Bài tập nâng cao · Bài 8
Bài tập nâng cao Lôgarit · Bài 8
Cho hàm số $f(x)=\log_{2}\left( x-\frac{1}{2}+\sqrt{x^{2}-x+\frac{17}{4}} \right)$. Tính $T=f\left( \frac{1}{2019} \right)+f\left( \frac{2}{2019} \right)+...+f\left( \frac{2018}{2019} \right)$
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Lời giải
Ta có: $f(1-x)=\log_{2}\left( 1-x-\frac{1}{2}+\sqrt{\left( 1-x \right)^{2}-\left( 1-x \right)+\frac{17}{4}} \right)=\log_{2}\left( \sqrt{x^{2}-x+\frac{17}{4}}-\left( x-\frac{1}{2} \right) \right)$
$f\left( x \right)+f\left( 1-x \right)=\log_{2}\left( x-\frac{1}{2}+\sqrt{x^{2}-x+\frac{17}{4}} \right)+\log_{2}\left( \sqrt{x^{2}-x+\frac{17}{4}}-\left( x-\frac{1}{2} \right) \right)$
$=\log_{2}\left[ \left( x-\frac{1}{2}+\sqrt{x^{2}-x+\frac{17}{4}} \right)\left( \sqrt{x^{2}-x+\frac{17}{4}}-\left( x-\frac{1}{2} \right) \right) \right]=\log_{2}4=2$
$\Rightarrow T=f\left( \frac{1}{2019} \right)+f\left( \frac{2}{2019} \right)+...+f\left( \frac{2018}{2019} \right)$
$=f\left( \frac{1}{2019} \right)+f\left( \frac{2018}{2019} \right)+f\left( \frac{2}{2019} \right)+f\left( \frac{2017}{2019} \right)+...+f\left( \frac{1009}{2019} \right)+f\left( \frac{1010}{2019} \right)$
$=1009.2=2018$
$f\left( x \right)+f\left( 1-x \right)=\log_{2}\left( x-\frac{1}{2}+\sqrt{x^{2}-x+\frac{17}{4}} \right)+\log_{2}\left( \sqrt{x^{2}-x+\frac{17}{4}}-\left( x-\frac{1}{2} \right) \right)$
$=\log_{2}\left[ \left( x-\frac{1}{2}+\sqrt{x^{2}-x+\frac{17}{4}} \right)\left( \sqrt{x^{2}-x+\frac{17}{4}}-\left( x-\frac{1}{2} \right) \right) \right]=\log_{2}4=2$
$\Rightarrow T=f\left( \frac{1}{2019} \right)+f\left( \frac{2}{2019} \right)+...+f\left( \frac{2018}{2019} \right)$
$=f\left( \frac{1}{2019} \right)+f\left( \frac{2018}{2019} \right)+f\left( \frac{2}{2019} \right)+f\left( \frac{2017}{2019} \right)+...+f\left( \frac{1009}{2019} \right)+f\left( \frac{1010}{2019} \right)$
$=1009.2=2018$