WBJEE 2012 Math Solved Paper

Given, f(x)=ax2+bx+c, g(x)=px2+qx+r
Since f(1)=g(1)
⇒a+b+c=p+q+r... (i)
f(2)=g(2)
⇒4a+2b+c=4p+2q+r....(ii)
Subtracting Eq. (ii) from Eq. (i), we get
3a+b=3p+q....(iii)
f(3)−g(3)=2
⇒c−r=2 ....(iv)
(3a+b=3p+q)
From Eq.(i)
(a−p)+(b−q)+(c−r)=0
⇒(a−p)+(b−q)+2=0....(v)
From Eq. (ii),
4(a−p)+2(b−q)+c−r=0
⇒2(a−p)+(b−q)+1=0....(vi)
Subtracting Eq.(v) from Eq. (vi), we get
(a−p)−1=0
a−p=1
∴ From Eq. (v), b−q=−3
Now,Substituting the values of
(a−p),(b−q),(c−r) from above in Eq. (vii),
we get
f(4)−g(4)=16×1+4(−3)+2
=16−12+2=6