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Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +2)} dx, 1< /><=4 .="">
Maximum value of f(x) is 53/4
Maximum value of f(x) is 63/4
Maximum value of f(x) is -1/2
Maximum value of f(x) is -1/3 ?
Maximum value of f(x) is 53/4
Maximum value of f(x) is 63/4
Maximum value of f(x) is -1/2
Maximum value of f(x) is -1/3 ?
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Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +...
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Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +...
Calculation of f(x)
To calculate the value of f(x), we need to evaluate the definite integral given in the definition of f(x). The integral is:
f(x) = ∫[x to 1] (x(x^2 - 3x - 2)) dx
Simplifying the integrand:
f(x) = ∫[x to 1] (x^3 - 3x^2 - 2x) dx
= [x^4/4 - x^3 + x^2] [x to 1]
= (1/4 - 1 + 1) - (x^4/4 - x^3 + x^2)
f(x) = 1/4 - 1 + 1 - (x^4/4 - x^3 + x^2)
= -x^4/4 + x^3 - x^2 + 5/4
Finding the Maximum Value of f(x)
To find the maximum value of f(x), we need to find the critical points of f(x) where the derivative of f(x) is zero or undefined.
Taking the derivative of f(x):
f'(x) = -4(x^3)/4 + 3x^2 - 2x
= -x^3 + 3x^2 - 2x
Setting f'(x) = 0:
-x^3 + 3x^2 - 2x = 0
Factoring out x:
x(-x^2 + 3x - 2) = 0
Using the quadratic formula to solve for x^2 - 3x + 2 = 0:
x = (3 ± √(9 - 4(1)(2))) / 2
x = (3 ± √1) / 2
x = (3 ± 1) / 2
The critical points are x = 1 and x = 2.
Evaluating f(x) at Critical Points
Now, we need to evaluate f(x) at the critical points to determine the maximum value.
For x = 1:
f(1) = -1/4 + 1 - 1 + 5/4
= 3/2
For x = 2:
f(2) = -2^4/4 + 2^3 - 2^2 + 5/4
= -16/4 + 8 - 4 + 5/4
= -4 + 8 - 4 + 5/4
= 5/4
Determining the Maximum Value
Comparing the values of f(x) at the critical points, we find:
f(1) = 3/2
f(2) = 5/4
The maximum value of f(x) is 3/2, which is greater than 5/4. Therefore, the correct answer is:
Maximum value of f(x) is 3/2.
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Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +2)} dx, 1Maximum value of f(x) is 53/4Maximum value of f(x) is 63/4Maximum value of f(x) is -1/2Maximum value of f(x) is -1/3 ?
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Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +2)} dx, 1Maximum value of f(x) is 53/4Maximum value of f(x) is 63/4Maximum value of f(x) is -1/2Maximum value of f(x) is -1/3 ? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +2)} dx, 1Maximum value of f(x) is 53/4Maximum value of f(x) is 63/4Maximum value of f(x) is -1/2Maximum value of f(x) is -1/3 ? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let f(x) be a defined by f(x) = integration from x to 1 { x(x^2 - 3x +2)} dx, 1Maximum value of f(x) is 53/4Maximum value of f(x) is 63/4Maximum value of f(x) is -1/2Maximum value of f(x) is -1/3 ?.
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