TS EAMCET 19-July-2022 Shift 2 Solved Paper

f(x)=Ax2+Bx,g(x)=Lx2+Mx+N
f(x)−g(x)=Ax2+Bx−Lx2−Mx−N
=(A−L)x2+(B−M)x−N
Let A−L=P,B−M=Q
f(x)−g(x)=Px2+Qx−N
‌f(2)−g(2)=1
‌4P+2Q−N=1.....(i)
Also, ‌f(3)−g(3)=4
‌9P+3Q−N=4......(ii)
Also, f(4)−g(4)=9
16P+4Q−N=9......(iii)
Now, Eq. (i) ×4− Eq. (iii), we get
4Q−3N=−5 .......(iv)
Eq. (i) ×9− Eq. (ii) ×4, we get
6Q−5N=−7 .......(v)
Eq. (iv) ×3− Eq. (v) ×2, we get
−9N+10N=−15+14⇒N=−1
From Eq. (iv),
‌4Q−3(−1)=−5
‌4Q+3=−5⇒4Q=−8
‌Q=−2
From Eq. (i),
‌4P+2(−2)−(−1)=1
‌4P−4+1=1
‌P=1
∴‌f(x)−g(x)=x2−2x+1=(x−1)2
f(x)−g(x)=0⇒(x−1)2=0
‌x=1
∴ The root of f(x)−g(x)=0 is 1 .