Cho hàm số \(y = f\left( x \right)\) liên tục trên khoảng \(\left( {0; + \infty } \right)\) và thỏa mãn \(f\left( {{x^2} + 1} \right) + \frac{{f\left( {\sqrt x } \right)}}{{4x\sqrt x }} = \frac{{2x + 1}}{{2x}}\ln \left( {x + 1} \right)\). Biết \(\int\limits_1^{17} {f\left( x \right){\rm{d}}x = a\ln 5 – 2\ln b + c} \) với \(a,{\rm{ }}b,{\rm{ }}c \in \mathbb{R}\). Tính \(T = a – 3b + 2c\).

Cho hàm số \(y = f\left( x \right)\) liên tục trên khoảng \(\left( {0; + \infty } \right)\) và thỏa mãn \(f\left( {{x^2} + 1} \right) + \frac{{f\left( {\sqrt x } \right)}}{{4x\sqrt x }} = \frac{{2x + 1}}{{2x}}\ln \left( {x + 1} \right)\). Biết \(\int\limits_1^{17} {f\left( x \right){\rm{d}}x = a\ln 5 – 2\ln b + c} \) với \(a,{\rm{ }}b,{\rm{ }}c \in \mathbb{R}\). Tính \(T = a – 3b + 2c\).

A. \(T = – 1\).

B. \(T = 2\).

C. \(T = 3\).

D. \(T = 0\).

Lời giải:

Ta có \(f\left( {{x^2} + 1} \right) + \frac{{f\left( {\sqrt x } \right)}}{{4x\sqrt x }} = \frac{{2x + 1}}{{2x}}\ln \left( {x + 1} \right) \Leftrightarrow xf\left( {{x^2} + 1} \right) + \frac{{f\left( {\sqrt x } \right)}}{{4\sqrt x }} = \frac{{2x + 1}}{2}\ln \left( {x + 1} \right)\).

Suy ra \(\int\limits_1^4 {\left[ {xf\left( {{x^2} + 1} \right) + \frac{{f\left( {\sqrt x } \right)}}{{4\sqrt x }}} \right]{\rm{d}}x = \int\limits_1^4 {\frac{{2x + 1}}{2}\ln \left( {x + 1} \right){\rm{d}}x} } \).

Mà \(\int\limits_1^4 {\left[ {xf\left( {{x^2} + 1} \right) + \frac{{f\left( {\sqrt x } \right)}}{{4\sqrt x }}} \right]} {\rm{d}}x = \int\limits_1^4 {f\left( {{x^2} + 1} \right)} \frac{{{\rm{d}}\left( {{x^2} + 1} \right)}}{2} + \int\limits_1^4 {f\left( {\sqrt x } \right)\frac{{{\rm{d}}\left( {\sqrt x } \right)}}{2}} \)\( = \int\limits_2^{17} {\frac{1}{2}f\left( x \right)} {\rm{d}}x + \int\limits_1^2 {\frac{1}{2}f\left( x \right)} {\rm{d}}x = \frac{1}{2}\int\limits_1^{17} {f\left( x \right)} {\rm{d}}x\).

Nên \(\int\limits_1^{17} {f\left( x \right){\rm{d}}x = 2\int\limits_1^4 {\frac{{2x + 1}}{2}\ln \left( {x + 1} \right){\rm{d}}x} } \)

Ta có \(\int\limits_1^4 {\frac{{2x + 1}}{2}\ln \left( {x + 1} \right){\rm{d}}x} = \frac{1}{2}\int\limits_1^4 {\ln \left( {x + 1} \right){\rm{d}}\left( {{x^2} + x} \right)} = \frac{1}{2}\left[ {\left. {\left( {{x^2} + x} \right)\ln \left( {x + 1} \right)} \right|_1^4 – \int\limits_1^4 {\left( {{x^2} + x} \right)\frac{1}{{x + 1}}{\rm{d}}x} } \right]\)

\( = \frac{1}{2}\left[ {20\ln 5 – 2\ln 2 – \left. {\frac{{{x^2}}}{2}} \right|_1^4} \right] = \frac{1}{2}\left[ {20\ln 5 – 2\ln 2 – \frac{{15}}{2}} \right]\)

Do đó \(\int\limits_1^{17} {f\left( x \right)} {\rm{d}}x = 20\ln 5 – 2\ln 2 – \frac{{15}}{2} \Rightarrow a = 20,b = 2,c = – \frac{{15}}{2}\).

Vậy \(T = a – 3b + 2c = 20 – 3.2 + 2.\left( { – \frac{{15}}{2}} \right) = – 1\).

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Đây là các câu ÔN THI TN THPT MÔN TOÁN 2023 – CHUYÊN ĐỀ NGUYÊN HAM – TICH PHÂN – ỨNG DỤNG.