Question

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 $[-1, 1]$

• A. $P(-1)$ is the minimum and $P(1)$ is the maximum of $P$
• B. $P(-1)$ is not minimum but $P(1)$ is the maximum of $P$
• C. $P(-1)$ is the minimum and $P(1)$ is not the maximum of $P$
• D. neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of $P$

Answer 1

Answer: (B)

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

Answer 2

$P\left(x\right)=x^{4}+ax^{3}+bx^{2}+cx+d$ $P'\left(x\right) = 4x^{3} + 3ax^{2} + 2bx + c$ $\because x = 0$ is a solution for $P'\left(x\right) = 0 , ? c = 0$ $\therefore P\left(x\right)=x^{4}+ax^{3}+bx^{2}+cx+d\,...\left(1\right)$ Also, we have $P\left(-1\right) < P\left(1\right)$ $?1- a + b + d < 1+ a + b + d? a >0$ $Q P'\left(x\right) = 0$ , only when $x = 0$ and $P\left(x\right)$ is differentiable in $\left( - 1, 1\right)$, we should have the maximum and minimum at the points $x = - 1, 0$ and $1$ only Also, we have $P\left(-1\right) < P\left(1\right)$ ? Max. of P\left(x\right) = Max. P\left(0\right), P\left(1\right) \& Min. of P\left(x\right) = Min. \left\\{P\left(-1\right), P\left(0\right)\right\\} In the interval $\left[ 0 , 1 \right],$ $P'\left(x\right) = 4x^{3} + 3ax^{2} + 2bx = x\left(4x^{2} + 3ax + 2b\right)$ $\because P'\left(x\right)$ has only one root $x = 0, 4x^{2} + 3ax + 2b = 0$ has no real roots $\therefore \left(3a\right)^{2}-2ab < 0 \Rightarrow \frac{3a^{2}}{32} < b$ $\therefore b < 0$ Thus, we have $a > 0$ and $b > 0$ $? P'\left(x\right) = 4x^{3} + 3ax^{2} + 2bx > 0, ?x ?\left(0, 1\right)$ Hence $P\left(x\right)$ is increasing in $\left[ 0, 1 \right]$ $?$ Max. of $P\left(x\right) = P\left(1\right)$ Similarly, $P(x)$ is decreasing in $[-1 , 0]$ Therefore Min. P(x) does not occur at $x = - 1$