7hmm,,, @Rishabh, what if the q asked for f(4) x g(4) or f(4)+g(4)..??? What would i do then???
By computing the values, it would take too much time, i guess
btw, ur solution was good.
Given:
f(x) =ax2+bx+c
g(x)= px2+qx+r such that f(1)=g(1), f(2)=g(2) and f(3)-g(3) = 2 . Find f(4)-g(4).
The q is easy but i want a shorter method...
-
UP 0 DOWN 0 0 7
7 Answers
1let P(x) = f(x)-g(x) = (a-p)x2+(b-q)x+(c-r)
so P(x) is a quadratic equation with roots 1,2
=> P(x) = (a-p) * (x-1)(x-2)
given that P(3) =2
=> 2= 2(a-p) => a-p = 1
.:. P(x) = (x-1)(x-2)
.:. f(4)-g(4) = P(4) = 6
7
71In such exams, they won't ask you such, and in exams where they are asked, you'll be given ample time/marks for them. Anyways, may be HS Bhatt sir has something to offer.
1I think that the solution of rishabh is correct, the assumpsion that if it is f(x)g(x) needs different strutgies.
21As far as I can see it is practically impossible to calculate f(4)+g(4) provided nothing is given for a,b,c and p,q,r.
since f(4)+g(4)=6+2[16p+4q+r]
it is not possible to calculate its values... Can you see Sigma?
21BTW..... The shortest method I can think of is this:
Just let h(x)=f(x)-g(x)
clearly h(x)=k(x-1)(x-2) for some real constant k.
we get k=1 computing h(3).
Hence the result follows.