Question of the week - 31 (How many non-positive integers satisfy ...)

Question

e-GMAT          Question of the Week #31

How many non-positive integers satisfy the inequality  \frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0 ?

A.  1
 B.  2
 C.  3
 D.  4
 E.  Infinite

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg

Archit3110 · Answer

Solution attached

Posted from my mobile device

Kem12 · Answer

hi  chetan2u  Pls please help me out with solving this using wavy line method
I have tried but i still keep getting B as my answer
Thanks
Also in the Explanation in the picture the writer wrote (9-x)^2 but the question says (9-x^2) they are not the same thing

chetan2u · Answer

\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0 ...

Hi..

You can get your answer by just looking at the extreme values of X for which the equation becomes 0..
Here the values will be 3 and -3, but since we are looking at only non positive integers, see the effect on the entire equation beyond x less than -3..
So substitute X as -4..
 \frac{(9 – (-4)^2)(-4+1)(-4 + 2)^2}{(-4+ 3)(-4– 2)} ≤ 0......\frac{(-7)*(-3)*4}{(-1)*(-6)}=14>0 ..
So beyond x<-3, the equation will be positive, so discard all the values..

Now, we are left with 0,-1,-2,-3
Discard -3 as the denominator becomes 0 and our equation becomes undefined..
-1 and -2 will give you a 0 as value as numerator has (x+1) and (x+2) in numerator, so ok.
When x is 0, term becomes 9*1*2/(3*(-2))=-3, so ok..

Therefore values will be 0, -1, -2--------3 values

C

SaladQueen · Answer

X cannot be equals to -3 or 2 since numerator is (3+X)(X-2).
Non positive integer starts from 0 all the way to negative infinity.
So just substitute X=0, X=-1, X=-2, X=-4 into the equation and see if the answer is a non positive value.
When X=0, the equation is negative.
When X=-1, the equation is negative.
When X=-2, the equation is negative.
When X>=-4, the equation is positive.
C.

Kem12 · Answer

Thanks so much  chetan2u

Posted from my mobile device

EgmatQuantExpert · Answer

Solution 
  Given:  
 •  \frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0

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

Approach and Working:   
 •  \frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0

The denominator must not be = 0. Thus, x ≠ {-3, 2}
 •  \frac{(3 – x)(3 + x)(x + 1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0

We can cancel out the common term (x + 3) in both numerator and denominator
 •  \frac{(3 – x)(x + 1)(x + 2)^2}{(x – 2)} ≤ 0

And, we know that  (x + 2)^2  is always ≥ 0. It is equal to 0, when x = -2
 • Thus, if x = -2, the inequality will be 0
• So, we get,  \frac{(3 – x)(x + 1)}{(x – 2)} ≤ 0

Now, if we multiply both numerator and denominator by (x – 2), we get,
 •  \frac{(3 – x)(x + 1)(x - 2)}{(x – 2)^2} ≤ 0 
 o Implies, (3 – x)(x + 1)(x - 2) ≤ 0
o Multiplying by -1 on both sides, we get, (x - 3)(x + 1)(x - 2) ≥ 0

• The zero points of the above inequality are x = {-1, 2, 3}

Let’s represent this on a number line and identify the regions, where this expression will give a non-negative value.
 
 https://e-gmat.com/blogs/wp-content/uploads/2019/01/QoW-30.png

• From the above diagram, we can see that (x - 3)(x + 1)(x - 2) ≥ 0 in the regions x ≥ 3 and -1 ≤ x ≤ 2.
• But, we are asked for non-positive values of x, they are x = {-1, 0}

Therefore, the non-positive values of x, which satisfy the inequality,  \frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0  are x = {-2, -1, 0}

Hence the correct answer is Option C.

Answer: C

https://e-gmat.com/blogs/wp-content/uploads/2018/07/FT-GC-FB-event-e1532524398949.jpg

bumpbot · Answer

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Question of the week - 31 (How many non-positive integers satisfy ...)

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Question of the week - 31 (How many non-positive integers satisfy ...)

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