5. Continuity and Differentiation
normal

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

A

no real roots

B

$10$ real roots

C

$6$ real roots

D

$8$ real roots

Solution

Let $f(\mathrm{x})=\mathrm{a}\left(\mathrm{x}-\mathrm{x}_{1}\right)\left(\mathrm{x}-\mathrm{x}_{2}\right) \ldots \ldots\left(\mathrm{x}-\mathrm{x}_{10}\right)$

$\ln f(x) = \ln a + \ln \left( {x – {x_1}} \right) + \ln \left( {x – {x_2}} \right) +  \ldots  \ldots  + $

${\rm{ln}}\left( {{\rm{x}} – {{\rm{x}}_{10}}} \right)$

$\frac{f(\mathrm{x})}{f^{\prime}(\mathrm{x})}=\frac{1}{\left(\mathrm{x}-\mathrm{x}_{1}\right)}+\frac{1}{\left(\mathrm{x}-\mathrm{x}_{2}\right)}+\ldots \ldots+\frac{1}{\left(\mathrm{x}-\mathrm{x}_{10}\right)}$

$\frac{{f\left( x \right)f''\left( x \right) – {{\left( {f'\left( x \right)} \right)}^2}}}{{{{\left( {f'\left( x \right)} \right)}^2}}} = 0$

$=-\left(\frac{1}{\left(x-x_{1}\right)^{2}}+\frac{1}{\left(x-x_{2}\right)^{2}}+\ldots . .+\frac{1}{\left(x-x_{10}\right)^{2}}\right)$

Standard 12
Mathematics

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