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प्रश्न
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] ______
विकल्प
P(-1) is the minimum and P(1) is the maximum of P
P(-1) is not minimum but P(1) is the maximum of P
P(-1) is the minimum but P(1) is not the maximum of P
Neither P(-1) is the maximum nor P(1) is the maximum of P
उत्तर
P(-1) is not minimum but P(1) is the maximum of P
Explanation:
P(x) = x4 + ax3 + bx2 + cx + d
⇒ P'(x) = `4x^3 + 3ax^2 + 2bx + c` .............(i)
Since, x = 0 is the only real root of P'(x) = 0.
∴ P'(0) = 0 ⇒ c = 0
Putting c = 0 in (i), we get
P'(x) = x(4x2 + 3ax + 2b)
Since x = 0 is the only real root of P'(x) = 0,
4x2 + 3ax + 2b = 0 has no real root.
⇒ `9a^2 - 32b < 0`
Given, P(-1) < P(1)
⇒ 1 - a + b - c + d < 1 + a + b + c + d
⇒ a > 0
But, `9a^2 - 32b < 0`.
∴ b > 0
∴ P'(x) = x(4x2 + 3ax + 2b) > 0 for all x ∈ [0, 1]
⇒ P(x) is increasing in [0, 1]
⇒ P(1) is the maximum value of P(x).
Also, P'(x) = x(4x2 + 3ax + 2b) < for all
x ∈ [-1, 0] ..........................[∵ 4x2 + 3ax + 2b > 0 for all x]
⇒ P(x) is decreasing in [-1, 0].
⇒ P(-1) is not the minimum value of P.
