In the given figure vertices of ∆ ABC lie on y = f (x) = ax 2 + bx + c. The ∆ ABC is…
Mathematics · JEE Advanced · NTA Exams — Complex Numbers and Quadratic Equations
In the given figure vertices of ∆ ABC lie on y = f (x)
= ax2 + bx + c. The ∆ ABC is right angled isosceles triangle whose hypotenuse AC = \(4 \sqrt{2}\) units, then
Minimum value of y = f (x) is
- \(2 \sqrt{2}\)
- \(-2 \sqrt{2}\)
- 2
- – 2
Answer
(B) -2 2
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