see hum sirf baatein nahi kar rahe hain

i will tell you all what i,ve done , no need to give any salute etc

let p(x)=Dx⁴+Kx³+Ex²+Hx+F , since p(-x)=p(x), so F=1 and the graph is symmetric in 1st and 2nd quadrant, so p(x) = Dx⁴+Ex²+1

also the 2 minima should be of the form {from the graph} (-t,0) and (t,0), since mod(a-b)=2, so the minima are (-1,0) and (1,0)

now, p'(-1or1)=0 this gives 2D+E=0    [ i didn't use this condition before ]

so p(x) = Dx⁴-2Dx²+1

for the pts of intersection of the curve and the graph, Dx⁴-2Dx²+1=1, this gives x=+-root(2)

now the area expression can be used, 2.integral { 1-( Dx⁴-2Dx²+1 ) } from 0 to root(2) = 8root(2)/15

this gives D = 1/2 , so the expn of p(x) = (1/2)x⁴-x²+1

min of p(x) is at x = -1 and 1, so p(x) min = 1/2

after this we have to use the given limit and i am not able to proceed