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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 $[-1,1]$, which of the following statements is true

$\begin {array} {1 1} \text{(A) P(-1) is the minimum and P(1) is the maximum of P}\\ \text{(B) P(-1) is not minimum but P(1) is the maximum of P} \\ \text{(C) P(-1) is the minimum and P(1) is not the maximum of P} \\ \text{(D) neither P(-1) is the minimum of P nor P(1) is the maximum of P} \end {array}$

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1 Answer

Ans : (B)
Given, $P(x)= x^4+ax^3+bx^2+cx+d$
Since, $x=0$ is a solution for $P’(x)=0$,
So, $P(x)= x^4+ax^3+bx^2+d$
Also we have $P(-1)< P(1)$
Since,$ P’(x)=0$, only when $x=0\: V \: and\: P(x)$ is differentiable in
$(-1,1)$, we should have the maximum and minimum at the points
$x= -1,\: 0\: and\: 1$ only
Also, we have $ P(-1)< P(1)$
So,Maximum of $P(x)= Max\{ P(0),P(1)\}$ and
Minimum of $P(x)=Min \{P(-1),P(0)\}$
In the interval $[0,1]$
$P’(x)= 4x^3+3ax^2+2bx=x(4x^2+3ax+2b)$
Since, $P’(x)$ has only one root $x=0$, then $4x^2+3ax+2b=0$ has no real roots.
So, $(3a)^2 – 32b<0$
$ 3a^2/32 > b $
So, $b>0$
Thus, we have $a>0$ and $b>0$
So, $P’(x)= 4x^3+3ax^2+2bx>0, \: for\: x \in (0,1)$
Hence, $P(x)$ is increasing in $[0,1]$ and $P(x)$ is decreasing in $[-1,0]$
Therefore, Maximum of $P(x)=P(1)$ and Minimum $P(x)$ does not occur at
$x=-1$ respectively


answered Jan 22, 2014 by thanvigandhi_1
edited Jan 22, 2014 by thanvigandhi_1

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