Question 54 of 82intermediate🔧 ApplyMCQ1 mark

The maximum value of [x(x-1)+1]^(1/3), 0 ≤\leq x ≤\leq 1 is

Exercise: EXERCISE 6.3 | Q: 29 | (Chapter: 31)
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Explanation

To find the maximum value of [x(x-1)+1]^(1/3) on [0,1], we first simplify the expression inside the cube root: x(x-1)+1 = x² - x + 1. Since the cube root function is strictly increasing, the maximum occurs where x² - x + 1 is maximum. Let g(x) = x² - x + 1. Finding critical points: g'(x) = 2x - 1 = 0 gives x = 1/2. Evaluating g(x) at critical point and endpoints: g(0) = 1, g(1/2) = 3/4, g(1) = 1. The maximum value of g(x) is 1, occurring at x = 0 and x = 1. Therefore, the maximum value of [x(x-1)+1]^(1/3) = 1^(1/3) = 1.

Solution Steps

  1. Step 1: Let f(x) = [x(x-1)+1]^(1/3) = (x² - x + 1)^(1/3)

  2. Step 2: Since cube root is strictly increasing, find max of g(x) = x² - x + 1

  3. Step 3: Find critical point: g'(x) = 2x - 1 = 0 → x = 1/2

  4. Step 4: Evaluate g(x) at endpoints and critical point: g(0) = 1, g(1/2) = 3/4, g(1) = 1

  5. Step 5: Maximum of g(x) is 1, so maximum of f(x) = 1^(1/3) = 1

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