Question:

For the function $f(x) = 2x^3 - 9x^2 + 12x - 5, x \in [0, 3]$, match List-I with List-II:
List-IList-II
(A) Absolute maximum value(I) 3
(B) Absolute minimum value(II) 0
(C) Point of maxima(III) -5
(D) Point of minima(IV) 4

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When finding the absolute maximum and minimum values of a function on a given interval, be sure to evaluate the function at the endpoints and at any critical points within the interval. Critical points are found by setting the first derivative equal to zero and solving for \( x \). Once you have these values, compare them to determine the absolute maximum and minimum.

Updated On: Mar 28, 2025
  • (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
  • (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
  • (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
  • (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
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The Correct Option is D

Approach Solution - 1

Differentiate \( f(x) = 2x^3 - 9x^2 + 12x - 5 \) to find \( f'(x) = 6x^2 - 18x + 12 \).
Solve \( f'(x) = 0 \) to find critical points within the interval \([0, 3]\).
Evaluate \( f(x) \) at the endpoints \( x = 0 \) and \( x = 3 \), and at the critical points, to determine the absolute maximum and minimum values.
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Approach Solution -2

Differentiate \( f(x) = 2x^3 - 9x^2 + 12x - 5 \) to find \( f'(x) \):

First, we differentiate the given function \( f(x) \): \[ f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x - 5) \] Using the power rule of differentiation, we get: \[ f'(x) = 6x^2 - 18x + 12 \]

Step 1: Solve \( f'(x) = 0 \) to find critical points within the interval \([0, 3]\):

We need to solve the equation \( f'(x) = 0 \): \[ 6x^2 - 18x + 12 = 0 \] Divide the entire equation by 6: \[ x^2 - 3x + 2 = 0 \] Factor the quadratic equation: \[ (x - 1)(x - 2) = 0 \] So, the solutions are: \[ x = 1 \quad \text{and} \quad x = 2 \] Thus, the critical points within the interval \([0, 3]\) are \( x = 1 \) and \( x = 2 \).

Step 2: Evaluate \( f(x) \) at the endpoints \( x = 0 \) and \( x = 3 \), and at the critical points:

Now, we evaluate \( f(x) \) at the endpoints and at the critical points: - At \( x = 0 \): \[ f(0) = 2(0)^3 - 9(0)^2 + 12(0) - 5 = -5 \] - At \( x = 3 \): \[ f(3) = 2(3)^3 - 9(3)^2 + 12(3) - 5 = 2(27) - 9(9) + 12(3) - 5 = 54 - 81 + 36 - 5 = 4 \] - At \( x = 1 \) (critical point): \[ f(1) = 2(1)^3 - 9(1)^2 + 12(1) - 5 = 2 - 9 + 12 - 5 = 0 \] - At \( x = 2 \) (critical point): \[ f(2) = 2(2)^3 - 9(2)^2 + 12(2) - 5 = 2(8) - 9(4) + 12(2) - 5 = 16 - 36 + 24 - 5 = -1 \]

Step 3: Determine the absolute maximum and minimum values:

From the evaluations: - \( f(0) = -5 \) - \( f(3) = 4 \) - \( f(1) = 0 \) - \( f(2) = -1 \) The absolute maximum value is \( f(3) = 4 \), and the absolute minimum value is \( f(0) = -5 \).

Conclusion: The absolute maximum value is \( 4 \) at \( x = 3 \), and the absolute minimum value is \( -5 \) at \( x = 0 \).

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