If f(x) = x² + 2bx + 2c² and g(x) = -x² - 2cx + b² such that min f(x) max g(x), then the relation

Q: Wait, why can't I just say $f(x) > g(x)$ for all $x$?

That is a common trap! The condition $f(x) > g(x)$ for all $x$ would imply that the entire curve of $f(x)$ lies above the entire curve of $g(x)$. However, the problem only asks for the minimum of $f(x)$ to be greater than the maximum of $g(x)$. This is a much weaker condition that allows the curves to exist in the same horizontal space, provided they are vertically separated.

Q: Why do we get $|c| > |b|\sqrt{2}$ instead of just $c > b\sqrt{2}$?

Excellent catch. When we take the square root of both sides of an inequality like $c^2 > 2b^2$, we must remember that $\sqrt{x^2}$ is defined as the absolute value $|x|$. Since $b$ and $c$ can be negative, simply writing $c > b\sqrt{2}$ would ignore the cases where $b$ or $c$ are negative. The absolute value ensures the relationship holds for all real numbers.

Q: Is there another way to find the minimum/maximum without the vertex formula?

Yes, you could use the discriminant method. For a quadratic $Ax^2 + Bx + C$, the extreme value is given by $-\frac{D}{4A}$, where $D = B^2 - 4AC$. Both methods are mathematically equivalent and equally valid. The vertex formula is often preferred for its intuitive connection to the geometry of the parabola.

Question

If $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ such that $\min f(x) > \max g(x)$, then the relation between $b$ and $c$, is Options: (A) no real value of $b$ & $c$ (B) $0 < c < b\sqrt{2}$ (C) $|c| < |b|\sqrt{2}$ (D) $|c| > |b|\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plane. In front of you, two distinct curves are being drawn by an invisible hand.

The first is $f(x) = x^2 + 2bx + 2c^2$, a graceful, upward-opening parabola. The second is $g(x) = -x^2 - 2cx + b^2$, a downward-opening parabola.

The problem requires that the lowest point of the upward curve is strictly higher than the highest point of the downward curve. This is a condition of **separation**.

### The Geometry of Extremes

To solve this, we must first master the peaks and valleys of these curves. For any quadratic expression of the form $Ax^2 + Bx + C$, the vertex is located at $x = -\frac{B}{2A}$.

For our upward-opening parabola $f(x)$, the coefficients are $A = 1$ and $B = 2b$. Plugging these into our vertex formula, we find the $x$-coordinate of the minimum:

$$x = -\frac{2b}{2(1)} = -b$$

Now, we find the actual minimum value by substituting $x = -b$ back into $f(x)$:

$$f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2$$

This value, $2c^2 - b^2$, represents the **floor** of our upward-opening parabola.

### The Peak of the Downward Curve

Now, let us turn our attention to the downward-opening parabola $g(x) = -x^2 - 2cx + b^2$. Here, $A = -1$ and $B = -2c$.

The vertex occurs at:

$$x = -\frac{-2c}{2(-1)} = -c$$

To find the maximum value, we substitute $x = -c$ into $g(x)$:

$$g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2$$

This value, $c^2 + b^2$, represents the **ceiling** of our downward-opening parabola.

### The Final Inequality

We are now ready to bridge the two. The problem demands that $\min f(x) > \max g(x)$.

Substituting our derived expressions, we obtain the following inequality:

$$2c^2 - b^2 > c^2 + b^2$$

By adding $b^2$ to both sides and subtracting $c^2$ from both sides, we isolate the relationship between $b$ and $c$:

$$c^2 > 2b^2$$

To reach the final form, we take the square root of both sides. Recalling that $\sqrt{x^2} = |x|$, we arrive at:

$$|c| > |b|\sqrt{2}$$

This is the elegant, final relation. It tells us that for these two parabolas to never cross, the magnitude of $c$ must be significantly larger than the magnitude of $b$ scaled by $\sqrt{2}$.

Animated Solution for Mathematics - Quadratic Equations: If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $min f (x) > max g (x)$ , then the relation between $b$ and $c$ , is

Select Answer:

Visualized Solution

Visualizing the Parabolas

The Vertex Formula for Quadratics

Finding the Minimum of $f (x)$

Simplifying $min f (x)$

Finding the Maximum of $g (x)$

Simplifying $max g (x)$

Setting up the Inequality

Solving the Inequality

Finding the Final Relation

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The Geometry of Extremes

The Peak of the Downward Curve

The Final Inequality

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x2+2bx+2c2 and g(x)=−x2−2cx+b2 such that minf(x)>maxg(x), then the relation between b and c, is

Select Answer:

Visualized Solution

Visualizing the Parabolas

The Vertex Formula for Quadratics

Finding the Minimum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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Finding the Maximum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } maxg(x)