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

• A. \( |c|>\dfrac{|b|}{\sqrt3} \)
• B. \( -1<c<\sqrt2\,b \)
• C. \( \dfrac{|c|}{\sqrt2}>|b| \)
• D. No real values of \(b\) and \(c\) exist \bigskip

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.

Answer 1

The Correct Option is C

Concept: Compare the least value of one quadratic with the greatest value of the other by using their vertex forms.

Step 1: \(f(x)=x^2+2bx+2c^2=(x+b)^2+2c^2-b^2\), so its minimum value is \(2c^2-b^2\).

Step 2: \(g(x)=-x^2-2cx+b^2=-(x+c)^2+b^2+c^2\), so its maximum value is \(b^2+c^2\).

Step 3: Since the minimum of \(f\) exceeds the maximum of \(g\), \(2c^2-b^2>c^2+b^2\Rightarrow c^2>2b^2\). Hence \(\boxed{\dfrac{|c|}{\sqrt2}>|b|}\).