MHT CET2009MathematicsFunctions
Find a polynomial f(x) of degree 2 where f(0)=8, f(1)=12, f(2)=18
Options
- Ax²+3 x-8
- Bx²-3 x+8
- C2 x²-x+3
- Dx²+3 x+8
Correct answer
D. x²+3 x+8
Step-by-step solution
Let the polynomial is a x²+b x+c . Now, f(0)=8 =8 Equation is a x²+b x+8 . Again, f(1)=12 a+b+8=12 a+b=4 ( i ) and f(2)=18 4 a+2 b+8=18 2 a+b=5 (ii) On solving Eqs. (i) and (ii), we get a=1, b=3 Required equation is x²+3 x+8 .