Question

[Đào Gia Huy]

co mik giao bai kho cua mn giup mik vs 

Answer 1

[Nguyễn Việt Lâm]

31.

\(\Leftrightarrow\int\limits^{\dfrac{\pi}{4}}_0log_a\left(1+tanx\right)dx-\dfrac{1}{4}\int\limits^{\dfrac{\pi}{4}}_0dx\ge0\)

\(\Leftrightarrow\int\limits^{\dfrac{\pi}{4}}_0log_a\dfrac{1+tanx}{\sqrt[4]{a}}dx\ge0\)

\(\Leftrightarrow log_a\dfrac{1+tanx}{\sqrt[4]{a}}\ge0\) ; \(\forall x\in\left[0;\dfrac{\pi}{4}\right]\)

\(\Leftrightarrow\dfrac{1+tanx}{\sqrt[4]{a}}\ge1;\forall x\in\left[0;\dfrac{\pi}{4}\right]\)

\(\Leftrightarrow\sqrt[4]{a}\le\min\limits_{\left[0;\dfrac{\pi}{4}\right]}\left(1+tanx\right)=1\)

\(\Rightarrow a\le1\)

Ko tồn tại a thuộc khoảng đã cho thỏa mãn

Answer 2

[Nguyễn Việt Lâm]

29.

\(f'\left(x\right)=\dfrac{f\left(x\right)+x+1}{x+1}\)

\(\Rightarrow\left(x+1\right).f'\left(x\right)-f\left(x\right)=x+1\)

\(\Rightarrow\dfrac{f'\left(x\right)}{x+1}-\dfrac{1}{\left(x+1\right)^2}.f\left(x\right)=\dfrac{1}{x+1}\)

\(\Rightarrow\left[\dfrac{f\left(x\right)}{x+1}\right]'=\dfrac{1}{x+1}\)

\(\Rightarrow\dfrac{f\left(x\right)}{x+1}=ln\left|x+1\right|+C\)

\(f\left(1\right)=ln4=2ln2\Rightarrow\dfrac{f\left(1\right)}{2}=ln2+C\Rightarrow C=0\)

\(\Rightarrow f\left(x\right)=\left(x+1\right).ln\left|x+1\right|\)

\(\Rightarrow f\left(3\right)=4ln4=8ln2\)