AP EAMCET · Maths · Quadratic Equation
Let \(f(x)=x^2+2 b x+2 c^2\) and \(g(x)=-x^2-2 c x+b^2, x \in R\). If \(b\) and \(c\) are nonzero real numbers such that \(\min f(x)>\max g(x)\), then \(\left|\frac{c}{b}\right|\) lies in the interval
- A \(\left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)\)
- B \(\left(\frac{1}{\sqrt{2}}, \sqrt{2}\right)\)
- C \((\sqrt{2}, \infty)\)
- D \((0,1)\)
Answer & Solution
Correct Answer
(C) \((\sqrt{2}, \infty)\)
Step-by-step Solution
Detailed explanation
\(\min f(x) = f(-b) = (-b)^2+2b(-b)+2c^2 = 2c^2-b^2\) \(\max g(x) = g(-c) = -(-c)^2-2c(-c)+b^2 = c^2+b^2\) \(2c^2-b^2 > c^2+b^2\) \(c^2 > 2b^2\) \(\frac{c^2}{b^2} > 2\) \(\left|\frac{c}{b}\right| > \sqrt{2}\) The interval is \((\sqrt{2}, \infty)\)
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