The absolute maximum value of y = x^3 - 3x + 2, 0 ≤ x ≤ 2, is Option 1: 4 Option 2: 6 Option 3: 2 Option 4: 0

Answer: 4. Solution: f'(x) = 3x^2 - 3 = 0 gives x = ± 1. Only x = 1 lies in [0,2]. Evaluating: f(0) = 2, f(1) = 1 - 3 + 2 = 0, f(2) = 8 - 6 + 2 = 4. The absolute maximum is 4 at x = 2.

Calculus

Application of Derivatives

Easy

The absolute maximum value of $y = x^3 - 3x + 2$ , $0 \leq x \leq 2$ , is

✅ Correct Option: 1

$f'(x) = 3x^2 - 3 = 0$ gives $x = \pm 1$ . Only $x = 1$ lies in $[0,2]$ . Evaluating: $f(0) = 2$ , $f(1) = 1 - 3 + 2 = 0$ , $f(2) = 8 - 6 + 2 = 4$ . The absolute maximum is $4$ at $x = 2$ .

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The absolute maximum value of y=x3−3x+2y = x^3 - 3x + 2y=x3−3x+2, 0≤x≤20 \leq x \leq 20≤x≤2, is

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CUET Mathematics Past Year Questions (Topic-Wise)

The absolute maximum value of $y = x^3 - 3x + 2$ , $0 \leq x \leq 2$ , is