Given $\mathrm{P}(\mathrm{x})=\mathrm{x}^{4}+\mathrm{ax}^{3}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}$ such that $\mathrm{x}=0$ is the only real root of $\mathrm{P}^{\prime}(\mathrm{x})=0$. If $\mathrm{P}(-1)<\mathrm{P}(1)$, then in the interval $[-1,1]$

(B) $P(-1)$ is not minimum but $P(1)$ is the maximum of $P$

(A) $\mathrm{P}(-1)$ is the minimum and $\mathrm{P}(1)$ is the maximum of P

(C) $\mathrm{P}(-1)$ is the minimum and $\mathrm{P}(1)$ is not the maximum of P

(D) neither $\mathrm{P}(-1)$ is the minimum nor $\mathrm{P}(1)$ is the maximum of P

JEE Main · 2009 · 12 Apr 2009 · Offline · Application of Derivatives

Given

P (x) = x^{4} + ax^{3} + bx^{2} + 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]

Answer: (B)

Since

x = 0

is the only real root of

P^{'} (x) = 4 x^{3} + 3 a x^{2} + 2 b x + c

, the constant

c

is zero and

P^{'} (x) = x (4 x^{2} + 3 a x + 2 b)

with the quadratic factor having no real root, so

9 a^{2} - 32 b < 0

and

b > 0

. The condition

P (- 1) < P (1)

reduces to

a > 0

. On

(0, 1)

then

P^{'} (x) > 0

, so

P

increases and

P (1)

is the maximum, while on

(- 1, 0)

P^{'} (x) < 0

, so

P

decreases to a minimum at

x = 0

rather than at

x = - 1

. Thus

P (1)

is the maximum but

P (- 1)

is not the minimum.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option