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f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, then
- a)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5
- b)f{x) is increasing for x ∈ [1, 2√5]
- c)f(x) has local minima at x = 1
- d)the value of f(0) is 5
Correct answer is option 'B,C'. Can you explain this answer?
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Most Upvoted Answer
f(x) is a cubic polynomial which has local maximum at x = -1. If f(2)...
Since f(x) has local maxima at x = -1 and f(x) has local minima at x =0
Agin integrating both sides we get
Using (1), (2) and (3) we get
Using number line rule
∴ f(x) is increasing for [1, 2√5] and f(x) has local maximum at x = -1 and f(x) has local minimum at x = 1
Hence options (b) and (c) are correct.
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Community Answer
f(x) is a cubic polynomial which has local maximum at x = -1. If f(2)...
To solve this question, let's analyze the given information step by step:
1. Local Maximum at x = -1:
Since f(x) has a local maximum at x = -1, the cubic polynomial must have a turning point at x = -1. Let's call this point (-1, f(-1)).
2. f(2) = 18:
We are given that f(2) = 18. This means that the cubic polynomial passes through the point (2, 18).
3. f(1) = -1:
We are given that f(1) = -1. This means that the cubic polynomial passes through the point (1, -1).
4. Local Minimum at x = 0 for f'(x):
We are given that f'(x) has a local minimum at x = 0. This means that the derivative of the cubic polynomial has a turning point at x = 0. Let's call this point (0, f'(0)).
Now, let's analyze each option:
a) The distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5:
To find the distance between (-1, 2) and [a, f(a)], we need to know the coordinates of the point (a, f(a)). However, we don't have enough information to determine the value of a or f(a). Therefore, option a cannot be concluded from the given information.
b) f(x) is increasing for x ∈ [1, 2√5]:
Since the cubic polynomial passes through the points (1, -1) and (2, 18), and it has a local maximum at x = -1, we can infer that the polynomial increases from x = -∞ to x = -1, then decreases from x = -1 to x = 1, and finally increases from x = 1 to x = +∞. Therefore, f(x) is increasing for x ∈ [1, 2√5]. This option is correct.
c) f(x) has local minima at x = 1:
From the given information, we know that f(x) has a local maximum at x = -1 and f'(x) has a local minimum at x = 0. However, we cannot determine whether f(x) has a local minimum at x = 1 based on this information alone. Therefore, option c cannot be concluded from the given information.
d) The value of f(0) is 5:
We do not have enough information to determine the value of f(0) based on the given information. Therefore, option d cannot be concluded from the given information.
In conclusion, based on the given information, options b and c can be concluded, while options a and d cannot be determined.
1. Local Maximum at x = -1:
Since f(x) has a local maximum at x = -1, the cubic polynomial must have a turning point at x = -1. Let's call this point (-1, f(-1)).
2. f(2) = 18:
We are given that f(2) = 18. This means that the cubic polynomial passes through the point (2, 18).
3. f(1) = -1:
We are given that f(1) = -1. This means that the cubic polynomial passes through the point (1, -1).
4. Local Minimum at x = 0 for f'(x):
We are given that f'(x) has a local minimum at x = 0. This means that the derivative of the cubic polynomial has a turning point at x = 0. Let's call this point (0, f'(0)).
Now, let's analyze each option:
a) The distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5:
To find the distance between (-1, 2) and [a, f(a)], we need to know the coordinates of the point (a, f(a)). However, we don't have enough information to determine the value of a or f(a). Therefore, option a cannot be concluded from the given information.
b) f(x) is increasing for x ∈ [1, 2√5]:
Since the cubic polynomial passes through the points (1, -1) and (2, 18), and it has a local maximum at x = -1, we can infer that the polynomial increases from x = -∞ to x = -1, then decreases from x = -1 to x = 1, and finally increases from x = 1 to x = +∞. Therefore, f(x) is increasing for x ∈ [1, 2√5]. This option is correct.
c) f(x) has local minima at x = 1:
From the given information, we know that f(x) has a local maximum at x = -1 and f'(x) has a local minimum at x = 0. However, we cannot determine whether f(x) has a local minimum at x = 1 based on this information alone. Therefore, option c cannot be concluded from the given information.
d) The value of f(0) is 5:
We do not have enough information to determine the value of f(0) based on the given information. Therefore, option d cannot be concluded from the given information.
In conclusion, based on the given information, options b and c can be concluded, while options a and d cannot be determined.
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f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer?
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f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. 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 f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer?.
f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. 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 f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer?.
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f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer?, a detailed solution for f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer? has been provided alongside types of f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice f(x) is a cubic polynomial which has local maximum at x = -1. If f(2) = 18, f(1) = -1 and f'(x) has local minimum at x = 0, thena)the distance between (-1, 2) and [a, f(a)], where x = a is the point of local minima, is 2√5b)f{x) is increasing for x ∈ [1, 2√5]c)f(x) has local minima at x = 1d)the value of f(0) is 5Correct answer is option 'B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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