2023 SAJC P1 Q6

2023 SAJC P1 Q6

12 marks
Prelims

The curve \({{C}_{1}}\) has equation \(y=\frac{x\left( x-a \right)}{x+a}\) and \(a\) is a positive real constant.

  1. Find algebraically, in terms of \(a\), the range of values that \(y\) can take.[3]

It is now given that \(a=1\) and that curve \({{C}_{2}}\) has equation \(y=-\frac{5}{2}+\frac{10}{x+3}\).

  1. Find the exact \(x\)-coordinates of the intersection points of \({{C}_{1}}\) and \({{C}_{2}}\).[2]
  2. Sketch \({{C}_{1}}\) and \({{C}_{2}}\) on the same graph, giving the coordinates of any points where \({{C}_{1}}\) and \({{C}_{2}}\) intersect and the equations of any asymptotes.[4]
  3. Find the exact area of the region bounded by the two curves \({{C}_{1}}\) and \({{C}_{2}}\), simplifying your answer in the form \(\left( a\ln 2+b\ln 5+c \right)\) units\(^{2}\), where \(a\), \(b\), \(c\) are constants to be determined.[3]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs a course plan.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs a course plan.

Answer:(i) \(y\ge \left( -3+2\sqrt{2} \right)a\) or \(y\le \left( -3-2\sqrt{2} \right)a\) (ii) \(x=1\) or \(x=-\frac{1}{2}\) or \(x=-5\) (iv) \(a=26\), \(b=-10\), \(c=-\frac{9}{8}\)

Need help? Join our JC Math tuition classes.

Learn more