Range Of Quadratic Expression
For all real x, the quadratic expression f(x) = x^2 - 6x + 11 attains a minimum value, and that minimum value of the expression equals which number?
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About This Question
- Subject
- mathematics
- Chapter
- complex numbers and quadratic equations
- Topic
- range of quadratic expression
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
2
A quadratic ax^2 + bx + c with positive leading coefficient opens upward and attains its minimum at the vertex, a basic JEE Advanced optimization fact. Completing the square, f(x) = x^2 - 6x + 11 = (x - 3)^2 + (11 - 9) = (x - 3)^2 + 2. Since (x - 3)^2 ≥ 0 with equality at x = 3, the minimum value of f is 2, achieved at x = 3. The vertex form makes the minimum immediately visible. Option 11 is the value of f at x = 0, not the minimum. Option -7 misapplies the constant correction with the wrong sign. Option 5 uses an incorrect vertex shift. Hence the minimum value is 2. Plausibility check: the discriminant of x^2 - 6x + (11 - 2) = x^2 - 6x + 9 is 36 - 36 = 0, meaning the line y = 2 touches the parabola at exactly one point, confirming 2 as the minimum value.
This easy difficulty mathematics question is from the chapter complex numbers and quadratic equations, covering the topic of range of quadratic expression. It appeared in the 2025 exam.
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