f(x)>0⇔4-2x>0⇔x<2⇒x∈(−∞;2) 

Lời giải:

\(f(x)=(-x+1)(x-2)>0\Leftrightarrow \left\{\begin{matrix} -x+1< 0\\ x-2< 0\end{matrix}\right.\) hay $1< x< 2$

hay $x\in (1;2)$

Đáp án D

Từ GT ta lấy tích phân 2 vế cận từ 0 đến 1 ; sẽ được : 

\(\int\limits^1_0f\left(x+1\right)dx+\int\limits^1_03f\left(3x+2\right)dx-\int\limits^1_04f\left(4x+1\right)dx-\int\limits^1_0f\left(2^x\right)dx=\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}\left(1\right)\)

\(\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}=\int\limits^1_03\left(\sqrt{x+2}-\sqrt{x+1}\right)dx\)  = 

\(2\left[\left(x+2\right)\sqrt{x+2}-\left(x+1\right)\sqrt{x+1}\right]\dfrac{1}{0}\)  = \(2+6\sqrt{3}-8\sqrt{2}\left(2\right)\)

Dễ thấy : \(\int\limits^1_0f\left(x+1\right)dx=\int\limits^2_1f\left(t\right)dt=\int\limits^2_1f\left(x\right)dx\)

\(\int\limits^1_03f\left(3x+2\right)dx=\int\limits^5_2f\left(t\right)dt=\int\limits^5_2f\left(x\right)dx\)  (3)

\(\int\limits^1_04f\left(4x+1\right)=\int\limits^5_1f\left(t\right)dt=\int\limits^5_1f\left(x\right)dx\left(4\right)\)

\(\int\limits^1_0f\left(2^x\right)dx=\int\limits^2_1\dfrac{f\left(t\right)dt}{tln2}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(t\right)dt}{t}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}\)  (5)

Thay (2) ; (3) ; (4) ; (5) vào (1) ta được : 

\(\int\limits^2_1f\left(x\right)dx+\int\limits^5_2f\left(x\right)dx-\int\limits^5_1f\left(x\right)dx-\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=2+6\sqrt{3}-8\sqrt{2}\)

\(\Leftrightarrow\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=\left(2+6\sqrt{3}-8\sqrt{2}\right)ln2\)

\(\text{f(x)}\)\(\text{>0}\)\(\text{⇔}\)\(\text{2x}\)²\(\text{-3x+1}\)\(>0\)⇔\(\left\{{}\begin{matrix}x>1\\x< \dfrac{1}{2}\end{matrix}\right.\)

⇒x∈(−∞;\(\dfrac{1}{2}\))∪(1;+∞)

\(a=1>0\) ; \(\Delta'=\left(m-2\right)^2-\left(m-2\right)=\left(m-2\right)\left(m-3\right)\)

a/ Để \(f\left(x\right)\le0\) \(\forall x\in\left(0;1\right)\)

\(\Leftrightarrow x_1\le0< 1\le x_2\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(0\right)\le0\\f\left(1\right)\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m-2\le0\\1-\left(m-2\right)\le0\end{matrix}\right.\) \(\Rightarrow\) ko tồn tại m thỏa mãn

Do đó các câu c, f cũng không tồn tại m thỏa mãn

b/ TH1: \(\Delta< 0\Rightarrow2< m< 3\)

TH2: \(\left\{{}\begin{matrix}\Delta=0\\-\frac{b}{2a}\notin\left(0;1\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=2\\m=3\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}0\le x_1< x_2\\x_1< x_2\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Delta>0\Rightarrow\left[{}\begin{matrix}m>3\\m< 2\end{matrix}\right.\)

\(0\le x_1< x_2\Leftrightarrow\left\{{}\begin{matrix}f\left(0\right)\ge0\\\frac{x_1+x_2}{2}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m-2\ge0\\m-2>0\end{matrix}\right.\) \(\Rightarrow m>2\) \(\Rightarrow m>3\)

\(x_1< x_2\le1\Leftrightarrow\left\{{}\begin{matrix}f\left(1\right)\ge0\\\frac{x_1+x_2}{2}< 1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3-m\ge0\\m-2< 1\end{matrix}\right.\) \(\Rightarrow\) ko tồn tại m

Kết hợp 3 TH \(\Rightarrow m\ge2\)

d/ Tương tự như câu b, nhưng

TH2: \(\left\{{}\begin{matrix}\Delta=0\\-\frac{b}{2a}\in\left[0;1\right]\end{matrix}\right.\) \(\Rightarrow\) không tồn tại m thỏa mãn

TH3: \(\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}0< x_1< x_2\\x_1< x_2< 1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow m>3\)

Kết hợp 3 TH \(\Rightarrow\left[{}\begin{matrix}2< m< 3\\m>3\end{matrix}\right.\)

e/

TH1: \(\Delta\le0\Rightarrow2\le m\le3\)

TH2: \(\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}0\le x_1< x_2\\x_1< x_2\le1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>3\)

\(\Rightarrow m\ge2\)

Ta có : \(\left\{{}\begin{matrix}a=1>0\\\Delta'=\left(m-1\right)^2-\left(m-2\right)=m^2-3m+3>0\end{matrix}\right.\)

Để \(f\left(x\right)\le0\forall x\in\left[0;1\right]\)

\(\Leftrightarrow x_1\le0< 1\le x_2\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(0\right)\le0\\f\left(1\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m-2\le0\\-m+1\le0\end{matrix}\right.\Rightarrow1\le m\le2\)

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