hàm số y=f(x) liên tục trên R thỏa mãn tích phân cotx.f(sin^2x)dx=tích phân f(căn x)/x dx = 1.

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hàm số \({\rm{y}} = {\rm{f}}\left( {\rm{x}} \right)\) liên tục trên \(\mathbb{R}\) thỏa mãn \[\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\cot x \cdot f\left( {{{\sin }^2}x} \right)dx = \int\limits_1^{16} {\frac{{{\rm{f}}\left( {\sqrt {\rm{x}} } \right)}}{{\rm{x}}}{\rm{dx}}} = 1} \]. Tích phân \({\rm{I}} = \int\limits_{\frac{1}{8}}^1 {\frac{{{\rm{f}}\left( {4{\rm{x}}} \right)}}{{\rm{x}}}{\rm{dx}}} \) bằng

\(\frac{5}{2}\).

\(\frac{{21}}{4}\).

Giải thích

f2x−m+4fx+2 m+4=0⇔f2x2−mfx−4fx+2 m+4=0

⇔fx−22−mfx−2=0⇔fx=2 (1)fx=m+2 (2)

Ta có \(\left\{ \begin{array}{l}x = 1 \Rightarrow {\rm{u}} = 1\\{\rm{x}} = 16 \Rightarrow {\rm{u}} = 4\end{array} \right.\).

Khi đó \[1 = \int\limits_1^{16} {\frac{{{\rm{f}}\left( {\sqrt {\rm{x}} } \right)}}{{\rm{x}}}{\rm{dx}}} = \int\limits_1^4 {\frac{{{\rm{2f}}\left( {\rm{u}} \right)}}{{\rm{u}}}{\rm{du}}} = 2\int\limits_1^4 {\frac{{{\rm{2f}}\left( {\rm{x}} \right)}}{{\rm{x}}}{\rm{dx}}} \Rightarrow \int\limits_1^4 {\frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{\rm{x}}}{\rm{dx}}} = \frac{1}{2}\].

Đặt \(v = 4x \Rightarrow dv = 4dx\). Ta có \(\left\{ \begin{array}{l}x = \frac{1}{8} \Rightarrow v = \frac{1}{2}\\x = 1 \Rightarrow v = 4\end{array} \right.\).

Vậy \[I = {\rm{I}} = \int\limits_{\frac{1}{8}}^1 {\frac{{{\rm{f}}\left( {4{\rm{x}}} \right)}}{{\rm{x}}}{\rm{dx}}} = \int\limits_{\frac{1}{8}}^1 {\frac{{{\rm{f}}\left( {4{\rm{x}}} \right)}}{{{\rm{4x}}}}{\rm{4dx}}} = \int\limits_{\frac{1}{2}}^4 {\frac{{{\rm{f}}\left( v \right)}}{{\rm{v}}}{\rm{dv}}} = \int\limits_{\frac{1}{2}}^4 {\frac{{{\rm{f}}\left( x \right)}}{{\rm{x}}}{\rm{dx}}} \]

\[ = \int\limits_{\frac{1}{2}}^1 {\frac{{{\rm{f}}\left( x \right)}}{{\rm{x}}}{\rm{dx}}} + \int\limits_1^4 {\frac{{{\rm{f}}\left( x \right)}}{{\rm{x}}}{\rm{dx}}} = 2 + \frac{1}{2} = \frac{5}{2}.\] Chọn A.