Google
Câu hỏi: Giá trị lớn nhất của hàm số $y={{\text{e}}^{x}}.\cos x$ trên $\left[ 0;\dfrac{\pi }{2} \right]$ là
A. $1$.
B. $\dfrac{1}{2}.{{\text{e}}^{\dfrac{\pi }{3}}}$.
C. $\dfrac{\sqrt{3}}{2}.{{\text{e}}^{\dfrac{\pi }{6}}}$.
D. $\dfrac{\sqrt{2}}{2}.{{\text{e}}^{\dfrac{\pi }{4}}}$.
A. $1$.
B. $\dfrac{1}{2}.{{\text{e}}^{\dfrac{\pi }{3}}}$.
C. $\dfrac{\sqrt{3}}{2}.{{\text{e}}^{\dfrac{\pi }{6}}}$.
D. $\dfrac{\sqrt{2}}{2}.{{\text{e}}^{\dfrac{\pi }{4}}}$.
Ta có $y={{\text{e}}^{x}}.\cos x\Rightarrow {y}'={{\text{e}}^{x}}.\cos x-{{\text{e}}^{x}}\sin x={{\text{e}}^{x}}\left( \cos x-\sin x \right)$.
${y}'=0\Rightarrow \cos x-\sin x=0\Leftrightarrow \sin \left( x-\dfrac{\pi }{4} \right)=0\Leftrightarrow x-\dfrac{\pi }{4}=k\pi \Leftrightarrow x=\dfrac{\pi }{4}+k\pi , k\in \mathbb{Z}$.
Trên $\left[ 0;\dfrac{\pi }{2} \right]$, ta được $x=\dfrac{\pi }{4}$.
Khi đó $y\left( 0 \right)=1; y\left( \dfrac{\pi }{2} \right)=0; y\left( \dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}.{{\text{e}}^{\dfrac{\pi }{4}}}$. Vậy $\underset{\left[ 0; \dfrac{\pi }{2} \right]}{\mathop{\max }} y=\dfrac{\sqrt{2}}{2}.{{\text{e}}^{\dfrac{\pi }{4}}}$.
${y}'=0\Rightarrow \cos x-\sin x=0\Leftrightarrow \sin \left( x-\dfrac{\pi }{4} \right)=0\Leftrightarrow x-\dfrac{\pi }{4}=k\pi \Leftrightarrow x=\dfrac{\pi }{4}+k\pi , k\in \mathbb{Z}$.
Trên $\left[ 0;\dfrac{\pi }{2} \right]$, ta được $x=\dfrac{\pi }{4}$.
Khi đó $y\left( 0 \right)=1; y\left( \dfrac{\pi }{2} \right)=0; y\left( \dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}.{{\text{e}}^{\dfrac{\pi }{4}}}$. Vậy $\underset{\left[ 0; \dfrac{\pi }{2} \right]}{\mathop{\max }} y=\dfrac{\sqrt{2}}{2}.{{\text{e}}^{\dfrac{\pi }{4}}}$.
Đáp án D.