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Question Number 29425 by puneet1789 last updated on 08/Feb/18

Commented by prof Abdo imad last updated on 12/Feb/18

$$f\left(x\right)=\lambda \mathrm{\Pi }{}_{o=1}^{10}(x-x_i)\Rightarrow \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=\sum _{i=1}^{10}\frac{1}{x-x_i}$$$$\Rightarrow \frac{f^{{}^{″}}\left(x\right)f\left(x\right)-{\left(f^{{}^{\prime }}\left(x\right)\right)}^2}{{\left(f\left(x\right)\right)}^2}=-\sum _{i=1}^{10}\frac{1}{{(x-x_i)}^2}\Rightarrow$$$${\left(f^{{}^{\prime }}\left(x\right)\right)}^2-f\left(x\right)f^{{}^{″}}\left(x\right)=\left(f{\left(x\right)}^2\right)\left(\sum _{i=1}^{10}\frac{1}{{(x-x_i)}^2}\right)$$$$anddueto\mathrm{\Sigma }(...)\ne 0\Rightarrow {\left(f^{{}^{\prime }}\left(x\right)\right)}^2-f\left(x\right)f^{{}^{″}}\left(x\right)=0$$$$\iff f\left(x\right)=0andthisequ.have10roots.$$

Answered by mrW2 last updated on 09/Feb/18

$$f\left(x\right)=(x-a_1)(x-a_2)...(x-a_{10})$$$$f^{\prime }\left(x\right)=f\left(x\right)(\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{10}})$$$$f^{″}\left(x\right)=f\left(x\right){(\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{10}})}^2-f\left(x\right)[\frac{1}{{(x-a_1)}^2}+\frac{1}{{(x-a_2)}^2}+...+\frac{1}{{(x-a_{10})}^2}]$$$$\Rightarrow f\left(x\right)f^{″}\left(x\right)=f^2\left(x\right){(\frac{1}{x-a_1}+\frac{1}{x-a_2}+...+\frac{1}{x-a_{10}})}^2-f^2\left(x\right)[\frac{1}{{(x-a_1)}^2}+\frac{1}{{(x-a_2)}^2}+...+\frac{1}{{(x-a_{10})}^2}]$$$$\Rightarrow f\left(x\right)f^{″}\left(x\right)={\left[f^{\prime }\left(x\right)\right]}^2-f^2\left(x\right)[\frac{1}{{(x-a_1)}^2}+\frac{1}{{(x-a_2)}^2}+...+\frac{1}{{(x-a_{10})}^2}]$$$$\Rightarrow f\left(x\right)f^{″}\left(x\right)-{\left[f^{\prime }\left(x\right)\right]}^2=-f^2\left(x\right)[\frac{1}{{(x-a_1)}^2}+\frac{1}{{(x-a_2)}^2}+...+\frac{1}{{(x-a_{10})}^2}]=0$$$$i.e.f\left(x\right)f^{″}\left(x\right)-{\left[f^{\prime }\left(x\right)\right]}^2=0hasthesame$$$$rootsasf\left(x\right)=0,i.e.10realroots.$$$$\Rightarrow answer\left(2\right)isright.$$

Commented by puneet1789 last updated on 08/Feb/18

vry nice sir.. thank you

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