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Câu hỏi: Giá trị lớn nhất và giá trị nhỏ nhất của hàm số $f\left( x \right)=x+{{\cos }^{2}}x$ trên đoạn $\left[ 0;\dfrac{\pi }{4} \right]$ là
A. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=-1$.
B. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=\dfrac{\pi }{6}$.
C. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}+\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=1$.
D. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}+\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=\dfrac{1}{2}$.
A. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=-1$.
B. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=\dfrac{\pi }{6}$.
C. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}+\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=1$.
D. $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}+\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=\dfrac{1}{2}$.
Ta có $f\left( x \right)=x+{{\cos }^{2}}x$ $\Rightarrow {f}'\left( x \right)=1-2\cos x. \sin x=1-\sin 2x$.
Cho ${f}'\left( x \right)=0\Leftrightarrow 1-\sin 2x=0\Leftrightarrow x=\dfrac{\pi }{4}+k\pi $. Do $x\in \left[ 0;\dfrac{\pi }{4} \right]\Rightarrow x=\dfrac{\pi }{4}$.
$f\left( 0 \right)=1; f\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{4}+\dfrac{1}{2}$ và hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;\dfrac{\pi }{4} \right]$
Suy ra $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}+\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=1$.
Cho ${f}'\left( x \right)=0\Leftrightarrow 1-\sin 2x=0\Leftrightarrow x=\dfrac{\pi }{4}+k\pi $. Do $x\in \left[ 0;\dfrac{\pi }{4} \right]\Rightarrow x=\dfrac{\pi }{4}$.
$f\left( 0 \right)=1; f\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{4}+\dfrac{1}{2}$ và hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;\dfrac{\pi }{4} \right]$
Suy ra $\underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\max }} f\left( x \right)=\dfrac{\pi }{4}+\dfrac{1}{2}; \underset{\left[ 0;\dfrac{\pi }{4} \right]}{\mathop{\min }} f\left( x \right)=1$.
Đáp án C.