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Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]
- a)p (-1) is the minimum and p (1) is the maximum of P
- b)p (-1) is not minimum but p (1) is the maximum of P
- c)p (-1) is the minimum and p (1) is not the maximum of P
- d)neither p (-1) is the minimum nor p (1) is the maximum of P
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real...
p ' (x) has only one change of sign. ⇒ x = 0 is a point of minima.
P(-1) =1- a + b + d P(0) = d
P(1) =1+ a + b + d
⇒ P (-1) < P (1) , P (0) < P (1) , P (-1) > P ( 0)
⇒ P (-1) is not minimum but P(1) is maximum
Most Upvoted Answer
Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real...
Understanding the Problem:
The given polynomial function is p(x) = x^4 + ax^3 + bx^2 + cx + d, where x = 0 is the only real root of p(x) = 0. We are also given that p(-1) = p(1). We need to determine the relationship between p(-1) and p(1) in the interval [-1, 1].
Understanding the Given Information:
1. x = 0 is the only real root of p(x) = 0. This means that the polynomial has no other real roots and is a perfect square.
2. p(-1) = p(1). This implies that the function is symmetric about the y-axis.
Analysis:
Since x = 0 is the only real root of p(x) = 0, the polynomial can be factored as p(x) = (x - 0)^2 * (x^2 + px + q), where p and q are constants.
Using the given information p(-1) = p(1), we can substitute x = -1 and x = 1 in the factored form of the polynomial to get two equations:
p(-1) = (-1)^2 * (-1^2 + p(-1) + q) ...(1)
p(1) = (1)^2 * (1^2 + p(1) + q) ...(2)
Simplifying the Equations:
1. Simplifying equation (1):
p(-1) = 1 * (1 - p(-1) + q)
p(-1) = 1 - p(-1) + q
2p(-1) = 1 + q
q = 2p(-1) - 1
2. Simplifying equation (2):
p(1) = 1 * (1 + p(1) + q)
p(1) = 1 + p(1) + q
2p(1) = 1 + q
q = 2p(1) - 1
From equations (1) and (2), we can conclude that q is the same for p(-1) and p(1).
Interpreting the Results:
1. If p(-1) < />
Since q is the same for both p(-1) and p(1), the value of p(1) will be greater than p(-1) due to the positive coefficient of p(1) in equation (2). Therefore, p(-1) cannot be the minimum value in the interval [-1, 1].
2. If p(-1) > p(1):
Similarly, if p(-1) is greater than p(1), p(1) cannot be the maximum value in the interval [-1, 1].
3. If p(-1) = p(1):
Since p(-1) = p(1), the function is symmetric about the y-axis. This means that the function has a minimum value at x = -1 and a maximum value at x = 1 in the interval [-1, 1].
Conclusion:
From the analysis, we can conclude that the correct answer is option B: p(-1) is not the minimum value, but p(
The given polynomial function is p(x) = x^4 + ax^3 + bx^2 + cx + d, where x = 0 is the only real root of p(x) = 0. We are also given that p(-1) = p(1). We need to determine the relationship between p(-1) and p(1) in the interval [-1, 1].
Understanding the Given Information:
1. x = 0 is the only real root of p(x) = 0. This means that the polynomial has no other real roots and is a perfect square.
2. p(-1) = p(1). This implies that the function is symmetric about the y-axis.
Analysis:
Since x = 0 is the only real root of p(x) = 0, the polynomial can be factored as p(x) = (x - 0)^2 * (x^2 + px + q), where p and q are constants.
Using the given information p(-1) = p(1), we can substitute x = -1 and x = 1 in the factored form of the polynomial to get two equations:
p(-1) = (-1)^2 * (-1^2 + p(-1) + q) ...(1)
p(1) = (1)^2 * (1^2 + p(1) + q) ...(2)
Simplifying the Equations:
1. Simplifying equation (1):
p(-1) = 1 * (1 - p(-1) + q)
p(-1) = 1 - p(-1) + q
2p(-1) = 1 + q
q = 2p(-1) - 1
2. Simplifying equation (2):
p(1) = 1 * (1 + p(1) + q)
p(1) = 1 + p(1) + q
2p(1) = 1 + q
q = 2p(1) - 1
From equations (1) and (2), we can conclude that q is the same for p(-1) and p(1).
Interpreting the Results:
1. If p(-1) < />
Since q is the same for both p(-1) and p(1), the value of p(1) will be greater than p(-1) due to the positive coefficient of p(1) in equation (2). Therefore, p(-1) cannot be the minimum value in the interval [-1, 1].
2. If p(-1) > p(1):
Similarly, if p(-1) is greater than p(1), p(1) cannot be the maximum value in the interval [-1, 1].
3. If p(-1) = p(1):
Since p(-1) = p(1), the function is symmetric about the y-axis. This means that the function has a minimum value at x = -1 and a maximum value at x = 1 in the interval [-1, 1].
Conclusion:
From the analysis, we can conclude that the correct answer is option B: p(-1) is not the minimum value, but p(
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Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer?
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Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. 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 Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer?.
Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. 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 Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer?.
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Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Given p (x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of p' (x) = 0 . If p (-1) < p (1) , then in the interval [-1,1]a)p (-1) is the minimum and p (1) is the maximum of Pb)p (-1) is not minimum but p (1) is the maximum of Pc)p (-1) is the minimum and p (1) is not the maximum of Pd)neither p (-1) is the minimum nor p (1) is the maximum of PCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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