Question

If the polynomial function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ has extreme at x=a and $x=2a$ then

• A. $a,b,c$ are in AP
• B. $a,b,c$ are in GP
• C. $6a,4b,9c$ are in AP
• D. $6a,4b,9c$ are in GP

Answer 1

The correct option is D $6a,4b,9c$ are in GP

$P\left(x\right)=a{x}^3+b{x}^2+cx+d$
$P^{\prime }\left(x\right)=3a{x}^2+2bx+c$
Given $x=a,x=2a$ are extremum value of $P\left(x\right)$
$\Rightarrow P\left(a\right)=0$ and $P\left(2a\right)=0$
$\Rightarrow 3a^3+2ab+c=0....\left(1\right)$
and $12a^3+4ab+c=0.......\left(2\right)$
Solving (1) and (2)
$8b^2=27ac\Rightarrow 16b^2=54ac\Rightarrow (4b{)}^2=(6a\left)\right(9c)$
Hence $6a,4b,9c$ are in G.P