Math topic中文

Modulus

Concept of ∣x∣|x|, and use of the relations ∣x−a∣<b⇔a−b<x<a+b|x - a| < b \Leftrightarrow a - b < x < a + b and ∣x−a∣>b⇔x<a−b or x>a+b|x - a| > b \Leftrightarrow x < a - b \text{ or } x > a + b.

Practice questions

  1. 1已知函数y=f(x)y=f(x)是定义在RR上的偶函数,且在区间[0,+∞)[0,+\infty)上单调递增,若实数mm满足f(log⁡3m)⩾f(1)f({{\log}_{3}}m)\geqslant f(1),则mm的取值范围为___.
  2. 2Sketch, on the same diagram, the graphs of y=∣2x+3∣y = |2x + 3| and y=−2x2−8x−3y = -2x^2 - 8x - 3. (You are not required to label any a…
  3. 3Find ∫3ex5−0.3ex dx\displaystyle \int \frac{3e^x}{5 - 0.3e^x} \, dx ∫cos⁡(ln⁡x) dx\displaystyle \int \cos(\ln x) \, dx, where x>0x > 0 the exact…
  4. 4已知函数f(x)=x2−ax(a≠0)f(x)=x^{2}-ax(a\neq 0),g(x)=ln⁡xg(x)=\ln x,f(x)f(x)图象与xx轴异于原点的交点MM处的切线为l1l_{1},g(x−1)g(x-1)与xx轴的交点NN处的切线为…
  5. 5On a single diagram, sketch the graphs of y=∣2x−p∣y = |2x - p| and y=qxy = qx where the following conditions are satisfied, indi…
  6. 6Prove by induction that, for all positive integers nn and for x∈Rx\in\mathbb{R}, x≠0, x≠1x\ne 0,\ x\ne 1, 1x−1−1x−1x2−⋯−1xn=1xn(x−1) .\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-\dots-\frac{1}{x^n}=\frac{1}{x^n(x-1)}\,.…
  7. 7Find the exact roots of the equation ∣4x−5x−2∣=3x.Onthesameaxes,sketchthecurveswithequations\left|\dfrac{4x-5}{x-2}\right| = 3x. On the same axes, sketch the curves with equations …
  8. 8A home interior designer is looking to design a pull-down wardrobe lift. To test his concept, the designer built a small…
  9. 9The points AA and BB have coordinates (0,2,0)(0, 2, 0) and (1,0,2)(1, 0, 2) respectively, and the plane π1\pi_1 has equation r⋅(221)=4\mathbf{r} \cdot \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = 4…
  10. 10Solve the inequality ∣y−5∣<1|y-5|<1. Hence solve the inequality ∣3x−5∣<1|3^{x}-5|<1, giving 3 significant figures in your answer.
  11. 11已知集合A={x∈z∣∣x∣⩽2}A=\{x\in z||x|\leqslant 2\},B={x∣x−2x+2⩽0}B=\left\{ x\left|\dfrac{x-2}{x+2}\leqslant 0\right.\right\},则()(\quad) A. A∩B=AA∩B=A …
  12. 12在数轴上,点MM,NN表示的数分别是x1x_{1},x2x_{2},我们把x1x_{1},x2x_{2}之差的绝对值叫做点MM,NN之间的距离,即MN=∣x1−x2∣.(1)MN=|x_{1}-x_{2}|.(1)已知数…
  13. 13Write down constants AA and BB such that for all values of xx, x−1=A(2x−4)+B.x - 1 = A(2x - 4) + B. By expressing x2−4x+5x^2 - 4x + 5…
  14. 14已知∣log⁡a34∣<1|\log _{a} \dfrac {3}{4}| < 1,求aa的取值集合.
  15. 15A function f is defined as f(x)=ln⁡(5−x),  x<5.f(x)=\ln(5-x),\; x<5. Find the volume of revolution when the region bounded by the curve of…
  16. 16已知函数f(x)f(x)是定义在RR上的偶函数,且当x⩾0x\geqslant 0时,f(x)=x2+x.f(x)=x^{2}+x.若f(a)f(a)十f(−a)<4f(-a) < 4,测实数aa的取值范围为___.
  17. 17已知f(x)=∣2x+1∣+∣x−1∣.(Ⅰ)f(x)=\left|2x+1\right|+\left|x-1\right|.(Ⅰ)求f(x)f(x)在[−1,1][-1,1]上的最大值mm及最小值nn; (Ⅱ)(Ⅱ)在(Ⅰ)(Ⅰ)的条件下,设a,b∈Ra,b\in R…
  18. 18Tetrahedral dice have four faces. Two fair tetrahedral dice, one red and one blue, have faces numbered 0, 1, 2, and 3 re…
  19. 19The function f\mathrm{f} is defined by f:x↦5+2x\mathrm{f} : x \mapsto \sqrt{5+2x}, for x∈Rx \in \mathbb{R}, x>−52x > -\frac{5}{2}…
  20. 20已知不等式a+3b⩽∣x2−1∣a+ \sqrt{3}b\leqslant \left|{x}^{2}-1\right| 对于满足条件a2+b2=1{a}^{2}+{b}^{2}=1 的任意实数aa,bb恒成立,实数xx的取值范围是 .
  21. 21不等式|x|>2的解集是( ) A. x>2 B. {x|x>2} C. {x|x>2或x<-2} D. x>2或x<-2
  22. 22The two planes p1p_{1} and p2p_{2}, given by the equations 2x+y=12x+y=1 and 8x+ay+z=48x+ay+z=4 respectively, meet at a line LL whi…
  23. 23大家知道|5|=|5-0|,它在数轴上表示5的点与原点(即表示0的点)之间的距离.又如式子|6-3|,它在数轴上的意义是表示6的点与表示3的点之间的距离.即点A、B在数轴上分别表示数a、b,则A、B两点的距离可表示为:|AB|=|a-b|.…
  24. 24已知函数f(x)f\left(x\right)是定义在RR上的偶函数,且f(0)=0f\left(0\right)=0,当x>0x>0时,f(x)=log⁡2(x+1).(1)f\left(x\right)={{\log}_{2}}\left(x+1\right).(1)…
  25. 25In this question you should state clearly the values of the parameters of any normal distributions that you use. The mas…
  26. 26Find, in terms of aa, the roots of the equation 1(x−a)2=∣x−a∣\dfrac{1}{(x-a)^2}=|x-a|. On the same axes, sketch the curves with eq…
  27. 27Do not use a calculator in answering this question. Solve the inequality 6x2+2x−32x−1≥2(x+1).\frac{6x^2 + 2x - 3}{2x - 1} \ge 2(x + 1).…
  28. 28Dendrologists are specialised scientists who study trees and woody plants. Their work is diverse and can encompass vario…
  29. 29The plane Π\Pi has equation 2x−3y+6z=72x-3y+6z=7 and the line ll has equation r=8i−2j+8k+t(3i−j+2k).\mathbf{r}=8\mathbf{i}-2\mathbf{j}+8\mathbf{k}+t(3\mathbf{i}-\mathbf{j}+2\mathbf{k}).…
  30. 30By using the substitution y=∣x∣y = |x|, solve the inequality ∣x∣+3x2+1>1.\frac{|x|+3}{x^2+1} > 1.
  31. 31Sketch the curve CC with equation y=x−2x−1y = \frac{x-2}{x-1}, stating the exact coordinates of any points of intersection wi…
  32. 32The diagram shows the graphs of y=∣a−bx∣y = |a - bx| and y=∣mx−6∣y = |mx - 6| where aa, bb and mm are positive constants. The gra…
  33. 33To build a Virtual Reality (VR) experience, a virtual world is created to contain geometric models. With reference to th…
  34. 34Section A: Pure Mathematics [40 marks] Sketch, on the same axes, the graphs of y=3xy = 3x and y=∣x2−a2∣y = |x^2 - a^2| where a>0a > 0…
  35. 35State a sequence of transformations that will transform the curve with equation y2−x2=1y^{2}-x^{2}=1 on to the curve with equ…
  36. 36f(x)f(x)是定义在[−2,2][-2,2]上的偶函数,且f(x)f(x)在[0,2][0,2]上单调递减,若f(1−m)<f(m)f(1-m) < f(m)成立,求实数mm的取值范围 ___ .
  37. 37The plane π1\pi_1 has equation r⋅(251)=11.\mathbf{r}\cdot\begin{pmatrix}2 \\ 5 \\ 1\end{pmatrix}=11. The plane π2\pi_2 passes t…
  38. 38已知点P的坐标为(2-a,3a+6),且点P到两坐标轴的距离相等,则a= .
  39. 39Given that a∈Ra \in \mathbb{R} and a>1a > 1, use an algebraic method to solve the inequality ax−a+2(x−a)(1−x)≤1.\frac{ax - a + 2}{(x - a)(1 - x)} \le 1 .…
  40. 40Given that both aa and bb are positive constants, on the same diagram, sketch the graphs of y=2∣x−a∣andy=b∣x−a∣.y = \frac{2}{|x - a|} \quad \text{and} \quad y = b|x - a|.…
  41. 41设集合S={x||x|<5},T={x|(x+7)(x-3)<0}.则S∩T=( ) A. {x|-7<x<5 } B. {x|3<x<5 } C. {x|-5<x<3 } D. {x|-7<x<-5 }
  42. 42关于x的方程||x-2|-1|=a恰有三个整数解,则a的值为 .
  43. 43已知实数a>0a > 0,设函数f(x)=∣x+1a∣+∣x−a∣f(x)=\left|x+ \dfrac{1}{a}\right|+\left|x-a\right| . ((Ⅰ))证明:f(x)⩾2f(x)\geqslant 2; ((Ⅱ))…
  44. 44The line ll has equation 1−x=z−α2,y=11 - x = \frac{z - \alpha}{2}, y = 1, where α\alpha is a constant. Given that the point of i…
  45. 45若|x-3|=2,则x的值为 .
  46. 46The plane Π1\Pi_1 has equation r=i+2j+k+θ(2j−k)+ϕ(3i+2j−2k).\mathbf{r} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} + \theta(2\mathbf{j} - \mathbf{k}) + \phi(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}).…
  47. 47设集合A={x||x-a|<1,x∈R},B={x|1<x<5,x∈R},若A∩B=∅,则实数a的取值范围是( ) A. {a|0≤a≤6} B. {a|a≤2,或a≥4} C. {a|a≤0,或a≥6} D. {a|2≤a≤4}
  48. 48Relative to the origin OO, two points AA and BB have position vectors given by a=2i−j\mathbf{a} = 2\mathbf{i} - \mathbf{j}…
  49. 49Without using a calculator, solve the inequality 4xx−3≥1.Hencefindtherangeofvaluesof\dfrac{4x}{x-3} \ge 1. Hence find the range of values of xthatsatisfiesthat satisfies…
  50. 50设全集为RR,A={x∣x2−5x−6>0}A=\{x|x^{2}-5x-6>0\},B={x∣∣x−5∣<a}(aB=\{x||x-5|< a\}(a为常数)),11∈B11\in B,则()(\quad) A. (∁RA)∪B=R(∁_{R}A)∪B=R B. A∪(∁RB)=RA∪(∁_{R}B)=R…
  51. 51In this question, kk is a positive constant. Using an algebraic method, solve the inequality 2x2−2k2x<x+k\frac{2x^2 - 2k^2}{x} < x + k…
  52. 52Without using a calculator, solve 1+6x≥x1 + \frac{6}{x} \ge x, 1+6∣x∣≥∣x∣1 + \frac{6}{|x|} \ge |x|.
  53. 53设不等式∣4x−a∣>∣1−2x+3∣\left|{{4}^{x}}-a\right|>\left|1-{{2}^{x\text{+3}}}\right|对所有x∈[1,2]x\in[1,2]均成立,则实数aa的取值范围是()(\quad) A. a<−15a< -15…
  54. 54Without the use of a calculator, solve the inequality (x2−2x+4)(x−3)(x+2)≥0.\frac{(x^2 - 2x + 4)(x - 3)}{(x + 2)} \ge 0. Hence solve the i…
  55. 55Without the use of a calculator, solve the inequality 3x2+12x2+3x−2>1\frac{3x^2 + 1}{2x^2 + 3x - 2} > 1. Hence solve the inequality 3x2+12x2+3∣x∣−2>1\frac{3x^2 + 1}{2x^2 + 3|x| - 2} > 1…
  56. 56Sketch, on the same diagram, the curves with equations y=ln⁡(x−2)y = \ln(x-2) and y=1ex+3y = \frac{1}{\text{e}^x} + 3, stating the eq…
  57. 57In a small town, there are two clinics, Lee's Clinic and Hope Clinic. At Lee's Clinic, the waiting time of a patient fol…
  58. 58By using the substitution y=∣x∣y = |x|, solve the inequality ∣x∣+3x2+1>1\frac{|x| + 3}{x^2 + 1} > 1.
  59. 59Show that 32r−1−42r+1+12r+3=8r+20(2r−1)(2r+1)(2r+3)\frac{3}{2r-1} - \frac{4}{2r+1} + \frac{1}{2r+3} = \frac{8r+20}{(2r-1)(2r+1)(2r+3)}. Hence show that ∑r=1n2r+5(2r−1)(2r+1)(2r+3)=23−n+2(2n+1)(2n+3)\sum_{r=1}^{n} \frac{2r+5}{(2r-1)(2r+1)(2r+3)} = \frac{2}{3} - \frac{n+2}{(2n+1)(2n+3)}…
  60. 60The planes Π1\Pi_1 and Π2\Pi_2 have equations r⋅(i−2j+k)=4\mathbf{r} \cdot (\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = 4 and r=3i−j+k+λ(−i+j+k)+μ(4i+j)\mathbf{r} = 3\mathbf{i} - \mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} + \mathbf{j} + \mathbf{k}) + \mu(4\mathbf{i} + \mathbf{j})…