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) lt P(1),` then in the interval `[-1,1]`

Options

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 not 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

2

Explanation

`P(x) =x^(4)+ax^(3)+bx^(2)+cx+d`
`P(x)=4x^(3)+3ax^(2)+2bx+c`
Since x=0 is solution to P(x) =0 we get c=0 Therefore
`P(x)=x^(4)+ax^(3)+bx^(2)+d`
Also we have
`P(-1)ltP(1)`
`rarr 1-a+b+dlt1+a+b+drarragt0`
Since P(x) =0 only when x=0 and P(x) is differentiable in
(-1,1) we should have the maximum and minimum points x=-1,0 and 1 only
Also we have f(-1)`lt`P(1)
Maximum of P(x) =maximum {P(0),P(1)}and
Minimum of f(x)=minimum {P(-21),P(0)}
In the interval [0,1]
`P(x)=4x^(3)+3ax^(2)+2bx{(p(1)}` and
`4x^(2)+3ax+2b=0` has no real roots .Therefore
`(3a)^(2)-32blt0`
`rarr (9a^(2))/(32)ltb`
`rarrbgt0`
Thus we have `agt0 and bv gt0` Thus
P(x) =`4x^(3)+3ax^(2)+2bxgt0 forall x in (0,1)`
Hence P(x) is increasing in [0,1]
Similarly P(x) id P(1)
Therefore the minimum of P(x) does not occur at x=1