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Câu hỏi: Cho hàm số $f\left( x \right)=\left( {{a}^{2}}+1 \right){{\ln }^{2017}}\left( x+\sqrt{1+{{x}^{2}}} \right)+bx{{\sin }^{2018}}+2$ với a, b là các số thực và $f\left( {{7}^{\log 5}} \right)=6$. Tính $f\left( -{{5}^{\log 7}} \right)$.
A. $f\left( -{{5}^{\log 7}} \right)=2$
B. $f\left( -{{5}^{\log 7}} \right)=4$
C. $f\left( -{{5}^{\log 7}} \right)=-2$
D. $f\left( -{{5}^{\log 7}} \right)=6$
A. $f\left( -{{5}^{\log 7}} \right)=2$
B. $f\left( -{{5}^{\log 7}} \right)=4$
C. $f\left( -{{5}^{\log 7}} \right)=-2$
D. $f\left( -{{5}^{\log 7}} \right)=6$
Ta có: $f\left( x \right)=\left( {{a}^{2}}+1 \right){{\ln }^{2017}}\left( x+\sqrt{1+{{x}^{2}}} \right)+bx{{\sin }^{2018}}x+2$
Và $\begin{aligned}
& f\left( -x \right)=\left( {{a}^{2}}+1 \right){{\ln }^{2017}}\left( -x+\sqrt{1+{{x}^{2}}} \right)-bx{{\sin }^{2018}}\left( -x \right)+2 \\
& \ \ \ \ \ \ \ \ \ =\left( {{a}^{2}}+1 \right){{\ln }^{2017}}{{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{-1}}-bx{{\sin }^{2018}}x+2 \\
& \ \ \ \ \ \ \ \ \ =-\left[ \left( {{a}^{2}}+1 \right){{\ln }^{2017}}\left( x+\sqrt{1+{{x}^{2}}} \right)+bx{{\sin }^{2018}}x+2 \right]+4=-f\left( x \right)+4 \\
\end{aligned}$
Vậy $f\left( -{{5}^{\log 7}} \right)=f\left( -{{7}^{\log 5}} \right)=-f\left( {{7}^{\log 5}} \right)+4=-6+4=-2$.
Và $\begin{aligned}
& f\left( -x \right)=\left( {{a}^{2}}+1 \right){{\ln }^{2017}}\left( -x+\sqrt{1+{{x}^{2}}} \right)-bx{{\sin }^{2018}}\left( -x \right)+2 \\
& \ \ \ \ \ \ \ \ \ =\left( {{a}^{2}}+1 \right){{\ln }^{2017}}{{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{-1}}-bx{{\sin }^{2018}}x+2 \\
& \ \ \ \ \ \ \ \ \ =-\left[ \left( {{a}^{2}}+1 \right){{\ln }^{2017}}\left( x+\sqrt{1+{{x}^{2}}} \right)+bx{{\sin }^{2018}}x+2 \right]+4=-f\left( x \right)+4 \\
\end{aligned}$
Vậy $f\left( -{{5}^{\log 7}} \right)=f\left( -{{7}^{\log 5}} \right)=-f\left( {{7}^{\log 5}} \right)+4=-6+4=-2$.
Đáp án C.