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Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Then the maximum possible value of f(x) is
Correct answer is '32'. Can you explain this answer?
Verified Answer
Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Th...
F(x) = min (2x2, 52 – 5x)
The maximum possible value of this function will be attained at the point in which 2x2 is equal to 52 – 5x.
2x2 = 52 − 5x
2x2 + 5x − 52 = 0
(2x + 13) (x-4) = 0
=>x = -13 / 2 or x = 4 It has been given that is a positive real number. Therefore, we can eliminate the case x = -13 / 2 . x = 4 is the point at which the function attains the maximum value. 4 is not the maximum value of the function. Substituting x=4 in the original function, we get, 2x2 = 2* 42 = 32.
f(x) = 32.
Therefore, 32 is the right answer.
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Most Upvoted Answer
Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Th...
Explanation:

To find the maximum possible value of f(x), we need to determine the value of x that results in the maximum value of f(x).

Step 1: Finding the critical points
To find the critical points of f(x), we need to determine the values of x where the derivative of f(x) is equal to zero.

Let's find the derivative of f(x) with respect to x:

f'(x) = d/dx(min(2x^2, 52 - 5x))

To find the derivative of min(2x^2, 52 - 5x), we need to consider two cases:

Case 1: When 2x^2 is smaller than 52 - 5x
In this case, the derivative of f(x) with respect to x is equal to the derivative of 2x^2:

f'(x) = d/dx(2x^2) = 4x

Case 2: When 2x^2 is greater than or equal to 52 - 5x
In this case, the derivative of f(x) with respect to x is equal to the derivative of (52 - 5x):

f'(x) = d/dx(52 - 5x) = -5

Step 2: Analyzing the critical points
To find the critical points, we set the derivatives equal to zero and solve for x:

Case 1: When 2x^2 is smaller than 52 - 5x:
4x = 0
x = 0

Case 2: When 2x^2 is greater than or equal to 52 - 5x:
-5 = 0
No solutions exist for this case.

Step 3: Determining the maximum value of f(x)
To determine the maximum value of f(x), we need to evaluate f(x) at the critical points and the endpoints of the domain.

Case 1: Evaluate f(x) at x = 0:
f(0) = min(2(0)^2, 52 - 5(0)) = min(0, 52) = 0

Case 2: Evaluate f(x) at the endpoint of the domain:
Since x is a positive real number, there is no endpoint of the domain.

Conclusion:
Therefore, the maximum possible value of f(x) is 0, which contradicts the given answer of 32. It is possible that there is an error in the given information or question.
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Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Then the maximum possible value of f(x) isCorrect answer is '32'. Can you explain this answer?
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Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Then the maximum possible value of f(x) isCorrect answer is '32'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Then the maximum possible value of f(x) isCorrect answer is '32'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f(x) = min (2x2, 52 – 5x) where x is any positive real number. Then the maximum possible value of f(x) isCorrect answer is '32'. Can you explain this answer?.
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