Question

Let $f$ be a polynomial function of degree $n$ such that $f\left(3x\right)=f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right),$ for all $x\in R.$ Then :

• A. $f\left(2\right)+f^{\prime }\left(2\right)=28$
• B. $f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)=0$
• C. $f^{\prime \prime }\left(2\right)-f\left(2\right)=4$
• D. $f^{\prime \prime }\left(2\right)-f\left(2\right)=-4$

Answer 1

The correct option is B $f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)=0$

$f\left(3x\right)=f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)-\left(i\right)$

Let $f\left(x\right)=n$ where $n$ is the power of polymomial

$\Rightarrow f^{\prime }\left(x\right)=n-1,f^{\prime \prime }\left(x\right)=n-2$

from (i)

$n=n-1+n-2$

$\Rightarrow n=3$

Hence,

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$

$f^{\prime }\left(x\right)=3a{x}^2+2bx+c$

$f^{\prime \prime }\left(x\right)=6ax+2b$

Put the values in (i)

$a(3x{)}^3+b(3x{)}^2+3xc+d=(3a{x}^2+2bx+c)(6ax+2b)$

$\Rightarrow 27a{x}^3+9b{x}^2+3cx+d=18a^2{x}^3+(6ab+12ab)x^2+(4b^2+6ac)x+2bc$

On comparing the like terms

$27a=18a^2;9b=18ab;3c=4b^2+6ac$

$\Rightarrow \frac{27}{18}=a9=18a\Rightarrow c=0$

$\Rightarrow a\ne \frac{1}{2}$

$\Rightarrow a=\frac{3}{2}\therefore b=0$

$\text{hence we get}f\left(x\right)=\frac{3}{2}{x}^3$

$f^{\prime }\left(x\right)=\frac{9}{2}{x}^2;f^{\prime \prime }\left(x\right)=9x$

Therefore the condition in option B is satisfied.

Hence, Option B is correct.