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

Options

A. No real value of b and c
B. `0ltcltbsqrt(2)`
C. `|c|lt|b|sqrt(2)`
D. `|c|gt|b|sqrt(2)`

Answer

To find the relation between \( b \) and \( c \) given the functions \( f(x) = x^2 + 2bx + 2c^2 \) and \( g(x) = -x^2 - 2cx + b^2 \), we need to analyze the minimum of \( f(x) \) and the maximum of \( g(x) \).

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

The function \( f(x) \) can be rewritten in a completed square form:

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

The minimum value of \( f(x) \) occurs when \( (x + b)^2 = 0 \) (i.e., \( x = -b \)). Thus, the minimum value is:

\[ \text{min } f(x) = 2c^2 - b^2 \]

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

The function \( g(x) \) can also be rewritten:

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

The maximum value of \( g(x) \) occurs when \( (x + c)^2 = 0 \) (i.e., \( x = -c \)). Thus, the maximum value is:

\[ \text{max } g(x) = b^2 + c^2 \]

### Step 3: Set up the inequality

According to the problem, we need to satisfy the condition:

\[ \text{min } f(x) > \text{max } g(x) \]

Substituting the values we found:

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

### Step 4: Simplify the inequality

Rearranging the inequality gives:

\[ 2c^2 - b^2 - b^2 - c^2 > 0 \]

This simplifies to:

\[ 2c^2 - 2b^2 > 0 \]

Dividing through by 2:

\[ c^2 > b^2 \]

### Step 5: Final relation

Taking the square root of both sides gives us the relation:

\[ |c| > |b| \]

This implies that the absolute value of \( c \) must be greater than the absolute value of \( b \).

### Conclusion

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

\[ |c| > |b| \]