Question

The sum of integers in the interval [-2, 5] satisfying the inequality ((x - 2)(x - 3) * (x - 5) ^ 3)/(x(x ^ 2 - 4)) >= 0 is​

Answer 1

Answer:

àns is 2

Step-by-step explanation:

step by step sol mark me brainlisit

Answer 2

Given: The interval [-2, 5] and the inequality ((x - 2)(x - 3) * (x - 5) ^ 3)/(x(x ^ 2 - 4)) >= 0.

To find: The sum of integers in the interval [-2, 5] satisfying the inequality ((x - 2)(x - 3) * (x - 5) ^ 3)/(x(x ^ 2 - 4)) >= 0.

Solution:

In the interval [-2,5], there are seven integers: -2, -1, 0, 1, 2, 3, 4 and 5. In order to satisfy the given inequation, the value of x must be such that it is greater than or equal to zero.

In case of the integer -2, the denominator becomes zero so the value of the inequation is undefined.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(x-2)(x-3) * (x-5)^{3}}{x((-2)^{2}-4)}$

                          $= \frac{(x-2)(x-3) * (x-5)^{3}}{x(0)}$

For -1, the value of the inequality is less than zero.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(-3)(-4) * (-125)}{3}$

                          $=-500$

For 0, the denominator is again zero so the inequation is undefined.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(x-2)(x-3) * (x-5)^{3}}{0(x^{2} -4)}$

                          $= \frac{(x-2)(x-3) * (x-5)^{3}}{0}$

For 1, the inequation is greater than zero since it gives a positive value.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(-1)(-2) * (-4)^{3}}{(-3)}$

For 2, the denominator is zero.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(x-2)(x-3) * (x-5)^{3}}{x((2)^{2}-4)}$

                          $= \frac{(x-2)(x-3) * (x-5)^{3}}{x(0)}$

For 3, the inequation becomes equal to zero.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(x-2)(0) * (x-5)^{3}}{x((2)^{2}-4)}$

                          $= \frac{0}{x((2)^{2}-4)}$

                          $= 0$

For 4, the inequation is less than zero because it's negative.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(2)(1) * (-1)}{48}$

For 5, the inequation is equal to zero.

$\frac{(x-2)(x-3) * (x-5)^{3}}{x(x^{2} -4)} = \frac{(x-2)(x-3) * (0)}{x(x^{2} -4)}$

                          $= \frac{0}{x((2)^{2}-4)}$

                          $= 0$

Hence, the integers in the interval satisfying the equation are 1, 3 and 5 and their sum is

$1+3+5 = 9$

Therefore, the sum of integers in the interval [-2, 5] satisfying the inequality ((x - 2)(x - 3) * (x - 5) ^ 3)/(x(x ^ 2 - 4)) >= 0 is 9.