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The number of integral values of ‘a’ so that the point of local minima of f(x) = x3 – 3ax2 + 3(a2 – 1)x + 1 is less than 4 and point of local maximum is greater than −2, is/are
Correct answer is '3'. Can you explain this answer?
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Given:
We are given a function f(x) = x^3 - 3ax^2 + 3(a^2 - 1)x - 1.
To find:
The number of integral values of a so that the point of local minima of f(x) is less than 4 and the point of local maximum is greater than 2.
Solution:
To find the number of integral values of a satisfying the given conditions, we need to analyze the behavior of the function f(x) and find the values of a that satisfy the conditions.
Finding the derivative:
First, let's find the derivative of f(x) to determine the critical points:
f'(x) = 3x^2 - 6ax + 3(a^2 - 1)
Finding the critical points:
For a function to have a local minimum or maximum, its derivative must be equal to zero. So, we set f'(x) = 0:
3x^2 - 6ax + 3(a^2 - 1) = 0
Simplifying the equation, we get:
x^2 - 2ax + a^2 - 1 = 0
Now, we can solve this quadratic equation for x using the quadratic formula:
x = [2a ± √(4a^2 - 4(a^2 - 1))] / 2
x = [2a ± √(4a^2 - 4a^2 + 4)] / 2
x = [2a ± √(4)] / 2
x = a ± 1
Therefore, the critical points are x = a + 1 and x = a - 1.
Analyzing the behavior of f(x):
To determine whether these critical points are local minima or maxima, we need to analyze the behavior of f(x) around these points.
Case 1: x = a + 1
Let's substitute x = a + 1 into f'(x) to determine the behavior:
f'(a + 1) = 3(a + 1)^2 - 6a(a + 1) + 3(a^2 - 1)
f'(a + 1) = 3(a^2 + 2a + 1) - 6a^2 - 6a + 3a^2 - 3
f'(a + 1) = 3a^2 + 6a + 3 - 6a^2 - 6a + 3a^2 - 3
f'(a + 1) = -3a^2 + 6a + 3
Since the coefficient of the quadratic term is negative, the critical point x = a + 1 corresponds to a local maximum.
Case 2: x = a - 1
Let's substitute x = a - 1 into f'(x) to determine the behavior:
f'(a - 1) = 3(a - 1)^2 - 6a(a - 1) + 3(a^2 - 1)
f'(a - 1) = 3(a^2 - 2a + 1) -
We are given a function f(x) = x^3 - 3ax^2 + 3(a^2 - 1)x - 1.
To find:
The number of integral values of a so that the point of local minima of f(x) is less than 4 and the point of local maximum is greater than 2.
Solution:
To find the number of integral values of a satisfying the given conditions, we need to analyze the behavior of the function f(x) and find the values of a that satisfy the conditions.
Finding the derivative:
First, let's find the derivative of f(x) to determine the critical points:
f'(x) = 3x^2 - 6ax + 3(a^2 - 1)
Finding the critical points:
For a function to have a local minimum or maximum, its derivative must be equal to zero. So, we set f'(x) = 0:
3x^2 - 6ax + 3(a^2 - 1) = 0
Simplifying the equation, we get:
x^2 - 2ax + a^2 - 1 = 0
Now, we can solve this quadratic equation for x using the quadratic formula:
x = [2a ± √(4a^2 - 4(a^2 - 1))] / 2
x = [2a ± √(4a^2 - 4a^2 + 4)] / 2
x = [2a ± √(4)] / 2
x = a ± 1
Therefore, the critical points are x = a + 1 and x = a - 1.
Analyzing the behavior of f(x):
To determine whether these critical points are local minima or maxima, we need to analyze the behavior of f(x) around these points.
Case 1: x = a + 1
Let's substitute x = a + 1 into f'(x) to determine the behavior:
f'(a + 1) = 3(a + 1)^2 - 6a(a + 1) + 3(a^2 - 1)
f'(a + 1) = 3(a^2 + 2a + 1) - 6a^2 - 6a + 3a^2 - 3
f'(a + 1) = 3a^2 + 6a + 3 - 6a^2 - 6a + 3a^2 - 3
f'(a + 1) = -3a^2 + 6a + 3
Since the coefficient of the quadratic term is negative, the critical point x = a + 1 corresponds to a local maximum.
Case 2: x = a - 1
Let's substitute x = a - 1 into f'(x) to determine the behavior:
f'(a - 1) = 3(a - 1)^2 - 6a(a - 1) + 3(a^2 - 1)
f'(a - 1) = 3(a^2 - 2a + 1) -
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The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer?
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The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer?.
The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of integral values of a so that the point of local minima of f(x) = x3 3ax2 + 3(a2 1)x + 1 is less than 4 and point of local maximum is greater than 2, is/are Correct answer is '3'. Can you explain this answer?.
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