Question

Let $f\left(x\right)$ be a polynomial function of the second degree. If $f\left(1\right)=f(-1)$ and $a_1,a_2,a_3$ are in AP, then $f\text{'}\left(a_1\right),f\text{'}\left(a_2\right),f\text{'}\left(a_3\right)$ are in

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is A

AP

Put the values in the given function:

Let $f\left(x\right)=p{x}^2+qx+r$

$$\begin{gathered} \Rightarrow f\left(1\right)=p+q+r \\ \Rightarrow f(-1)=p-q+r \end{gathered}$$

It is given that $f\left(1\right)=f(-1)$.

$$\begin{gathered} \Rightarrow p+q+r=p-q+r \\ \Rightarrow 2q=0 \\ \Rightarrow q=0 \end{gathered}$$

Now,

$$\begin{gathered} f\text{'}\left(x\right)=2px+q \\ =2px \end{gathered}$$

$$\begin{gathered} f\text{'}\left(a_1\right)=2p{a}_1, \\ f\text{'}\left(a_2\right)=2p{a}_2, \\ f\text{'}\left(a_3\right)=2p{a}_3 \end{gathered}$$

$a_1,a_2,a_3$ are in AP.

Therefore, $2p{a}_1,2p{a}_2,2p{a}_3$ are in AP.

Thus $f\text{'}\left(a_1\right),f\text{'}\left(a_2\right),f\text{'}\left(a_3\right)$ are in AP.

Hence, the correct answer is option (A).