Let f(x)=(x2−1,if0<x<22x+3,if2≤x<3, a quadratic equation whose roots are limx→2−f(x)andlimx→2+f(x) is
A
x2−6x+9=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x2−7x+8=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2−14x+49=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x2−10x+21=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is Dx2−10x+21=0 Letusfirstfindtherootsoftherequiredquadraticequationi.e.limx→2−f(x)=limx→2−x2−1=4−1=3limx→2+f(x)=limx→2+2x+3=2(2)+3=7Now,letusfindthequdraticequationwithrootsasx=3andx=7(x−3)(x−7)=0⇒x2−10x+21=0