this video to learn the Basics of Absolute Values ]
=> x+2 > 4 or x+2 < -4
=> x > 2 or x < -6 => NOT a finite range of values of x
C. \(9x - 7 < 3x + 14\)
=> 9x - 3x < 14 + 7
=> 6x < 21
=> x < \(\frac{21}{6}\)
=> x < 3.5 => NOT a finite range of values of x
D. \(x^2 - 4x + 3 < 0\)
=> \(x^2 - x - 3x + 3 < 0\)
=> \(x(x - 1) - 3(x - 1) < 0\)
=> \((x - 1)*(x - 3) < 0\)
Using, Sine Wave/Wavy Method [ Watch this video to learn about Sine Wave/Wavy method ]
image 6.jpg [ 10.41 KiB | Viewed 6645 times ]
As question is asking us about < 0 so we will take the values which are in the "-" zone
=> 1 < x < 3 => FINITE Range of values of x
In exam we will not proceed, but just to complete the problem solving E also
E. \(|x^2 - 5x - 6| > 0\)
We know that Absolute value of a number is always Non-Negative
=> \(|x^2 - 5x - 6| > 0\) will be true for all values of x except the values which can make it zero.
So, lets find the values which will make \(x^2 - 5x - 6 = 0\)
=> \(x^2 + x - 6x - 6 = 0\)
=> \(x(x + 1) - 6(x + 1) = 0\)
=> \((x + 1) * (x - 6) = 0\)
=> x = -1, 6
So, \(|x^2 - 5x - 6| > 0\) for all Real values of x except x = -1, 6 => NOT a finite range of values of x
So, answer will be D
Hope it helps!
Watch the following video to learn How to Solve an Inequality Problem using Algebra, Sine Wave or Wavy Method
A. \(x^2 + 5x + 6 > 0\)
=> \(x^2 + 2x + 3x + 6 > 0\)
=> \(x(x + 2) + 3(x + 2) > 0\)
=> \((x + 2) * (x + 3) > 0\)
Using, Sine Wave/Wavy Method [ Watch this video to learn about Sine Wave/Wavy method ]
Attachment:
image 5.jpg [ 9.79 KiB | Viewed 6901 times ]
As question is asking us about > 0 so we will take the values which are in the "+" zone
Solution will be x < 2 or x > 3 => NOT a finite range of values of x (because x can go from 2 to negative infinity and from 3 to positive infinity)
B. \(|x + 2| > 4\)
Theory: |x| > a => x > a or x < -a [ Watch this video to learn the Basics of Absolute Values ]
=> x+2 > 4 or x+2 < -4
=> x > 2 or x < -6 => NOT a finite range of values of x
C. \(9x - 7 < 3x + 14\)
=> 9x - 3x < 14 + 7
=> 6x < 21
=> x < \(\frac{21}{6}\)
=> x < 3.5 => NOT a finite range of values of x
D. \(x^2 - 4x + 3 < 0\)
=> \(x^2 - x - 3x + 3 < 0\)
=> \(x(x - 1) - 3(x - 1) < 0\)
=> \((x - 1)*(x - 3) < 0\)
Using, Sine Wave/Wavy Method [ Watch this video to learn about Sine Wave/Wavy method ]
Attachment:
image 6.jpg [ 10.41 KiB | Viewed 6645 times ]
As question is asking us about < 0 so we will take the values which are in the "-" zone
=> 1 < x < 3 => FINITE Range of values of x
In exam we will not proceed, but just to complete the problem solving E also
E. \(|x^2 - 5x - 6| > 0\)
We know that Absolute value of a number is always Non-Negative
=> \(|x^2 - 5x - 6| > 0\) will be true for all values of x except the values which can make it zero.
So, lets find the values which will make \(x^2 - 5x - 6 = 0\)
=> \(x^2 + x - 6x - 6 = 0\)
=> \(x(x + 1) - 6(x + 1) = 0\)
=> \((x + 1) * (x - 6) = 0\)
=> x = -1, 6
So, \(|x^2 - 5x - 6| > 0\) for all Real values of x except x = -1, 6 => NOT a finite range of values of x
So, answer will be D
Hope it helps!
Watch the following video to learn How to Solve an Inequality Problem using Algebra, Sine Wave or Wavy Method
