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805+ (Hard)|   Algebra|      
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How many non-negative integers satisfy the inequality x^2 – 7x – 18 . 16 Jan 2019, 02:14
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B
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95% (hard)

Question Stats:

47% (02:11) correct 53% (02:18) wrong based on 358 sessions

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How many non-negative integers satisfy the inequality \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)?

    A. 6
    B. 7
    C. 8
    D. 9
    E. Infinite

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B
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How many non-negative integers satisfy the inequality x^2 – 7x – 18 . Updated on: 18 Jan 2019, 08:31
Originally posted by Dare Devil on 16 Jan 2019, 02:25.
Last edited by Dare Devil on 18 Jan 2019, 08:31, edited 1 time in total.
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\((x2–7x–18)(x+5)/(x2–4)≤0\)

= \(((x+2)(x-9)(x+5))/((x-2)(x+2))\)
= ((x-9)(x+5))/(x-2))
non-negative implies 0,1,2,3,4.....

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
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How many non-negative integers satisfy the inequality x^2 – 7x – 18 . 16 Jan 2019, 02:47
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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

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How many non-negative integers satisfy the inequality \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)?

    A. 6
    B. 7
    C. 8
    D. 9
    E. Infinite

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How many non-negative integers satisfy the inequality x^2 – 7x – 18 . 18 Jan 2019, 01:48
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Solution


Given:
    • An inequality, \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)

To find:
    • The number of non-negative integer values of x that satisfy the given inequality

Approach and Working:
    • \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)

Let’s factorise the quadratic expressions in the numerator and denominator
    • \(\frac{(x – 9)(x + 2)(x + 5)}{(x + 2)(x - 2)} ≤ 0\)

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}

Hence the correct answer is Option B.

Answer: B

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How many non-negative integers satisfy the inequality x^2 – 7x – 18 . 18 Jan 2019, 08:13
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Solution


Given:
    • An inequality, \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)

To find:
    • The number of non-negative integer values of x that satisfy the given inequality

Approach and Working:
    • \(\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0\)

Let’s factorise the quadratic expressions in the numerator and denominator
    • \(\frac{(x – 9)(x + 2)(x + 5)}{(x + 2)(x - 2)} ≤ 0\)

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}

Hence the correct answer is Option B.

Answer: B

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@EgmatQuantExpert"

the expression
[m]\frac{(x^2 – 7x – 18)(x + 5)}{(x^2 – 4)} ≤ 0

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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How many non-negative integers satisfy the inequality x^2 – 7x – 18 . 31 May 2024, 23:15
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