If `f(x)=x^2+2bx+2c^2` and `g(x)=-x^2-2cx+b^2` such that min `f(x)gt` max g(x), then the relation between b and c is

Options

A. No real value of b and c
B. `0ltbltsqrt2`
C. `|c|lt|b|sqrt2`
D. `|c|gt|b|sqrt2`

Answer

To solve the problem, we need to find the relationship between \( b \) and \( c \) given the functions \( f(x) \) and \( g(x) \) and the condition that the minimum of \( f(x) \) is greater than the maximum of \( g(x) \).

### Step 1: Find the minimum of \( f(x) \)

The function \( f(x) \) is given by:

\[ f(x) = x^2 + 2bx + 2c^2 \]

To find the minimum, we first calculate the derivative \( f'(x) \):

\[ f'(x) = 2x + 2b \]

Setting the derivative equal to zero to find critical points:

\[ 2x + 2b = 0 \implies x = -b \]

Now, we substitute \( x = -b \) back into \( f(x) \) to find the minimum value:

\[ f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2 \]

### Step 2: Find the maximum of \( g(x) \)

The function \( g(x) \) is given by:

\[ g(x) = -x^2 - 2cx + b^2 \]

To find the maximum, we calculate the derivative \( g'(x) \):

\[ g'(x) = -2x - 2c \]

Setting the derivative equal to zero to find critical points:

\[ -2x - 2c = 0 \implies x = -c \]

Now, we substitute \( x = -c \) back into \( g(x) \) to find the maximum value:

\[ g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = b^2 + c^2 \]

### Step 3: Set up the inequality

According to the problem, we have the condition:

\[ \text{Minimum of } f(x) > \text{Maximum of } g(x) \]

Substituting the values we found:

\[ 2c^2 - b^2 > b^2 + c^2 \]

### Step 4: Simplify the inequality

Now, we simplify the inequality:

\[ 2c^2 - b^2 > b^2 + c^2 \]

Rearranging gives:

\[ 2c^2 - c^2 > 2b^2 \]

This simplifies to:

\[ c^2 > 2b^2 \]

### Conclusion

Thus, the relationship between \( b \) and \( c \) is:

\[ c^2 > 2b^2 \]