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
⇒(9a+3b+c)−(9p+3q+r)=2 ⇒3(3a+b)+c−3(3p+q)−r=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,