Bộ 10 Đề kiểm tra học kì I Chuyên đề toán 11: Kiểm tra 15 phút có đáp án (Đề 10)

$y^{\text{'}}=2$

$y^{\text{'}}=-\frac{1}{{\left(x-1\right)}^2}$

$y^{\text{'}}=-\frac{3}{{\left(x-1\right)}^2}$

$y^{\text{'}}=\frac{1}{{\left(x-1\right)}^2}$

$4y-y^{\text{'}}=0$

$4y+y^{\text{'}\text{'}}=0$

$y=y^{\text{'}}\tan 2x$

$y^2={\left(y^{\text{'}}\right)}^2=4$

$y^{\text{'}}=\frac{x\cos x+\sin }{x^2}$

$y^{\text{'}}=\frac{x\cos -\sin x}{x^2}$

$y^{\text{'}}=\frac{x\sin x+\cos x}{x^2}$

$y^{\text{'}}=\frac{x\sin x-\cos x}{x^2}$

$y^{\text{'}}=\frac{1}{\sqrt{\sin x}}-\frac{1}{\sqrt{\cos x}}$

$y^{\text{'}}=\frac{1}{\sqrt{\sin x}}+\frac{1}{\sqrt{\cos x}}$

$y^{\text{'}}=\frac{\cos x}{\sqrt{\sin x}}-\frac{\sin x}{\sqrt{\cos x}}$

$y^{\text{'}}=\frac{\cos x}{\sqrt{\sin x}}+\frac{\sin x}{\sqrt{\cos x}}$

$\left[-1;2\right]$

$\left(-\infty ;0\right]$

$\left\{-1\right\}$

$Ø$

$dy=-\frac{x^2-2x-2}{{\left(x-1\right)}^2}dx$

$dy=\frac{2x+1}{{\left(x-1\right)}^2}dx$

$dy=-\frac{2x+1}{{\left(x-1\right)}^2}dx$

$dy=\frac{x^2-2x-2}{{\left(x-1\right)}^2}dx$

$k=1$

$k=\frac{1}{2}$

$k=\frac{\sqrt{2}}{2}$

k=2

$a=1,b=1$

$a=2,b=1$

$a=1,b=2$

$a=2,b=2$

$y^{\left(5\right)}=-\frac{120}{{\left(x+1\right)}^6}$

$y^{\left(5\right)}=\frac{120}{{\left(x+1\right)}^6}$

$y^{\left(5\right)}=\frac{1}{{\left(x+1\right)}^6}$

$y^{\left(5\right)}=-\frac{1}{{\left(x+1\right)}^6}$

$y=-x-1;y=\frac{1}{4}x+\frac{7}{2}$

$y=-x-1;y=-\frac{1}{4}x+\frac{7}{2}$

$y=-x+1;y=-\frac{1}{4}x+\frac{7}{2}$

$y=-x+1;y=-\frac{1}{4}x-\frac{7}{2}$