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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ thỏa mãn ${{f}^{3}}\left( x \right)+3f\left( x \right)=\sin \left( 2{{x}^{3}}-3{{x}^{2}}+x \right),\forall x\in \mathbb{R}$. Tích phân $I=\int\limits_{0}^{1}{f\left( x \right)dx}$ thuộc khoảng nào?
A. $\left( -3;-2 \right)$
B. $\left( -2;-1 \right)$
C. $\left( -1;1 \right)$
D. $\left( 1;2 \right)$
A. $\left( -3;-2 \right)$
B. $\left( -2;-1 \right)$
C. $\left( -1;1 \right)$
D. $\left( 1;2 \right)$
Phương pháp:
- Từ giả thiết ${{f}^{3}}\left( x \right)+3f\left( x \right)=\sin \left( 2{{x}^{3}}-3{{x}^{2}}+x \right)$ thay $x$ bởi $1-x$ và chứng minh $f\left( x \right)=-f\left( 1-x \right).$
- Chứng minh $\int\limits_{0}^{1}{f\left( 1-x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx},$ từ đó tính $I.$
Cách giải:
Theo bài ra ta có:
${{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=\sin \left[ 2{{\left( 1-x \right)}^{3}}-3{{\left( 1-x \right)}^{2}}+1-x \right]$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=\sin \left( 2-6x+6{{x}^{2}}-2{{x}^{3}}-3+6x-3{{x}^{2}}+1-x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=\sin \left( -2{{x}^{3}}+3{{x}^{2}}-x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=-\sin \left( 2{{x}^{3}}-3{{x}^{2}}+x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=-{{f}^{3}}\left( x \right)-3f\left( x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+{{f}^{3}}\left( x \right)+3f\left( 1-x \right)+3f\left( x \right)=0$
$\Leftrightarrow \left[ f\left( 1-x \right)+f\left( x \right) \right]\left[ {{f}^{2}}\left( 1-x \right)-f\left( 1-x \right)f\left( x \right)+{{f}^{2}}\left( x \right) \right]+3\left[ f\left( 1-x \right)+f\left( x \right) \right]=0$
$\Leftrightarrow \left[ f\left( 1-x \right)+f\left( x \right) \right]\left[ {{f}^{2}}\left( 1-x \right)-f\left( 1-x \right)f\left( x \right)+{{f}^{2}}\left( x \right)+3 \right]=0$
$\Leftrightarrow f\left( x \right)=-f\left( 1-x \right)$
Suy ra $\int\limits_{0}^{1}{f\left( x \right)dx}=-\int\limits_{0}^{1}{f\left( 1-x \right)dx}$.
Ta lại có $\int\limits_{0}^{1}{f\left( 1-x \right)dx}=-\int\limits_{0}^{1}{f\left( 1-x \right)d\left( 1-x \right)}=-\int\limits_{1}^{0}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx}.$
Do đó $I=\int\limits_{0}^{1}{f\left( x \right)dx}=-\int\limits_{0}^{1}{f\left( x \right)dx}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=0.$
Vậy $I\left( -1;1 \right).$
- Từ giả thiết ${{f}^{3}}\left( x \right)+3f\left( x \right)=\sin \left( 2{{x}^{3}}-3{{x}^{2}}+x \right)$ thay $x$ bởi $1-x$ và chứng minh $f\left( x \right)=-f\left( 1-x \right).$
- Chứng minh $\int\limits_{0}^{1}{f\left( 1-x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx},$ từ đó tính $I.$
Cách giải:
Theo bài ra ta có:
${{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=\sin \left[ 2{{\left( 1-x \right)}^{3}}-3{{\left( 1-x \right)}^{2}}+1-x \right]$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=\sin \left( 2-6x+6{{x}^{2}}-2{{x}^{3}}-3+6x-3{{x}^{2}}+1-x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=\sin \left( -2{{x}^{3}}+3{{x}^{2}}-x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=-\sin \left( 2{{x}^{3}}-3{{x}^{2}}+x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+3f\left( 1-x \right)=-{{f}^{3}}\left( x \right)-3f\left( x \right)$
$\Leftrightarrow {{f}^{3}}\left( 1-x \right)+{{f}^{3}}\left( x \right)+3f\left( 1-x \right)+3f\left( x \right)=0$
$\Leftrightarrow \left[ f\left( 1-x \right)+f\left( x \right) \right]\left[ {{f}^{2}}\left( 1-x \right)-f\left( 1-x \right)f\left( x \right)+{{f}^{2}}\left( x \right) \right]+3\left[ f\left( 1-x \right)+f\left( x \right) \right]=0$
$\Leftrightarrow \left[ f\left( 1-x \right)+f\left( x \right) \right]\left[ {{f}^{2}}\left( 1-x \right)-f\left( 1-x \right)f\left( x \right)+{{f}^{2}}\left( x \right)+3 \right]=0$
$\Leftrightarrow f\left( x \right)=-f\left( 1-x \right)$
Suy ra $\int\limits_{0}^{1}{f\left( x \right)dx}=-\int\limits_{0}^{1}{f\left( 1-x \right)dx}$.
Ta lại có $\int\limits_{0}^{1}{f\left( 1-x \right)dx}=-\int\limits_{0}^{1}{f\left( 1-x \right)d\left( 1-x \right)}=-\int\limits_{1}^{0}{f\left( x \right)dx}=\int\limits_{0}^{1}{f\left( x \right)dx}.$
Do đó $I=\int\limits_{0}^{1}{f\left( x \right)dx}=-\int\limits_{0}^{1}{f\left( x \right)dx}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=0.$
Vậy $I\left( -1;1 \right).$
Đáp án C.