Porisma

Discussion questions

All chapter practice
  1. 1

    Without using a calculator, find the solution sets of the following inequalities.

    (a) (x+2)2x2−1≤0\dfrac{(x+2)^2}{x^2-1}\leq0,

    (b) x2+2x+2x2−x+2<0\dfrac{x^2+2x+2}{x^2-x+2}<0,

    (c) 1(x−1)(x−3)<−1\dfrac1{(x-1)(x-3)}<-1,

    (d) 1≤13x−2<31\leq\dfrac1{3x-2}<3.

    CJC · Discussion · Question 1

  2. 2

    Without using a calculator, solve the following inequalities.

    (a) ∣2x−3∣<11|2x-3|<11,

    (b) ∣x2−8∣≥4|x^2-8|\geq4.

    CJC · Discussion · Question 2

  3. 3

    Solve ∣x2+3∣>3−x2\left|\dfrac{x}{2}+3\right|>3-x^2. Hence, solve ∣ex2−3∣>3−e2x\left|\dfrac{e^x}{2}-3\right|>3-e^{2x}.

    [2007/NJC/I/2]

    CJC · Discussion · Question 3

  4. 4

    It is given that f(x)=x−52x−3f(x)=\dfrac{x-5}{2x-3}. Using an algebraic method, solve the inequality f(x)≤23f(x)\leq\dfrac23.

    Hence find the exact range of values of xx for which

    (i) f(ln⁡x)≤23f(\ln x)\leq\dfrac23,

    (ii) f(∣x+12∣)≤23f\left(\left|x+\dfrac12\right|\right)\leq\dfrac23.

    [2013/RI/I/10]

    CJC · Discussion · Question 4

  5. 5

    It is given that y=x2+x+1x−1y=\dfrac{x^2+x+1}{x-1}, x∈Rx\in\mathbb R, x≠1x\ne1.

    Without using a calculator, find the set of values that yy can take.

    [2013/I/2]

    CJC · Discussion · Question 5

  6. 6

    (a) Show that if x2>k(x+1)x^2>k(x+1) for all real xx, then −4<k<0-4<k<0.

    (b) Find the range of values of kk for which x2+2kx+15−2kx^2+2kx+15-2k is always positive for all real xx. Deduce the range of values of kk for which k(x2+2kx+15−2k)>0k(x^2+2kx+15-2k)>0 for all real xx.

    CJC · Discussion · Question 6

  7. 7

    Given that aa is a positive real number, solve the inequality x2−x+2x2+(1−a)x−a≤0\dfrac{x^2-x+2}{x^2+(1-a)x-a}\leq0, leaving your answer in terms of aa.

    [2013/HCI/I/1]

    CJC · Discussion · Question 7

  8. 8

    Without using a calculator, solve 3x−11−2x2≥1\dfrac{3x-1}{1-2x^2}\geq1.

    Hence find the solution of 3sin⁡θ−1cos⁡2θ≥1\dfrac{3\sin\theta-1}{\cos2\theta}\geq1, where −π2≤θ≤π2-\dfrac\pi2\leq\theta\leq\dfrac\pi2.

    [2016/AJC/Promo/5]

    CJC · Discussion · Question 8