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Chọn A.
Đặt u = ln x + x 2 + 1 , d v = d x ta được
F(x)=x ln x + x 2 + 1 - x 2 + 1 + C
Vì F(0) = 1 nên C = 2
Vậy
![](https://rs.olm.vn/images/avt/0.png?1311)
\(f\left(1-x\right)+f\left(x\right)=\dfrac{9^{1-x}}{9^{1-x}+3}+\dfrac{9^x}{9^x+3}=\dfrac{9}{9+3.9^x}+\dfrac{9^x}{9^x+3}=\dfrac{3}{9^x+3}+\dfrac{9^x}{9^x+3}=1\)
\(\Rightarrow f\left(x\right)=1-f\left(1-x\right)\)
\(\Rightarrow f\left(cos^2x\right)=1-f\left(sin^2x\right)\)
Do đó:
\(f\left(3m+\dfrac{1}{4}sinx\right)+f\left(cos^2x\right)=1\)
\(\Leftrightarrow f\left(3m+\dfrac{1}{4}sinx\right)=f\left(sin^2x\right)\) (1)
Hàm \(f\left(x\right)=\dfrac{9^x}{9^x+3}\) có \(f'\left(x\right)=\dfrac{3.9^x.ln9}{\left(9^x+3\right)^2}>0\Rightarrow f\left(x\right)\) đồng biến trên R
\(\Rightarrow\left(1\right)\Leftrightarrow3m+\dfrac{1}{4}sinx=sin^2x\)
Đến đây chắc dễ rồi, biện luận để pt \(sin^2x-\dfrac{1}{4}sinx=3m\) có 8 nghiệm trên khoảng đã cho
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(sinx=t\Rightarrow cosx.dx=dt\) ; \(\left\{{}\begin{matrix}x=\dfrac{\pi}{6}\Rightarrow t=\dfrac{1}{2}\\x=\dfrac{\pi}{2}\Rightarrow t=1\end{matrix}\right.\)
\(I=\int\limits^1_{\dfrac{1}{2}}\dfrac{dt}{1+t}=ln\left|1+t\right||^1_{\dfrac{1}{2}}=ln2-ln\left(\dfrac{3}{2}\right)=-ln3+2ln2\)
\(\Rightarrow ab=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Biến đổi :
\(5\sin x=a\left(2\sin x-\cos x+1\right)+b\left(2\cos x+\sin x\right)+c\)
= \(\left(2a+b\right)\sin x+\left(2b-a\right)\cos x+a+c\)
Đồng nhất hệ số hai tử số :
\(\begin{cases}2a+b=5\\2b-a=0\\a+c=0\end{cases}\)
\(\Rightarrow\) \(\begin{cases}a=2\\b=1\\c=-2\end{cases}\)
Khi đó :
\(f\left(x\right)=\frac{2\left(2\sin x-\cos x+1\right)+\left(2\cos x+\sin x\right)-2}{2\sin x-\cos x+1}\)
= \(2+\frac{2\cos x+\sin x}{2\sin x-\cos x+1}-\frac{2}{2\sin x-\cos x+1}\)
Do vậy :
\(I=2\int dx+\int\frac{\left(2\cos x+\sin x\right)dx}{2\sin x-\cos x+1}-2\int\frac{dx}{2\sin x-\cos x+1}\)
=\(2x+\ln\left|2\sin x-\cos x+1\right|-2J+C\)
Với
\(J=\int\frac{dx}{2\sin x-\cos x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
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a.
\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)
\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)
\(\Rightarrow y_{min}=y\left(1\right)=0\)
\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)
b.
\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]
\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)
c.
\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)
Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)
\(y=f\left(t\right)=-t^2-t+2\)
\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)
\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)
\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)
d.
Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)
\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)
\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)
\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)
Chọn A.
Biến đổi F x = ∫ d x cos 2 x + ∫ x sin x cos 2 x d x = tan x + I ( x )
Tính I(x) bằng cách đặt u=x; d v = sin x cos 2 x d x ⇒ I ( x ) = x cos x - ∫ d x cos x
Tính
J x = - ∫ d x cos x = ∫ cos x sin 2 x - 1 = ∫ d sin x sin x - 1 sin x + 1 = ln sin x - 1 sin x + 1 + C
Kết quả![](http://cdn.hoc24.vn/bk/WA9J7WV1PgFm.png)