17N.1.SL.TZ0.S_5

pestleMathematicsAASLPaper 117N· sl-2-5-composite-functions-identity-finding-inversesource ↗

Let $f(x) = 1 + {{\text{e}}^{ - x}}$ and $g(x) = 2x + b$ , for $x \in \mathbb{R}$ , where $b$ is a constant.

Find $(g \circ f)(x)$ .

[2]

Given that $\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = - 3$ , find the value of $b$ .

[4]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to form composite (M1)

eg $\,\,\,\,\,$ $g(1 + {{\text{e}}^{ - x}})$

correct function A1 N2

eg $\,\,\,\,\,$ $(g \circ f)(x) = 2 + b + 2{{\text{e}}^{ - x}},{\text{ }}2(1 + {{\text{e}}^{ - x}}) + b$

[2 marks]

evidence of $\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ - x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ - x}})$ (M1)

eg $\,\,\,\,\,$ $2 + b + 2{{\text{e}}^{ - \infty }}$ , graph with horizontal asymptote when $x \to \infty$

Note: Award M0 if candidate clearly has incorrect limit, such as $x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}$ .

evidence that ${{\text{e}}^{ - x}} \to 0$ (seen anywhere) (A1)

eg $\,\,\,\,\,$ $\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ - x}}) = 0,{\text{ }}1 + {{\text{e}}^{ - x}} \to 1,{\text{ }}2(1) + b = - 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0$ , graph of $y = {{\text{e}}^{ - x}}$ or

$y = 2{{\text{e}}^{ - x}}$ with asymptote $y = 0$ , graph of composite function with asymptote $y = - 3$

correct working (A1)

eg $\,\,\,\,\,$ $2 + b = - 3$

$b = - 5$ A1 N2

[4 marks]

Examiners’ report

[N/A]