Google
Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{1}{1+\sqrt{{{\pi }^{1-2x}}}}$ Tính giá trị của biểu thức sau:
$Q=f\left( {{\sin }^{2}}\dfrac{\pi }{2020} \right)+f\left( {{\sin }^{2}}\dfrac{2\pi }{2020} \right)+...+f\left( {{\sin }^{2}}\dfrac{1009\pi }{2020} \right)$
A. 1009
B. 504
C. $\dfrac{1009}{2}$
D. 505
$Q=f\left( {{\sin }^{2}}\dfrac{\pi }{2020} \right)+f\left( {{\sin }^{2}}\dfrac{2\pi }{2020} \right)+...+f\left( {{\sin }^{2}}\dfrac{1009\pi }{2020} \right)$
A. 1009
B. 504
C. $\dfrac{1009}{2}$
D. 505
Biến đổi: $f\left( x \right)=\dfrac{1}{1+\sqrt{{{\pi }^{1-2x}}}}=\dfrac{1}{1+\dfrac{\sqrt{\pi }}{{{\pi }^{x}}}}=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}$
Ta thấy
$f\left( x \right)+f\left( 1-x \right)=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}+\dfrac{{{\pi }^{1-x}}}{{{\pi }^{1-x}}+\sqrt{\pi }}=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}+\dfrac{\dfrac{\pi }{{{\pi }^{x}}}}{\dfrac{\pi }{{{\pi }^{x}}}+\sqrt{\pi }}=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}+\dfrac{\sqrt{\pi }}{\sqrt{\pi }+{{\pi }^{x}}}=1, \forall x\in R$
Vậy
$Q=\left[ f\left( {{\sin }^{2}}\dfrac{\pi }{2020} \right)+f\left( {{\sin }^{2}}\dfrac{1009\pi }{2020} \right) \right]+...+\left[ f\left( {{\sin }^{2}}\dfrac{504\pi }{2020} \right)+f\left( {{\sin }^{2}}\dfrac{506\pi }{2020} \right) \right]+f\left( {{\sin }^{2}}\dfrac{505\pi }{2020} \right)$
$=504+f\left( {{\sin }^{2}}\dfrac{\pi }{4} \right)=\dfrac{1009}{2}$
Ta thấy
$f\left( x \right)+f\left( 1-x \right)=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}+\dfrac{{{\pi }^{1-x}}}{{{\pi }^{1-x}}+\sqrt{\pi }}=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}+\dfrac{\dfrac{\pi }{{{\pi }^{x}}}}{\dfrac{\pi }{{{\pi }^{x}}}+\sqrt{\pi }}=\dfrac{{{\pi }^{x}}}{{{\pi }^{x}}+\sqrt{\pi }}+\dfrac{\sqrt{\pi }}{\sqrt{\pi }+{{\pi }^{x}}}=1, \forall x\in R$
Vậy
$Q=\left[ f\left( {{\sin }^{2}}\dfrac{\pi }{2020} \right)+f\left( {{\sin }^{2}}\dfrac{1009\pi }{2020} \right) \right]+...+\left[ f\left( {{\sin }^{2}}\dfrac{504\pi }{2020} \right)+f\left( {{\sin }^{2}}\dfrac{506\pi }{2020} \right) \right]+f\left( {{\sin }^{2}}\dfrac{505\pi }{2020} \right)$
$=504+f\left( {{\sin }^{2}}\dfrac{\pi }{4} \right)=\dfrac{1009}{2}$
Đáp án C.