CUET · MATHS · PYQ PAPER 2023
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined as \(f(x) = 4x^2 - 16x + 5\).
- A Maximum value of f(x) is - 11.
- B Minimum value of f(x) is - 11.
- C Minimum value of f(x) is 20.
- D No maximum value of f(x).
Answer & Solution
Correct Answer
(B) Minimum value of f(x) is - 11.
Step-by-step Solution
Detailed explanation
\(a=4 > 0\), so \(f(x)\) has a minimum value. \(x_{vertex} = -\frac{b}{2a} = -\frac{-16}{2(4)} = \frac{16}{8} = 2\) \(f(2) = 4(2)^2 - 16(2) + 5 = 16 - 32 + 5 = -11\)
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