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when x = 0,1 -ve sign cancels at x= 2 denominator is zero implies the value is infinty. So, x!= 2 x = 3,4,5,6,7,8,9 numerator is negative OR Zero and denominator is positive Rest of the values x >9 given equation is positive So, Given equation satisfies, implies for 7 integers it satisfies the given condition
Option B is correct
Originally posted by Dare Devil on 16 Jan 2019, 01:25.
Last edited by Dare Devil on 18 Jan 2019, 07:31, edited 1 time in total.
How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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16 Jan 2019, 01:47
simplify relation we get (x-9) * ( x+5)/ ( x-2) <=0
values of x would satisfy relation <=0 x at 2,3,4,5,6,7,8,9 satisfies the relation as for 0,1, we get >0 values and for integers beyond 9 as well we get >0 values
IMO C ; 8
EgmatQuantExpert wrote:
How many non-negative integers satisfy the inequality \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)?
We can cancel out the common term, x + 2, in both numerator and denominator
• \(\frac{(x – 9)(x + 5)}{(x - 2)} ≤ 0\)
Now, multiplying and dividing by (x – 2), we get,
• \(\frac{(x – 2)(x - 9)(x + 5)}{(x - 2)^2} ≤ 0\)
In the above expression, the denominator is always greater than 0. Thus, the numerator must be ≤ 0.
• (x – 2)(x - 9)(x + 5) ≤ 0
Representing the zero points {-5, 2, 9} on a number line, we can find the regions in which the expression is ≤ 0
• From the above figure, we can see that (x – 2)(x - 9)(x + 5) ≤ 0 in the regions, 2 ≤ x ≤ 9 and x ≤ -5 • And, we know that x ≠ 2, since, the denominator of the given expression cannot be = 0
We are asked to find the number of non-negative integer values of x.
• Therefore, the values of x can be {3, 4, 5 ,6 ,7 ,8, 9}
We can cancel out the common term, x + 2, in both numerator and denominator
• \(\frac{(x – 9)(x + 5)}{(x - 2)} ≤ 0\)
Now, multiplying and dividing by (x – 2), we get,
• \(\frac{(x – 2)(x - 9)(x + 5)}{(x - 2)^2} ≤ 0\)
In the above expression, the denominator is always greater than 0. Thus, the numerator must be ≤ 0.
• (x – 2)(x - 9)(x + 5) ≤ 0
Representing the zero points {-5, 2, 9} on a number line, we can find the regions in which the expression is ≤ 0
• From the above figure, we can see that (x – 2)(x - 9)(x + 5) ≤ 0 in the regions, 2 ≤ x ≤ 9 and x ≤ -5 • And, we know that x ≠ 2, since, the denominator of the given expression cannot be = 0
We are asked to find the number of non-negative integer values of x.
• Therefore, the values of x can be {3, 4, 5 ,6 ,7 ,8, 9}
so wont it be =0 and should be valid at x=2 would be a valid to count as an integer ... considering this relation only I opted for option C '8' over 7 option b ..
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Re: How many non-negative integers satisfy the inequality x^2 – 7x – 18 .
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18 Jan 2019, 07:13
close
46% (02:08) correct 
