Đáp án: `38) C: x ∈ [-1; 1) ∪ (1; 3]`
`39) D: -1`
Giải thích các bước giải:
`Câu` `38:` `f'(x)=((x^2+x+2)/(x-1))^'=((x^2+x+2)'(x-1)-(x^2+x+2)(x-1)')/((x-1)^2)`
`=((2x+1)(x-1)-(x^2+x+2).1)/((x-1)^2)=(2x^2-x-1-x^2-x-2)/((x-1)^2)`
`=(x^2-2x-3)/((x-1)^2)`
`=> f'(x)≤0 `
`<=> (x^2-2x-3)/((x-1)^2)≤0 `
`Ta` `thấy` `ĐKXĐ: x#1`
`Cho` `PT = 0 => x=-1; x=3`
`Ta` `có` `bảng`
$\text{ x -∞ -1 1 3 +∞ }$
$\text{ PT + 0 - || - 0 +}$
`=> x ∈ [-1; 1) ∪ (1; 3]`
`=> Chọn` `C`
.
`Câu` `39)` `f'(x)=(sin(cos(x))/sin(x))^'`
`=(sin(cos(x))'.sin(x)-sin(cos(x)).sin(x)']/sin^2(x)`
`=(cos(cos(x)).(cosx)'.sin(x)-sin(cos(x)).cos(x)]/sin^2(x)`
`=-(cos(cos(x)).sin(x).sin(x)-sin(cos(x)).cos(x)]/sin^2(x)`
`=-(cos(cos(x)).sin^2(x)-sin(cos(x)).cos(x)]/sin^2(x)`
`=> f'(π/2)=-(cos(cos(π/2)).sin^2(π/2)-sin(cos(π/2)).cos(π/2)]/sin^2(π/2)`
`=-(cos(0).1^2-sin(0).cos(0)]/(1^2)=-(1.1-0.1]/(1)=-1`
`=> Chọn` `D`
