2022 H2 Mathematics Paper 1 Question 10

Graphs and Transformations

Answers

(a)

a<−2b.{a < -2b.}

(b)

(d)

x≤−2{x \leq - 2} or x>1.{x > 1.}

Full solutions

(a)

dydx=a−a+2b(x−1)2\frac{\mathrm{d}y}{\mathrm{d}x} = a - \frac{a+2b}{(x-1)^2}
At stationary points,
dydx=0a−a+2b(x−1)2=0a(x−1)2−a−2b=0ax2−2ax+a−a−2b=0ax2−2ax−2b=0\begin{gather*} \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \\ a - \frac{a+2b}{(x-1)^2} = 0 \\ a(x-1)^2 - a - 2b = 0 \\ ax^2 - 2ax + a - a - 2b = 0 \\ ax^2 - 2ax - 2b = 0 \end{gather*}
Since C{C} has no stationary points,
Discriminant < 0(−2a)2−4(a)(−2b)<04a2+8ab<0Since a >0,a+2b<0a<−2b  ■\begin{gather*} \textrm{Discriminant < 0} \\ (-2a)^2 - 4(a)(-2b) < 0 \\ 4a^2 + 8ab < 0 \\ \textrm{Since a >0,} \\ a + 2b < 0 \\ a < -2b \; \blacksquare \end{gather*}

(b)

(d)

Let a=1.{a=1.}
From GC, intersection between the curve and the line occurs at x=−2.{x=- 2.}

From the sketch, the solution to the inequality is

x≤−2 or x>1  ■x \leq - 2 \textrm{ or } x > 1 \; \blacksquare

Question Commentary

For part (a), the phrase "no stationary points" hints us to differentiate, equate dydx{\frac{\mathrm{d}y}{\mathrm{d}x}} to zero and then use the discriminant.

For part (b), we can pick a random positive value of a{a} to use in our graphing calculator to get the shape of the graph. However, we must be careful in finding the axial intercepts. In particular, the y{y}-intercept is related to a,{a,} so it will be wrong to use the graphing calculator blindly here and report a numeric value.

For part (d), notice that just 2 marks are given along with the "hence" keyword. We should use the graph we have already drawn and our graphing calculator (applying the value of a{a} in it) instead of going through the rational inequalities algebraic approach.