If non-zero real number b and c are such that min `f(x) gt` max g(x) where f(x) `=x^(2) + 2bx + 2c^(2) and g(x) = -x^(2) - 2cx + b^(2) (x in R) " then " |(c )/(b)|` lies in the interval

Options

A. `[(1)/(sqrt2), sqrt2]`
B. `[0, (1)/(2)]`
C. `[(1)/(2), (1)/(sqrt2)]`
D. `[sqrt2, oo]`

Answer

To solve the problem, we need to find the interval in which \(|\frac{c}{b}|\) lies, given that the minimum of \(f(x)\) is greater than the maximum of \(g(x)\).

### Step 1: Define the functions
We have:
- \(f(x) = x^2 + 2bx + 2c^2\)
- \(g(x) = -x^2 - 2cx + b^2\)

### Step 2: Find the minimum of \(f(x)\)
To find the minimum of \(f(x)\), we first calculate its derivative:
\[ f'(x) = 2x + 2b \]
Setting the derivative equal to zero to find critical points:
\[ 2x + 2b = 0 \implies x = -b \]
Now, we evaluate \(f(x)\) at this critical point:
\[ f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = -b^2 + 2c^2 \]
Thus, the minimum value of \(f(x)\) is:
\[ \text{min } f(x) = 2c^2 - b^2 \]

### Step 3: Find the maximum of \(g(x)\)
Next, we find the maximum of \(g(x)\). We calculate its derivative:
\[ g'(x) = -2x - 2c \]
Setting the derivative equal to zero:
\[ -2x - 2c = 0 \implies x = -c \]
Now, we evaluate \(g(x)\) at this critical point:
\[ g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2 \]
Thus, the maximum value of \(g(x)\) is:
\[ \text{max } g(x) = c^2 + b^2 \]

### Step 4: Set up the inequality
We are given that:
\[ \text{min } f(x) > \text{max } g(x) \]
This translates to:
\[ 2c^2 - b^2 > c^2 + b^2 \]

### Step 5: Solve the inequality
Rearranging the inequality:
\[ 2c^2 - b^2 - c^2 - b^2 > 0 \implies c^2 - 2b^2 > 0 \]
This simplifies to:
\[ c^2 > 2b^2 \]
Taking the square root of both sides (and considering both positive and negative roots):
\[ |c| > \sqrt{2}|b| \]
Dividing both sides by \(|b|\) (since \(b\) is non-zero):
\[ \left|\frac{c}{b}\right| > \sqrt{2} \]

### Step 6: Determine the interval
The above inequality implies:
\[ \frac{c}{b} < -\sqrt{2} \quad \text{or} \quad \frac{c}{b} > \sqrt{2} \]
Thus, \(|\frac{c}{b}|\) lies in the intervals:
\[ (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \]

### Final Answer
The intervals in which \(|\frac{c}{b}|\) lies are:
\[ (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \]