Kritisood wrote:
How many integrals value of x satisfy the inequality (1-x^2)(4-x^2)(9-x^2) > 0 ?
A. 0
B. 1
C. 3
D. 5
E. Greater than 5
\((1-x^2)(4-x^2)(9-x^2)\) 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
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