Question
Let \(f(x) = 1 + {{\text{e}}^{ – x}}\) and \(g(x) = 2x + b\), for \(x \in \mathbb{R}\), where \(b\) is a constant.
Find \((g \circ f)(x)\).
Given that \(\mathop {\lim }\limits_{x \to + \infty } (g \circ f)(x) = – 3\), find the value of \(b\).
Answer/Explanation
Markscheme
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]