Question:
If the minimum value of
\[
f(x)=x^2+2bx+2c^2
\]
is greater than the maximum value of
\[
g(x)=-x^2-2cx+b^2,
\]
\(x\) being real, then
If the minimum value of
\[
f(x)=x^2+2bx+2c^2
\]
is greater than the maximum value of
\[
g(x)=-x^2-2cx+b^2,
\]
\(x\) being real, then
Show Hint
For quadratic functions, completing the square is the fastest method to find maximum or minimum values. After obtaining the extreme values, directly apply the given condition and simplify.
Updated On: Jul 9, 2026
- \( |c|>\dfrac{|b|}{\sqrt3} \)
- \( -1<c<\sqrt2\,b \)
- \( \dfrac{|c|}{\sqrt2}>|b| \)
- No real values of \(b\) and \(c\) exist \bigskip
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For a quadratic function
\[
ax^2+bx+c,
\]
• If \(a>0\), the parabola opens upward and has a minimum value.
• If \(a<0\), the parabola opens downward and has a maximum value.
• The extreme value occurs at \[ x=-\frac{b}{2a}. \]
Step 1: Find the minimum value of \(f(x)\). Given \[ f(x)=x^2+2bx+2c^2. \] Completing the square, \[ f(x) = (x+b)^2+2c^2-b^2. \] Since \[ (x+b)^2\ge 0, \] the minimum value of \(f(x)\) is \[ 2c^2-b^2. \]
Step 2: Find the maximum value of \(g(x)\). Given \[ g(x)=-x^2-2cx+b^2. \] Completing the square, \[ g(x) = -\left(x+c\right)^2+c^2+b^2. \] Since \[ -\left(x+c\right)^2\le 0, \] the maximum value of \(g(x)\) is \[ c^2+b^2. \]
Step 3: Use the given condition. The minimum value of \(f(x)\) is greater than the maximum value of \(g(x)\). Therefore, \[ 2c^2-b^2 > c^2+b^2. \] \[ c^2-2b^2>0. \] \[ c^2>2b^2. \]
Step 4: Simplify the inequality. Taking positive square roots, \[ |c|>\sqrt2\,|b|. \] Dividing by \(\sqrt2\), \[ \frac{|c|}{\sqrt2}>|b|. \]
Step 5: Write the final answer. \[ \boxed{\frac{|c|}{\sqrt2}>|b|} \]
• If \(a>0\), the parabola opens upward and has a minimum value.
• If \(a<0\), the parabola opens downward and has a maximum value.
• The extreme value occurs at \[ x=-\frac{b}{2a}. \]
Step 1: Find the minimum value of \(f(x)\). Given \[ f(x)=x^2+2bx+2c^2. \] Completing the square, \[ f(x) = (x+b)^2+2c^2-b^2. \] Since \[ (x+b)^2\ge 0, \] the minimum value of \(f(x)\) is \[ 2c^2-b^2. \]
Step 2: Find the maximum value of \(g(x)\). Given \[ g(x)=-x^2-2cx+b^2. \] Completing the square, \[ g(x) = -\left(x+c\right)^2+c^2+b^2. \] Since \[ -\left(x+c\right)^2\le 0, \] the maximum value of \(g(x)\) is \[ c^2+b^2. \]
Step 3: Use the given condition. The minimum value of \(f(x)\) is greater than the maximum value of \(g(x)\). Therefore, \[ 2c^2-b^2 > c^2+b^2. \] \[ c^2-2b^2>0. \] \[ c^2>2b^2. \]
Step 4: Simplify the inequality. Taking positive square roots, \[ |c|>\sqrt2\,|b|. \] Dividing by \(\sqrt2\), \[ \frac{|c|}{\sqrt2}>|b|. \]
Step 5: Write the final answer. \[ \boxed{\frac{|c|}{\sqrt2}>|b|} \]
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