AP EAMCET · Maths · Quadratic Equation
If the minimum value of \(f(x)=x^2+2 b x+2 c^2\) is greater than the maximum value of \(g(x)=-x^2-2 c x+b^2\) for all real \(x\) then
- A \(|c|>\sqrt{2}|b|\)
- B \(|c| \sqrt{3}>|b|\)
- C \(-1 < c < \sqrt{2} b\)
- D \(\frac{1}{\sqrt{2}} < c < |b|\)
Answer & Solution
Correct Answer
(A) \(|c|>\sqrt{2}|b|\)
Step-by-step Solution
Detailed explanation
Min \(f(x)\) at \(x=-b\): \(f(-b) = (-b)^2 + 2b(-b) + 2c^2 = b^2 - 2b^2 + 2c^2 = 2c^2 - b^2\). Max \(g(x)\) at \(x=-c\): \(g(-c) = -(-c)^2 - 2c(-c) + b^2 = -c^2 + 2c^2 + b^2 = c^2 + b^2\). \(2c^2 - b^2 > c^2 + b^2\) \(c^2 > 2b^2\) \(|c| > \sqrt{2}|b|\)
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