24anyone ?????????
There exists a plane f(a).x+f(0).y+f(c).z=1 which is passing through the point (0,1,1) where y=f(x) is a second degree polynomial which cuts the line x+y=1 at 2 distinct points x=1,x=c where c=rational number.It is also given that there exists a positive integral value of a for which the product of roots of f(x)=0 becomes -1
Q1 Find value of f(a)+f(0)+f(c)
Q2 Find hte equation of plane parallel to given plane for negative value of c and passing through (1,1,-1)
Q3 Find hte eqn of line on 3D space such that line os perpendicular to a plane for c>1 and passing through (1,1,1)
Q4 If there exists two such planes ,then find hte nagle between them.
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4 Answers
62f(1) = 1-1 = 0
and f(c) = 1-c
also, f(a).0+f(0).1+f(c) = 1
so f(0)+1-c = 1
so f(0) = c
also f is a 2nd degree polynomial
let f(x)= mx2+nx+l
so f(0) = l
so l=c
f(1) =m+n = -c
reading further.. i did not understand anything at all about this phrase It is also given that there exists a positive integral value of a for which the product of roots of f(x)=0 becomes -1
because f is a quadratic.. so the product will be independent of a!!