Đáp án:

11B

12C

13B

14B

15A

Giải thích các bước giải:

Câu 11:

`f'(x)=(x^2-1)^2(x^2-3x+2)x^{2021}`

`f'(x)=(x+1)^2(x-1)^3(x-2)x^{2021}`

`f'(x)=0 ⇒` \(\left[ \begin{array}{l}x=-1\ (\text{nghiệm kép})\\x=1\\x=2\\x=0\end{array} \right.\) 

Bỏ qua nghiệm bội chẵn, ta có:

\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}0&\text{}&\text{}1&\text{}&\text{}2&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}-&\text{0}&\text{}+&\text{0}&\text{}-&\text{}0&\text{}+&\\\hline \end{array}\)

Vậy HS có 3 cực trị

Câu 12:

`f'(x)=x^2(x^2-1)`

`f'(x)=0 ⇒` \(\left[ \begin{array}{l}x=0\ (\text{nghiệm kép})\\x=1\\x=-1\end{array} \right.\) 

Bỏ qua nghiệm bội chẵn, ta có:

\(\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{}-1&\text{}&\text{}1&\text{}&\text{$+\infty$}\\\hline \text{$y'$}&\text{}&\text{}+&\text{0}&\text{}-&\text{0}&\text{}+&\text{}\\\hline \end{array}\)

Vậy HS có 1 điểm cực đại

Câu 13:

`f(x)=\frac{2x-1}{x+1}`

TXĐ: `D=\mathbb{R} \\ {-1}`

`f'(x)=\frac{3}{(x+1)^2}`

`f'(x) > 0 \forall x \in D`

Ta có: 

`f(0)=-1`

`f(4)=7/5`

`⇒ min_{[0;4]} f(x)=-1,max_{[0;4]}=7/5`

Vậy `5M-3m=5.7.5-3.(-1)=10`

Câu 14:`

`f(x)=\sqrt{x-1}+\sqrt{5-x}`

TXĐ: `D=[1;5]`

`f'(x)=\frac{1}{2\sqrt{x-1}}-\frac{1}{2\sqrt{5-x}}`

`f'(x)=0 ⇒ x=3`

Ta có: 

`f(1)=2`

`f(3)=2\sqrt{2}`

`f(5)=2`

Vậy `min_{[1;5]} f(x)=2,max_{[1;5]}=2\sqrt{2}`

Câu 15:

`y=x\sqrt{1-x^2}`

TXĐ: `D=[-1;1]`

`f'(x)=\frac{-2x^2+1}{\sqrt{1-x^2}}`

`f'(x)=0 ⇒`\(\left[ \begin{array}{l}x=-\dfrac{\sqrt{2}}{2}\\x=\dfrac{\sqrt{2}}{2}\end{array} \right.\) \(\in [-1;1]\)

Ta có: 

`f(-1)=0`

`f(-\frac{\sqrt{2}}{2})=-1/2`

`f(\frac{\sqrt{2}}{2})=1/2`

`f(1)=0`

`⇒ min_{[-1;1]} f(x)=-1/2,max_{[-1;1]}=1/2`

Vậy `M+m=1/2+(-1/2)=0`