2023 H2 Mathematics Paper 1 Question 7

Functions

Answers

(a)

Asymptotes: y=2y=2 and x=3.x=3.

(b)

Rf=[0,∞). R_f = \left[ 0, \infty \right).

(c)

Rf⊈Df. R_f \not \subseteq D_f.

(d)

Greatest value of a=−2.a = -2.

(e)

f−1(x)=4+3xx−2. f^{-1}(x) = \frac{4+3x}{x-2}.
Df−1=[0,2). D_{f^{-1}} = \left[ 0, 2 \right).

Full solutions

(a)

q1-diagram
Asymptotes: y=2y=2 and x=3  ■x=3 \; \blacksquare

(b)

Rf=[0,∞)  ■{ R_f = \left[ 0, \infty \right) \; \blacksquare }

(c)

Rf⊈Df\begin{gather*} R_f \not \subseteq D_f \end{gather*}

Hence the function f2f^2 does not exist   ■\; \blacksquare

(d)

Greatest value of a=−2  ■{ \text{Greatest value of } a = -2 \; \blacksquare }

(e)

y=−(2x+43−x)−y=2x+43−x−3y+yx=2x+4yx−2x=4+3yx(y−2)=4+3yx=4+3yy−2\begin{gather*} y = - \left( \frac{2x+4}{3-x} \right) \\ - y = \frac{2 x + 4}{3 - x} \\ - 3 y + y x = 2 x + 4 \\ y x - 2 x = 4 + 3 y \\ x \left( y - 2 \right) = 4 + 3 y \\ x = \frac{4 + 3 y}{y-2} \end{gather*}
f−1(x)=4+3xx−2  ■Df−1=Rf=[0,2)  ■\begin{gather*} f^{-1}(x) = \frac{4+3x}{x-2} \; \blacksquare \\ D_{f^{-1}} = R_f = \left[ 0, 2 \right) \; \blacksquare \end{gather*}