Q.

If  f(x) is a polynomial with positive leading coefficient satisfying  f(0)=0  f(f(x))=x0xf(t)dtxR, then locus of point of intersection of two perpendicular tangents can be drawn to the curve  y=f(x) can be given by  ax2+2hxy+by2+2gx+2fy+3=0, then the value of  a25h3+2b4+3g5+f=

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answer is 2.

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Detailed Solution

Let the degree of  f(x)  be  'n'
So, we get  n2=1+n+1n=2
Let  f(x)=ax2+bx as  f(0)=0
Given relation f(f(x))=x0xf(t)dt

a(ax2+bx)2+b(ax2+bx)=x(ax33+bx22)

a3=a3 & 2a2b=b2 & ab2+ab=0 & b2=0

b=0 & a=13

So, we get  f(x)=x23
So, the directrix will be  y=43

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