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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 not minimum but P(1) is the maximum of P
- b)P(–1) is the minimum but P(1) is not the maximum of P
- c)Neither P(–1) is the minimum nor P(1) is the maximum of P
- d)P(–1) is the minimum and P(1) is the maximum of P
Correct answer is option 'A'. 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 ...
We have P (x) = x4 + ax3 + bx2 + cx + d
⇒ P' (x) = 4x3 + 3ax2 + 2bx + c
But P' (0) = 0 ⇒ c = 0
∴ P(x) = x4 + ax3 + bx2 + d
As given that P (– 1) < P (a)
⇒ 1 – a + b + d < 1 + a + b + d ⇒ a > 0
Now P ' (x) = 4x3 + 3ax2 +2bx = x (4x2 + 3ax + 2b)
As P' (x) = 0, there is only one solution x = 0, therefore 4x2 + 3ax + 2b = 0 should not have any real roots i.e. D < 0
⇒ P' (x) = 4x3 + 3ax2 + 2bx + c
But P' (0) = 0 ⇒ c = 0
∴ P(x) = x4 + ax3 + bx2 + d
As given that P (– 1) < P (a)
⇒ 1 – a + b + d < 1 + a + b + d ⇒ a > 0
Now P ' (x) = 4x3 + 3ax2 +2bx = x (4x2 + 3ax + 2b)
As P' (x) = 0, there is only one solution x = 0, therefore 4x2 + 3ax + 2b = 0 should not have any real roots i.e. D < 0
∴ P (x) is an increasing function on (0,1) ∴ P (0) < P (a)
Similarly we can prove P (x) is decreasing on (– 1, 0)
∴ P (– 1) > P (0)
So we can conclude that
Max P (x) = P (1) and Min P (x) = P (0) ⇒ P(–1) is not minimum but P (1) is the maximum of P.
Similarly we can prove P (x) is decreasing on (– 1, 0)
∴ P (– 1) > P (0)
So we can conclude that
Max P (x) = P (1) and Min P (x) = P (0) ⇒ P(–1) is not minimum but P (1) is the maximum of P.
Most Upvoted Answer
Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real ...
We know that x = 0 is the only real root of P(x) = 0. This means that P(0) = 0.
P(0) = 0^4 + a(0)^3 + b(0)^2 + c(0) + d = d = 0
Therefore, d = 0.
So, P(x) = x^4 + ax^3 + bx^2 + cx.
Since x = 0 is the only real root, we can conclude that P(x) cannot be factored further.
P(0) = 0^4 + a(0)^3 + b(0)^2 + c(0) + d = d = 0
Therefore, d = 0.
So, P(x) = x^4 + ax^3 + bx^2 + cx.
Since x = 0 is the only real root, we can conclude that P(x) cannot be factored further.
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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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. 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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. 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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. 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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. 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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. 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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. 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 not minimum but P(1) is the maximum of Pb)P(–1) is the minimum but P(1) is not the maximum of Pc)Neither P(–1) is the minimum nor P(1) is the maximum of Pd)P(–1) is the minimum and P(1) is the maximum of PCorrect answer is option 'A'. Can you explain this answer?.
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