Google
Câu hỏi: Cho hàm số $y=f(x)$ có đạo hàm không âm trên $\left[ 0;1 \right],$ thỏa mãn $f(x)>0$ với mọi $x\in \left[ 0;1 \right]$ và ${{\left[ f(x) \right]}^{2}}.{{\left[ f'(x) \right]}^{2}}{{\left( {{x}^{2}}+1 \right)}^{2}}=1+{{\left[ f(x) \right]}^{2}}$. Nếu $f(0)=\sqrt{3}$ thì giá trị $f(1)$ thuộc khoảng nào sau đây?
A. $\left( 3;\dfrac{7}{2} \right)$.
B. $\left( 2;\dfrac{5}{2} \right)$.
C. $\left( \dfrac{5}{2};3 \right)$.
D. $\left( \dfrac{3}{2};2 \right)$.
A. $\left( 3;\dfrac{7}{2} \right)$.
B. $\left( 2;\dfrac{5}{2} \right)$.
C. $\left( \dfrac{5}{2};3 \right)$.
D. $\left( \dfrac{3}{2};2 \right)$.
Ta có: ${{\left[ f(x) \right]}^{2}}.{{\left[ f'(x) \right]}^{2}}{{\left( {{x}^{2}}+1 \right)}^{2}}=1+{{\left[ f(x) \right]}^{2}}$ $\Leftrightarrow \dfrac{{{\left[ f(x) \right]}^{2}}.{{\left[ f'(x) \right]}^{2}}}{1+{{\left[ f(x) \right]}^{2}}}=\dfrac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$
$\Leftrightarrow \dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}=\dfrac{1}{{{x}^{2}}+1}$ $\Rightarrow \int\limits_{0}^{1}{\dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}dx}=\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+1}dx}$
$\Rightarrow \int\limits_{0}^{1}{\dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}dx}=\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+1}dx}$
+ Nếu đặt $t=\sqrt{1+{{\left[ f(x) \right]}^{2}}}\Rightarrow dt=\dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}dx$ $\Rightarrow $ VT = $\int\limits_{2}^{\sqrt{1+{{f}^{2}}\left( 1 \right)}}{dt}=\sqrt{1+{{f}^{2}}\left( 1 \right)}-2$
+ Nếu đặt $x=\tan u$ $\Rightarrow dx=\left( 1+{{\tan }^{2}}u \right)du$ $\Rightarrow $ VP = $\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{1}{1+{{\tan }^{2}}u}\left( 1+{{\tan }^{2}}u \right)dx}=\dfrac{\pi }{4}$
$\Rightarrow \sqrt{1+{{f}^{2}}\left( 1 \right)}-2$ $=\dfrac{\pi }{4}$ $\Rightarrow f\left( 1 \right)=\sqrt{\dfrac{{{\pi }^{2}}}{16}+\pi +3}\approx 2,6$ $\in \left( \dfrac{5}{2};3 \right)$.
$\Leftrightarrow \dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}=\dfrac{1}{{{x}^{2}}+1}$ $\Rightarrow \int\limits_{0}^{1}{\dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}dx}=\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+1}dx}$
$\Rightarrow \int\limits_{0}^{1}{\dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}dx}=\int\limits_{0}^{1}{\dfrac{1}{{{x}^{2}}+1}dx}$
+ Nếu đặt $t=\sqrt{1+{{\left[ f(x) \right]}^{2}}}\Rightarrow dt=\dfrac{f(x).f'(x)}{\sqrt{1+{{\left[ f(x) \right]}^{2}}}}dx$ $\Rightarrow $ VT = $\int\limits_{2}^{\sqrt{1+{{f}^{2}}\left( 1 \right)}}{dt}=\sqrt{1+{{f}^{2}}\left( 1 \right)}-2$
+ Nếu đặt $x=\tan u$ $\Rightarrow dx=\left( 1+{{\tan }^{2}}u \right)du$ $\Rightarrow $ VP = $\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{1}{1+{{\tan }^{2}}u}\left( 1+{{\tan }^{2}}u \right)dx}=\dfrac{\pi }{4}$
$\Rightarrow \sqrt{1+{{f}^{2}}\left( 1 \right)}-2$ $=\dfrac{\pi }{4}$ $\Rightarrow f\left( 1 \right)=\sqrt{\dfrac{{{\pi }^{2}}}{16}+\pi +3}\approx 2,6$ $\in \left( \dfrac{5}{2};3 \right)$.
Đáp án C.