How many integrals value of x satisfy the inequality (1-x2)(4-x2)(9-x2

to use the wavy line approach we need to get the factors in the form of (x+a) or (x-a). this can be done by taking -1 common.
(x^2-1)(x^2-4)(x^2-9) < 0 (flip the sign as multiplying by -1 on both sides)

draw the wavy line at points -3,-2,-1,1,2,3
only 0 is the integral value.

That approach doesn't always work. A more efficient way would be to first identify all the zeroes/critical points of the equation, then pick the largest range available from these critical points (in this case that range is x>3), substitute a relevant value, say 4, into the equation and see whether the resulting value is greater than 0 or lesser. If it is lesser than 0, start with a negative from the RHS and alternate the signs, based on the powers of roots, as you traverse left; if it is greater than 0, do the opposite, start with a positive from the RHS.

 

You have performed each computation in isolation and have ignored the fact that the obtained values for one computation can make the other term zero. For example, for the second term, \((4 - x^2) > 0\), we have obtained that the valid values can be -1, 0, and 1. However, among these three values, only 0 is valid for the inequality to hold true. This is because both -1 and 1 result in \((1-x^2) = 0 \) and the inequality then doesn't hold true. For the approach you've used, you should take the common terms. 

P.S. There is also a broader issue with this approach. It may not be applicable to this question, but thought of sharing it in general. The only criteria that you've used is when all three terms are positive. However, we can have other possible cases as well. Ex: when two terms are negative and one is positive. ­

\((1-x^2)(4-x^2)(9-x^2)\) has 6 zero solutions, two for each bracket we have currently. They are -3, -2, -1, 1, 2, 3.

Let's start from when x > 3, we would have (-) * (-) * (-) = (-), a negative result so none of the integers bigger than 3 satisfy this inequality.

The same applies for x < -3. Another quick way to see this is, each zero solution flips the sign of the result. We can start from the x > 3 case which gives a negative result, and to reach x < -3 we need to pass 6 zero solutions. Then the sign of \((1-x^2)(4-x^2)(9-x^2)\) would still remain negative after 6 flips.

Then we only have x = -3 through x = 3 as possible solutions. None of the zero's will give a positive result so we can only try x = 0. Plugging in that gives 1 * 4 * 9 > 0, our only solution.

Ans: B

(1-x^2)(4-x^2)(9-x^2) > 0

The critical values are x (for which function is zero) = {-3, -2, -1, 1, 2, 3}

All these values of consecutive integers except integer 0 among them so effectively we have 3 ranges

Range 1: x < -3
Range 2: x = 0
Range 3: x > 3

Only x = 0 satisfies for the function to be positive hence

Answer: Option B

I don't get why you start with negative on the RHS on your number line.

https://www.youtube.com/watch?v=j-jGsD2dweI

In this video it is clearly shown that we have to start with + on the right hand side, when solving those inequality.

We can check zero value:
x=0 at the points:
x=-3,-2,-1,1,2,3

For x>3 -> (-)(-)(-)<0
For 2<x<3 -> (-)(-)(+)>0 -> no integers between 2 and 3
for 1<x<2 (-)(+)(+)<0
for -1<x<1 (+)(+)(+)>0 -> only integer is x=0
for -2<x<-1 (-)(+)(+)<0
for -3<x<-2 (-)(-)(+)>0 -> no integers between -3 and -2
for x<-3 (-)(-)(-)<0

only 1 solution, x=0
B

egmat KarishmaB gmatophobia Kindly explain whats wrong with my approach­