Question

Let $f\left(x\right)=a{x}^2+bx+c$ such that $f\left(1\right)=f(-1)$ and a, b, c are in Arithmetic Progression.

Then find what kind of progression $f^{\prime \prime }\left(a\right),f^{\prime \prime }\left(b\right),f^{\prime \prime }\left(c\right)$ form.

• A. In A.P. only
• B. In G.P. only
• C. In both A.P. and G.P.
• D. Neither in A.P. nor in G.P.

Answer 1

The correct option is C In both A.P. and G.P.

The given equation is:

$y=f\left(x\right)=a{x}^2+bx+c$

Differentiating w.r.t to x we get,

$\Rightarrow y^{\prime }=2ax+b$

Since a, b, c are in A.P. we have, $2b=a+c$

$\Rightarrow b-a=c-b$

Again differentiating w.r.t. to x we get,

$\Rightarrow y^{\prime \prime }=2a$

Finding the respective derivative values we get,

$y^{\prime \prime }\left(a\right)=2a$

$y^{\prime \prime }\left(b\right)=2a$

$y^{\prime \prime }\left(c\right)=2a$

Finding the common difference between the terms we get,

$y^{\prime \prime }\left(b\right)-y^{\prime \prime }\left(a\right)=2a-2a$ $\frac{y^{\prime \prime }\left(b\right)}{y^{\prime \prime }\left(a\right)}=1$

$\Rightarrow y^{\prime \prime }\left(b\right)-y^{\prime \prime }\left(a\right)=0$

$y^{\prime \prime }\left(c\right)-y^{\prime \prime }\left(b\right)=2a-2a$ $\frac{y^{\prime \prime }\left(c\right)}{y^{\prime \prime }\left(b\right)}=1$

$\Rightarrow y^{\prime \prime }\left(c\right)-y^{\prime \prime }\left(b\right)=0$

Since the common difference is same and the common ratio is also same $y^{\prime \prime }\left(a\right),y^{\prime \prime }\left(b\right),y^{\prime \prime }\left(c\right)$can be both $A.P$ and $G.P$. .....Answer