**2. How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200**

To begin with - 0, -2, -4 and -6 work. These are the values for which the left-hand side goes to zero.

There are 4 terms in the product. If all 4 are positive or all 4 are negative the product will be positive

The
product can be negative only if exactly 1 or exactly 3 are negative. When 1 or 3 terms are negative, the product is clearly less than 200.

When x = -1, one term is negative

When x = -5, three terms are negative

So, adding these two numbers also to the set of solutions {-6, -5, -4, -2, -1, 0} satisfy the inequality.

Beyond this it is just trial and error.

Let us try x = 1. Product is 1 * 3 * 5 * 7 = 105. This works

x = -7 gives the same product. So, that also works.

So, the solution set is now refined to {-7, -6, -5, -4, -2, -1, 0, 1}

x = 2 => Product is 2 * 4 * 6 * 8 = 8 * 48. Not possible. Any x greater than 1 does not work

x = -8 is also not possible. Any value of x less than -7 does not work.

So, the solution set stays as {-7, -6, -5, -4, -2, -1, 0, 1}

The one missing value in this sequence is -3. When x = -3, product becomes -3 * -1 * 1 * 3. = 9. This also holds good.

So, values {-7, -6,-5, -4, -3, -2, -1, 0, 1} hold good. 9 different values satisfy this inequality

**3. Find the range of x where ||x - 3| - 4| > 3**

If we have an inequality |y| > 3, this will be satisfied if => y > 3 or y < -3.

So, the above inequality simplifies to two inequalities

**Inequality I: | x - 3| - 4 > 3 **

| x - 3| - 4 > 3 => | x - 3 | > 7
x - 3 > 7 or x - 3 < -7
Or, x > 10 or x < -4

Of, x lies outside
of -4 and 10. Or, x can lie in the range ( - infinity, -4) or ( 10,
infinity)

**Inequality II: |x -3| - 4 < -3**

|x - 3 | - 4 < - 3
=> | x - 3 | < 1
=> -1 < x - 3 < 1 or x lies between 2 and 4

So, final solution is given by the range

( - infinity, -4) or (2,4) or ( 10, infinity)