Mathematics - Differential Equation Question with Solution | TestHub

Degree of$f\left(x\right)$will be$3$

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

$f\left(3x\right)=27a{x}^3+9b{x}^2+3cx+d$

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

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

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

Comparing the coefficient, we get

$27a=18a^2$$\Rightarrow a=\frac{3}{2}$

Also$b=0, c=0, d=0$

$f\left(x\right)=\frac{3}{2}{x}^3$,$f\left(2\right)=12$

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

Hence,$f^{\text{'}}\left(2\right)=18, f{}^{\text{'}\text{'}}\left(2\right)=18$

Hence,$f^{\text{'}\text{'}}\left(2\right)-f\text{'}\left(2\right)=0$