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)the maximum of P
B. (-1) is not minimum but p(1) the maximum of P
C. p(-1) is the minimum but P(1) is not the maximum of P
D. Niether p(-1) is the maxiumum nor P(1) is the maximum of P

Answer

B

Explanation

We have
`p(x) =x^4+ax^3+bx^2+cx+d`
`rArr P'(x) =4x^3+3ax^2+2bx+c`
It is gien that x=0 is a root of p'(x)=0
` therefore P'(0)=0rArrc=0`
Putting c=0 , we get
`P'(x)=x(4x^2+3ax+2b)`
Since x=0 is only real root of P'(x)=0. Therefore,
`4x^2+3ax+2b=0` has no real root
`rArr 9a^2-32blt0`
It is given that
`P(-1)ltP(1)`
`rArr1-a+b-c+dlt1+a+b+c+d`
`rArr agt0`
But , `9a^2-32b lt 0` . Therefore ,`bgt0`
`therefore P'(x)=x(4x^2+3ax+2b)gt0 " for all " x in (0,1]`
`rArr` P(x) is increasing in (0,1]
`rArr` P(1) is the maximum value of P(x)
Also `P'(x)=x(4x^2+3ax+2b)lt0 " for all " x in [-1,0] "         " [therefore4x^2+3ax+2bgt0 " for all " x]`
`rArr` P(x) is decreasing in [-1,0]
`rArr` P(-1) is not the minimum value of P