Question

Let $f\left(x\right)$ be a polynomial function of second degree . Given that $f\left(1\right)=f(-1)$ and $a,b,c$ are in A.P. Also $f\left(a\right),f\left(b\right),f\left(c\right)$ are in A.P., then find the values of $a,b,c$

• A. $f\left(x\right)$ of second degree cannot exist.
• B. $a=2,b=0,c=1$
• C. $a=3,b=5,c=0$
• D. none of the above

Answer 1

The correct option is A $f\left(x\right)$ of second degree cannot exist.

Let $f\left(x\right)=\alpha x^2+\beta x+\gamma =0,\alpha \ne 0$
$f(-1)=f\left(1\right)$ gives us $\beta =0$
$f\left(a\right)=\alpha a^2+\gamma$
$f\left(b\right)=\alpha b^2+\gamma$
$f\left(c\right)=\alpha c^2+\gamma$
Given $2b=a+c$.....(1)
$f\left(c\right)-f\left(b\right)=f\left(b\right)-f\left(a\right)$ (Given $f\left(a\right),f\left(b\right),f\left(c\right)$ are in A.P.$)$
$⟺c^2-b^2=b^2-a^2$
$⟺(c-b)(c+b)=(b-a)(b+a)$
$⟺c+b=b+a$ or $c=b=a$

In both cases, $a=c=b$