Cho hàm số $\Large f(x)=\left(a^{2}+1\right) \ln ^{2019}\left(x+\sqrt{

Cho hàm số $\Large f(x)=\left(a^{2}+1\right) \ln ^{2019}\left(x+\sqrt{x^{2}+1}\right)+b x \sin ^{2020} x+3$ với a, b là các số thực và $\Large f\left(2^{\log 3}\right)=9$. Tính $\Large f\left(-3^{\log 2}\right)$

• A. $\Large f\left(-3^{\log 2}\right)=3$
• B. $\Large f\left(-3^{\log 2}\right)=-3$
• C. $\Large f\left(-3^{\log 2}\right)=2$
• D. $\Large f\left(-3^{\log 2}\right)=-2$

Chọn B

Ta có $\Large f(-x)=\left(a^{2}+1\right) \ln ^{2019}\left(-x+\sqrt{(-x)^{2}+1}\right)-b x \sin ^{2020}(-x)+3$

$\Large \Rightarrow f(-x)=\left(a^{2}+1\right) \ln ^{2019}\left(-x+\sqrt{x^{2}+1}\right)-b x \sin ^{2020} x+3$

Do đó

$\Large f(x)+f(-x)$$\Large =\left(a^{2}+1\right) \ln ^{2019}\left(x+\sqrt{x^{2}+1}\right)+b x \sin ^{2020} x+3$

$\Large +\left(a^{2}+1\right) \ln ^{2019}\left(-x+\sqrt{x^{2}+1}\right)-b x \sin ^{2020} x+3$

$\Large \Leftrightarrow f(x)+f(-x)=\left(a^{2}+1\right) \ln ^{2019}\left(x+\sqrt{x^{2}+1}\right)+\left(a^{2}+1\right) \ln ^{2019} \frac{1}{x+\sqrt{x^{2}+1}}+6$

$\Large \Leftrightarrow f(x)+f(-x)=\left(a^{2}+1\right) \ln ^{2019}\left(x+\sqrt{x^{2}+1}\right)-\left(a^{2}+1\right) \ln ^{2019}\left(x+\sqrt{x^{2}+1}\right)+6$

$\Large \Leftrightarrow f(x)+f(-x)=6$

Suy ra

$\Large f\left(2^{\log 3}\right)+f\left(-2^{\log 3}\right)=6$$\Large \Leftrightarrow f\left(2^{\log 3}\right)+f\left(-3^{\log 2}\right)=6 \Leftrightarrow f\left(-3^{\log 2}\right)=6-f\left(2^{\log 3}\right)$

$\Large \Leftrightarrow f\left(-3^{\log 2}\right)=6-9 \Leftrightarrow f\left(-3^{\log 2}\right)=-3$