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Câu hỏi: Gọi $M$ và $m$ tương ứng là giá trị lớn nhất, giá trị nhỏ nhất của hàm số $y={{2}^{\left| \sin x \right|}}+{{2}^{\left| \cos x \right|}}$. Tính tổng $T=1010M+2021m$.
A. $T=1010\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+6063$.
B. $T=2020\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+2021$.
C. $T=1010\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+2021$.
D. $T=2020\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+6063$.
A. $T=1010\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+6063$.
B. $T=2020\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+2021$.
C. $T=1010\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+2021$.
D. $T=2020\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+6063$.
Đặt $t=\left| \sin x \right|,t\in \left[ 0;1 \right]$ suy ra $\sqrt{1-{{t}^{2}}}=\left| \cos x \right|$.
Khi đó $y=f\left( t \right)={{2}^{t}}+{{2}^{\sqrt{1-{{t}^{2}}}}}$, với $t\in \left[ 0;1 \right]$.
Ta có ${f}'\left( t \right)={{2}^{t}}\cdot \ln 2-{{2}^{\sqrt{1-{{t}^{2}}}}}\cdot \ln 2\cdot \dfrac{t}{\sqrt{1-{{t}^{2}}}}=0$
$\Leftrightarrow {{2}^{t}}\cdot \ln 2={{2}^{\sqrt{1-{{t}^{2}}}}}\cdot \ln 2\cdot \dfrac{t}{\sqrt{1-{{t}^{2}}}}\Leftrightarrow \dfrac{{{2}^{t}}}{t}=\dfrac{{{2}^{\sqrt{1-{{t}^{2}}}}}}{\sqrt{1-{{t}^{2}}}} \left( * \right)$.
Đặt $g\left( u \right)=\dfrac{{{2}^{u}}}{u}$ với $u\in \left( 0;1 \right)$ ; ${g}'\left( u \right)=\dfrac{{{2}^{u}}\cdot \ln 2-{{2}^{u}}}{{{u}^{2}}}>0,\forall u\in \left( 0;1 \right)$.
Do đó $g$ đồng biến trên $\left( 0;1 \right)$.
Nên $\left( * \right)\Leftrightarrow t=\sqrt{1-{{t}^{2}}}\Leftrightarrow {{t}^{2}}=1-{{t}^{2}}\Leftrightarrow {{t}^{2}}=\dfrac{1}{2}\Rightarrow t=\dfrac{\sqrt{2}}{2}$.
Ta có $f\left( 0 \right)=3$, $f\left( \dfrac{\sqrt{2}}{2} \right)={{2}^{\dfrac{\sqrt{2}}{2}}}+{{2}^{\dfrac{\sqrt{2}}{2}}}=2\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}$, $f\left( 1 \right)=3$.
Do đó $M=\max y=\underset{\left[ 0;1 \right]}{\mathop{\max }} f\left( t \right)=3$, $m=\min y=\underset{\left[ 0;1 \right]}{\mathop{\min }} f\left( t \right)=2\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}$.
Vậy $T=1010M+2021m=2020\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+6063$.
Khi đó $y=f\left( t \right)={{2}^{t}}+{{2}^{\sqrt{1-{{t}^{2}}}}}$, với $t\in \left[ 0;1 \right]$.
Ta có ${f}'\left( t \right)={{2}^{t}}\cdot \ln 2-{{2}^{\sqrt{1-{{t}^{2}}}}}\cdot \ln 2\cdot \dfrac{t}{\sqrt{1-{{t}^{2}}}}=0$
$\Leftrightarrow {{2}^{t}}\cdot \ln 2={{2}^{\sqrt{1-{{t}^{2}}}}}\cdot \ln 2\cdot \dfrac{t}{\sqrt{1-{{t}^{2}}}}\Leftrightarrow \dfrac{{{2}^{t}}}{t}=\dfrac{{{2}^{\sqrt{1-{{t}^{2}}}}}}{\sqrt{1-{{t}^{2}}}} \left( * \right)$.
Đặt $g\left( u \right)=\dfrac{{{2}^{u}}}{u}$ với $u\in \left( 0;1 \right)$ ; ${g}'\left( u \right)=\dfrac{{{2}^{u}}\cdot \ln 2-{{2}^{u}}}{{{u}^{2}}}>0,\forall u\in \left( 0;1 \right)$.
Do đó $g$ đồng biến trên $\left( 0;1 \right)$.
Nên $\left( * \right)\Leftrightarrow t=\sqrt{1-{{t}^{2}}}\Leftrightarrow {{t}^{2}}=1-{{t}^{2}}\Leftrightarrow {{t}^{2}}=\dfrac{1}{2}\Rightarrow t=\dfrac{\sqrt{2}}{2}$.
Ta có $f\left( 0 \right)=3$, $f\left( \dfrac{\sqrt{2}}{2} \right)={{2}^{\dfrac{\sqrt{2}}{2}}}+{{2}^{\dfrac{\sqrt{2}}{2}}}=2\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}$, $f\left( 1 \right)=3$.
Do đó $M=\max y=\underset{\left[ 0;1 \right]}{\mathop{\max }} f\left( t \right)=3$, $m=\min y=\underset{\left[ 0;1 \right]}{\mathop{\min }} f\left( t \right)=2\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}$.
Vậy $T=1010M+2021m=2020\cdot {{2}^{\dfrac{\sqrt{2}}{2}}}+6063$.
Đáp án D.