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 $f^{\prime }\left(a\right),f^{\prime }\left(b\right),f^{\prime }\left(c\right)$ form.

• A. A.P.
• B. G.P.
• C. H.P.
• D. Arithmetico-geometric progression

Answer 1

The correct option is A A.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, $2b=a+c$

$\Rightarrow b-a=c-b$

Finding the respective derivative values we get,

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

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

$y^{\prime }\left(c\right)=2ac+b$

Finding the common difference between the terms we get,

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

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

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

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

$y^{\prime }\left(c\right)-y^{\prime }\left(b\right)=(2ac+b)-(2ab+b)$

$\Rightarrow y^{\prime }\left(c\right)-y^{\prime }\left(b\right)=2ac-2ab$

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

We got the same common difference between the three terms hence the derivatives are in A.P. .....Answer