If minimum value of `f(x)=(x^(2)+2bx+2c^(2))` is greater than the maximum value of `g(x)=-x^(2)-2cx+b^(2)`, then `(x in R)`

Options

A. `|c|gt (|b|)/(sqrt(3))`
B. `(|c|)/(sqrt(2))gt|b|`
C. `-1lt c lt sqrt(2)b`
D. no real values of b and c exist

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the question.

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

The function is given as:
\[ f(x) = x^2 + 2bx + 2c^2 \]

This is a quadratic function in the standard form \( ax^2 + bx + c \) where \( a = 1 \), \( b = 2b \), and \( c = 2c^2 \). The minimum value of a quadratic function occurs at \( x = -\frac{B}{2A} \).

Calculating the vertex:
\[ x = -\frac{2b}{2 \cdot 1} = -b \]

Now, substituting \( x = -b \) 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: Finding the maximum value of \( g(x) \)

The function is given as:
\[ g(x) = -x^2 - 2cx + b^2 \]

This is also a quadratic function where \( a = -1 \), \( b = -2c \), and \( c = b^2 \). The maximum value occurs at the vertex:
\[ x = -\frac{-2c}{2 \cdot -1} = c \]

Now, substituting \( x = c \) back into \( g(x) \) to find the maximum value:
\[ g(c) = -c^2 - 2c(c) + b^2 = -c^2 - 2c^2 + b^2 = b^2 - 3c^2 \]

### Step 3: Setting up the inequality

According to the problem, we need to find when the minimum value of \( f(x) \) is greater than the maximum value of \( g(x) \):
\[ 2c^2 - b^2 > b^2 - 3c^2 \]

### Step 4: Simplifying the inequality

Rearranging the inequality:
\[ 2c^2 - b^2 > b^2 - 3c^2 \]
\[ 2c^2 + 3c^2 > 2b^2 \]
\[ 5c^2 > 2b^2 \]

Dividing both sides by 5:
\[ c^2 > \frac{2}{5}b^2 \]

Taking the square root of both sides:
\[ |c| > \sqrt{\frac{2}{5}}|b| \]

### Conclusion

Thus, the condition we derived is:
\[ |c| > \frac{\sqrt{2}}{\sqrt{5}} |b| \]

This inequality gives us the relationship between \( c \) and \( b \) that satisfies the original condition of the problem.