Question

The smallest value of M such that |x² – 3x + 2| ≤ M for all x in the interval [1, 5/2] is –

• A. ¼
• B. ¾
• C. 5/4
• D. 5/16

Answer 1

Answer: (B)

¾

Answer 2

Consider ƒ(x) = x² – 3x + 2 on [1, 5/2]. Now

ƒ′(x) = 2x – 3 so the only critical point is 3/2. Since

ƒ(1) = 0, ƒ(3/2) = –5/4 and ƒ(5/2) = 3/4. Hence the max

{ƒ(x) : x ∈ [1, 5/2]} = 3/4.Thus M = 3/4.