Question

If $f:R\to R$ and $f\left(x\right)$ is a polynomial function of degree eleven and $f\left(x\right)=0$ has all real and distinct roots. Then the equation $\left(f^{\prime }\right(x){)}^2-f(x\left)f^{\prime \prime }\right(x)=0$ has

• A. no real roots
• B. 10 real roots
• C. 11 real roots
• D. 21 real roots

Answer 1

The correct option is A no real roots

$f\left(x\right)=a(x-x_1)(x-x_2)...(x-x_{11})$
$\ln f\left(x\right)=\ln a+\ln (x-x_1)+\ln (x-x_2)...+\ln (x-x_{11})$
$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{x-x_1}+\frac{1}{x-x_2}...+\frac{1}{x-x_{11}}$
$\frac{f^{\prime \prime }\left(x\right)f\left(x\right)-\left(f^{\prime }\right(x){)}^2}{\left(f\right(x){)}^2}$ $=-[\frac{1}{(x-x_1{)}^2}+\frac{1}{(x-x_2{)}^2}.......+\frac{1}{(x-x_{11}{)}^2}]$
Therefore
$f^{\prime \prime }\left(x\right)f\left(x\right)-\left(f^{\prime }\right(x){)}^2$ $=-\left(f\right(x){)}^2.[\frac{1}{(x-x_1{)}^2}+.......+\frac{1}{(x-x_{11}{)}^2}]$
$\Rightarrow \left(f^{\prime }\right(x){)}^2-f(x\left)f^{\prime \prime }\right(x)=0$ has no real roots.