Question:
If \( A = \left\{ \frac{x}{9x^2 + 20} \right\} \) and \( f: A \to \mathbb{R} \) is defined by \( f(x) = 2x^3 - 15x^2 + 36x - 48 \), then the maximum value of \( f(x) \) is
If \( A = \left\{ \frac{x}{9x^2 + 20} \right\} \) and \( f: A \to \mathbb{R} \) is defined by \( f(x) = 2x^3 - 15x^2 + 36x - 48 \), then the maximum value of \( f(x) \) is
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Updated On: May 28, 2025
- \( 7 \)
- \( -20 \)
- \( 20 \)
- \( -16 \)
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The Correct Option is B
Approach Solution - 1
Given:
\( A = \left\{ \frac{x}{9x^2 + 20} \right\} \)
Function defined by:
\( f(x) = 2x^3 - 15x^2 + 36x - 48 \)
We need to find the maximum value of \( f(x) \).
First, we'll find the derivative \( f'(x) \) to identify the critical points.
Calculate:
\( f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x - 48) \)
\( f'(x) = 6x^2 - 30x + 36 \)
Set \( f'(x) = 0 \) to find critical points:
\( 6x^2 - 30x + 36 = 0 \)
Divide throughout by 6:
\( x^2 - 5x + 6 = 0 \)
Factorize:
\( (x - 2)(x - 3) = 0 \)
Thus, \( x = 2 \) and \( x = 3 \) are critical points.
Examine changes in \( f'(x) \) around these points to find local max/min:
- If \( x < 2 \), \( f'(x) > 0 \)
- If \( 2 < x < 3 \), \( f'(x) < 0 \)
- If \( x > 3 \), \( f'(x) > 0 \)
\( x = 2 \) is a local maximum, and \( x = 3 \) is a local minimum.
Calculate function values at these points:
\( f(2) = 2(2)^3 - 15(2)^2 + 36(2) - 48 \)
\( = 16 - 60 + 72 - 48 \)
\( = -20 \)
\( f(3) = 2(3)^3 - 15(3)^2 + 36(3) - 48 \)
\( = 54 - 135 + 108 - 48 \)
\( = -21 \)
The maximum value of \( f(x) \) occurring at \( x = 2 \) is therefore:
Maximum Value: \( -20 \)
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Approach Solution -2
To find the maximum value of \( f(x) = 2x^3 - 15x^2 + 36x - 48 \), we'll first compute its derivative to locate critical points.
Calculate \( f'(x) \):
\( f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x - 48) = 6x^2 - 30x + 36 \).
Set \( f'(x) = 0 \) to find critical points:
\( 6x^2 - 30x + 36 = 0 \).
Divide the equation by 6:
\( x^2 - 5x + 6 = 0 \).
Factor the quadratic:
\( (x - 2)(x - 3) = 0 \).
This gives \( x = 2 \) and \( x = 3 \) as critical points.
To determine which point is a maximum, evaluate the second derivative \( f''(x) \):
\( f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30 \).
Evaluate \( f''(x) \) at the critical points:
\( f''(2) = 12(2) - 30 = 24 - 30 = -6 \) (negative, local maximum)
\( f''(3) = 12(3) - 30 = 36 - 30 = 6 \) (positive, local minimum)
Since \( x = 2 \) corresponds to a local maximum, calculate \( f(2) \):
\( f(2) = 2(2)^3 - 15(2)^2 + 36(2) - 48 \)
\( = 16 - 60 + 72 - 48 \)
\( = -20 \).
Therefore, the maximum value of \( f(x) \) is \( -20 \).
Calculate \( f'(x) \):
\( f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x - 48) = 6x^2 - 30x + 36 \).
Set \( f'(x) = 0 \) to find critical points:
\( 6x^2 - 30x + 36 = 0 \).
Divide the equation by 6:
\( x^2 - 5x + 6 = 0 \).
Factor the quadratic:
\( (x - 2)(x - 3) = 0 \).
This gives \( x = 2 \) and \( x = 3 \) as critical points.
To determine which point is a maximum, evaluate the second derivative \( f''(x) \):
\( f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30 \).
Evaluate \( f''(x) \) at the critical points:
\( f''(2) = 12(2) - 30 = 24 - 30 = -6 \) (negative, local maximum)
\( f''(3) = 12(3) - 30 = 36 - 30 = 6 \) (positive, local minimum)
Since \( x = 2 \) corresponds to a local maximum, calculate \( f(2) \):
\( f(2) = 2(2)^3 - 15(2)^2 + 36(2) - 48 \)
\( = 16 - 60 + 72 - 48 \)
\( = -20 \).
Therefore, the maximum value of \( f(x) \) is \( -20 \).
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