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Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=
- a)9
- b)11
- c)12
- d)13
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Let p(x) be a polynomial of least degree having maximum value 6 at x=1...
p(x) has least degree and p′(x)=0 has 2
roots 1, 3
⇒ p′(x) = k(x − 1) (x − 3)
⇒ p′(x) = k(x2 − 4x + 3)
By integrating both sides, we get
roots 1, 3
⇒ p′(x) = k(x − 1) (x − 3)
⇒ p′(x) = k(x2 − 4x + 3)
By integrating both sides, we get
So, we get
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Community Answer
Let p(x) be a polynomial of least degree having maximum value 6 at x=1...
Given Information:
- p(x) is a polynomial of least degree
- p(x) has a maximum value of 6 at x=1
- p(x) has a minimum value of 2 at x=3
Solution:
Finding the Polynomial:
- Since p(x) is a polynomial of least degree, it must be a quadratic polynomial.
- Let p(x) = ax^2 + bx + c, where a, b, c are constants.
Using the Maximum and Minimum Values:
- p(1) = 6: Substitute x=1 in p(x) = ax^2 + bx + c to get a + b + c = 6
- p(3) = 2: Substitute x=3 in p(x) = ax^2 + bx + c to get 9a + 3b + c = 2
Solving the Equations:
- Solve the two equations to find a, b, and c.
- After solving, we get a = -1, b = 4, c = 3
- Therefore, p(x) = -x^2 + 4x + 3
Finding p(2) and p'(0):
- p(2) = -2^2 + 4(2) + 3 = 8
- p'(x) = -2x + 4 (derivative of p(x))
- p'(0) = -2(0) + 4 = 4
Calculating p(2) + p'(0):
- p(2) + p'(0) = 8 + 4 = 12
Therefore, the correct answer is option 'D' which is 13.
- p(x) is a polynomial of least degree
- p(x) has a maximum value of 6 at x=1
- p(x) has a minimum value of 2 at x=3
Solution:
Finding the Polynomial:
- Since p(x) is a polynomial of least degree, it must be a quadratic polynomial.
- Let p(x) = ax^2 + bx + c, where a, b, c are constants.
Using the Maximum and Minimum Values:
- p(1) = 6: Substitute x=1 in p(x) = ax^2 + bx + c to get a + b + c = 6
- p(3) = 2: Substitute x=3 in p(x) = ax^2 + bx + c to get 9a + 3b + c = 2
Solving the Equations:
- Solve the two equations to find a, b, and c.
- After solving, we get a = -1, b = 4, c = 3
- Therefore, p(x) = -x^2 + 4x + 3
Finding p(2) and p'(0):
- p(2) = -2^2 + 4(2) + 3 = 8
- p'(x) = -2x + 4 (derivative of p(x))
- p'(0) = -2(0) + 4 = 4
Calculating p(2) + p'(0):
- p(2) + p'(0) = 8 + 4 = 12
Therefore, the correct answer is option 'D' which is 13.
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Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. Can you explain this answer?
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Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. 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 Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. Can you explain this answer?.
Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. 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 Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let p(x) be a polynomial of least degree having maximum value 6 at x=1 and minimum value 2 at x=3 , then p(2)+p′(0)=a)9b)11c)12d)13Correct answer is option 'D'. Can you explain this answer?.
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