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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?
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) 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.
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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.
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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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