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CUET · MATHS · PYQ PAPER 2023

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined as \(f(x) = 4x^2 - 16x + 5\).

  1. A Maximum value of f(x) is - 11.
  2. B Minimum value of f(x) is - 11.
  3. C Minimum value of f(x) is 20.
  4. D No maximum value of f(x).
Verified Solution

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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