Let `f:R rarr R` be a function as
`f(x)=(x-1)(x+2)(x-3)(x-6)-100`. If `g(x)` is a polynomial of degree `le 3` such that `int (g(x))/(f(x))dx` does not contain any logarithm function and `g(-2)=10`. Then
The equation `f(x)=0` has

Options

A. all four distinct roots
B. three distinct real roots
C. two real and two imaginary
D. all four imaginary roots

Answer

To solve the problem, we need to analyze the function \( f(x) = (x-1)(x+2)(x-3)(x-6) - 100 \) and determine how many roots the equation \( f(x) = 0 \) has.

### Step 1: Expand the function \( f(x) \)

First, we expand the polynomial:

\[ f(x) = (x-1)(x+2)(x-3)(x-6) - 100 \]

We can first multiply the pairs:

\[ (x-1)(x-6) = x^2 - 7x + 6 \]
\[ (x+2)(x-3) = x^2 - x - 6 \]

Now, we multiply these two results:

\[ f(x) = (x^2 - 7x + 6)(x^2 - x - 6) - 100 \]

### Step 2: Expand the product

Now we expand \( (x^2 - 7x + 6)(x^2 - x - 6) \):

\[ = x^4 - x^3 - 6x^2 - 7x^3 + 7x^2 + 42x + 6x^2 - 6x - 36 \]

Combining like terms:

\[ = x^4 - 8x^3 + 7x^2 + 36x - 36 \]

So, we have:

\[ f(x) = x^4 - 8x^3 + 7x^2 + 36x - 36 - 100 \]
\[ = x^4 - 8x^3 + 7x^2 + 36x - 136 \]

### Step 3: Analyze the roots of \( f(x) = 0 \)

To find the roots of \( f(x) = 0 \), we can use the discriminant method on the quadratic factors of \( f(x) \).

Let’s consider the quadratic terms in \( f(x) \):

1. Set \( t = x^2 - 4x \). Then we can rewrite \( f(x) \) in terms of \( t \):

\[ f(x) = (t + 3)(t - 12) - 100 \]
\[ = t^2 - 9t - 136 \]

### Step 4: Find the discriminant of the quadratic

The discriminant \( D \) of \( t^2 - 9t - 136 \) is given by:

\[ D = b^2 - 4ac = (-9)^2 - 4(1)(-136) = 81 + 544 = 625 \]

Since \( D > 0 \), this quadratic has two distinct real roots.

### Step 5: Analyze the quadratic \( x^2 - 4x + k \)

Next, we need to analyze the quadratic equations formed from the roots of \( t \):

1. For \( t + 3 = 0 \) and \( t - 12 = 0 \):

- \( t + 3 = 0 \) leads to \( x^2 - 4x + 3 = 0 \)
- \( t - 12 = 0 \) leads to \( x^2 - 4x - 17 = 0 \)

### Step 6: Find the discriminants of the quadratics

1. For \( x^2 - 4x + 3 = 0 \):

\[ D_1 = (-4)^2 - 4(1)(3) = 16 - 12 = 4 \quad (D_1 > 0 \text{, two real roots}) \]

2. For \( x^2 - 4x - 17 = 0 \):

\[ D_2 = (-4)^2 - 4(1)(-17) = 16 + 68 = 84 \quad (D_2 > 0 \text{, two real roots}) \]

### Conclusion

Since both quadratics yield two real roots each, the total number of roots for \( f(x) = 0 \) is:

\[ 2 \text{ (from } x^2 - 4x + 3\text{)} + 2 \text{ (from } x^2 - 4x - 17\text{)} = 4 \text{ real roots.} \]

Thus, the equation \( f(x) = 0 \) has **four distinct real roots**.