Cho hàm số \(y = f\left( x \right)\) có đạo hàm liên tục trên đoạn \(\left[ {0;1} \right]\) và \(f\left( 0 \right) + f\left( 1 \right) = 0\). Biết \(\int\limits_0^1 {{f^2}\left( x \right){\rm{d}}x} = \dfrac{1}{2},{\rm{ }}\int\limits_0^1 {f'\left( x \right){\rm{cos}}\left( {\pi x} \right){\rm{d}}x} = \dfrac{\pi }{2}\). Tính \(\int\limits_0^1 {f\left( x \right){\rm{d}}x} \).

A. \(\pi \).

B. \(\dfrac{{3\pi }}{2}\).

C. \(\dfrac{2}{\pi }\).

D. \(\dfrac{1}{\pi }\).

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Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

Ta có :

\(\begin{array}{l}{\rm{ }}\int\limits_0^1 {{\rm{cos}}\left( {\pi x} \right){\rm{d}}\left( {f\left( x \right)} \right)} = \left. {\left( {{\rm{cos}}\left( {\pi x} \right).f\left( x \right)} \right)} \right|_0^1 - \int\limits_0^1 {f\left( x \right){\rm{d}}\left( {{\rm{cos}}\left( {\pi x} \right)} \right)} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left. {\left( {{\rm{cos}}\left( {\pi x} \right).f\left( x \right)} \right)} \right|_0^1 + \pi \int\limits_0^1 {f\left( x \right){\rm{.sin}}\left( {\pi x} \right){\rm{d}}x} \\ \Rightarrow - f\left( 1 \right) - f\left( 0 \right) + \pi \int\limits_0^1 {f\left( x \right){\rm{.sin}}\left( {\pi x} \right){\rm{d}}x} = \dfrac{\pi }{2}\\ \Leftrightarrow - 0 + \pi \int\limits_0^1 {f\left( x \right){\rm{.sin}}\left( {\pi x} \right){\rm{d}}x} = \dfrac{\pi }{2} \Leftrightarrow \int\limits_0^1 {f\left( x \right){\rm{.sin}}\left( {\pi x} \right){\rm{d}}x} = \dfrac{1}{2}\\ \Rightarrow \int\limits_0^1 {{f^2}\left( x \right){\rm{d}}x} - \int\limits_0^1 {f\left( x \right){\rm{.sin}}\left( {\pi x} \right){\rm{d}}x} = 0 \Leftrightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) - f\left( x \right){\rm{.sin}}\left( {\pi x} \right)} \right]{\rm{d}}x} = 0\\ \Rightarrow {f^2}\left( x \right) - f\left( x \right){\rm{.sin}}\left( {\pi x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}f\left( x \right) = 0\\f\left( x \right) = \sin \left( {\pi x} \right)\end{array} \right.\end{array}\)

+) \(f\left( x \right) = 0\) mâu thuẫn với \(\int\limits_0^1 {{f^2}\left( x \right){\rm{d}}x} = \dfrac{1}{2}\)

+) \(f\left( x \right) = \sin \left( {\pi x} \right)\)\( \Rightarrow \int\limits_0^1 {f\left( x \right){\rm{d}}x} = \int\limits_0^1 {\sin \left( {\pi x} \right){\rm{d}}x} = \left. {\dfrac{{ - \cos \left( {\pi x} \right)}}{\pi }} \right|_0^1 = \dfrac{{1 + 1}}{\pi } = \dfrac{2}{\pi }\).

Chọn: C

Câu hỏi này thuộc đề thi trắc nghiệm dưới đây, bấm vào Bắt đầu thi để làm toàn bài