if `x^(4)+3cos(ax^(2)+bx+c)=2(x^(2)-2)` has two solution with a,b,c `in` (2,5) then value of `(ac)/(b^(2))` can not be

Options

A. a. 4
B. b. 2
C. c. 1
D. d. 3

Explanation

To solve the problem, we need to analyze the given equation and the conditions provided. The equation is:

\[ x^4 + 3 \cos(ax^2 + bx + c) = 2(x^2 - 2) \]

We can rearrange this to:

\[ x^4 - 2x^2 + 6 = 3 \cos(ax^2 + bx + c) \]

### Step 1: Analyze the left-hand side

The left-hand side can be simplified as:

\[ f(x) = x^4 - 2x^2 + 6 \]

To find the critical points, we differentiate \( f(x) \):

\[ f'(x) = 4x^3 - 4x \]

Setting the derivative to zero gives:

\[ 4x(x^2 - 1) = 0 \]

This implies:

\[ x = 0, \, x = 1, \, x = -1 \]

### Step 2: Evaluate the function at critical points

Now we evaluate \( f(x) \) at these critical points:

1. **At \( x = 0 \)**:
\[ f(0) = 0^4 - 2(0^2) + 6 = 6 \]

2. **At \( x = 1 \)**:
\[ f(1) = 1^4 - 2(1^2) + 6 = 1 - 2 + 6 = 5 \]

3. **At \( x = -1 \)**:
\[ f(-1) = (-1)^4 - 2(-1)^2 + 6 = 1 - 2 + 6 = 5 \]

### Step 3: Determine the range of \( f(x) \)

The minimum value of \( f(x) \) occurs at \( x = 0 \), which is 6. The maximum values at \( x = 1 \) and \( x = -1 \) are both 5. Therefore, the function \( f(x) \) has a minimum value of 5 and a maximum value of 6.

### Step 4: Analyze the right-hand side

The right-hand side of the equation is:

\[ 3 \cos(ax^2 + bx + c) \]

The range of \( 3 \cos(ax^2 + bx + c) \) is between -3 and 3, since the cosine function oscillates between -1 and 1.

### Step 5: Set conditions for solutions

For the equation to have two solutions, the left-hand side must intersect the right-hand side at two points. This means that the maximum value of \( f(x) \) (which is 6) must be equal to or greater than the maximum value of the right-hand side (which is 3), and the minimum value of \( f(x) \) (which is 5) must be equal to or less than the minimum value of the right-hand side (which is -3).

However, since the maximum of \( f(x) \) is 6 and the minimum is 5, and the right-hand side oscillates between -3 and 3, there are no values of \( a, b, c \) that can satisfy this condition.

### Step 6: Determine the value of \( \frac{ac}{b^2} \)

Given that \( a, b, c \) are in the range (2, 5), we can analyze the expression \( \frac{ac}{b^2} \):

- The maximum value of \( a \) and \( c \) is 5, and the minimum value of \( b \) is 2.
- Thus, the maximum value of \( \frac{ac}{b^2} \) can be calculated as:

\[ \frac{5 \cdot 5}{2^2} = \frac{25}{4} = 6.25 \]

- The minimum value of \( a \) and \( c \) is 2, and the maximum value of \( b \) is 5.
- Thus, the minimum value of \( \frac{ac}{b^2} \) can be calculated as:

\[ \frac{2 \cdot 2}{5^2} = \frac{4}{25} = 0.16 \]

### Conclusion

The possible values of \( \frac{ac}{b^2} \) range from \( 0.16 \) to \( 6.25 \). The value that cannot be achieved from the options provided (4, 2, 1, and 3) is:

**Answer: 4**