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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
no real roots
$10$ real roots
$6$ real roots
$8$ 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 + $
${\rm{ln}}\left( {{\rm{x}} – {{\rm{x}}_{10}}} \right)$
$\frac{f(\mathrm{x})}{f^{\prime}(\mathrm{x})}=\frac{1}{\left(\mathrm{x}-\mathrm{x}_{1}\right)}+\frac{1}{\left(\mathrm{x}-\mathrm{x}_{2}\right)}+\ldots \ldots+\frac{1}{\left(\mathrm{x}-\mathrm{x}_{10}\right)}$
$\frac{{f\left( x \right)f''\left( x \right) – {{\left( {f'\left( x \right)} \right)}^2}}}{{{{\left( {f'\left( x \right)} \right)}^2}}} = 0$
$=-\left(\frac{1}{\left(x-x_{1}\right)^{2}}+\frac{1}{\left(x-x_{2}\right)^{2}}+\ldots . .+\frac{1}{\left(x-x_{10}\right)^{2}}\right)$
Similar Questions
Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
$x=-1$ | $x=0$ | $x=2$ | |
$f(x)$ | $3$ | $6$ | $0$ |
$g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$