(a) Sketch the graph of \( y = f_1(x) \):

Coordinates of local maximum: \( (1, \frac{1}{e}) \) or \( (1, 0.368) \quad (A1) \).

Graph passes through the origin \( (0, 0) \quad (A1) \).

Correct domain \( x \geq 0 \quad (A1) \).

Correct shape with single maximum and asymptotic behavior to x-axis (equation not required) or point of inflection \quad (A1) \).

[4 marks]

(b) Show the area is \( \frac{e^b – b – 1}{e^b} \):

Area = \( \int_0^b x e^{-x} \, dx \quad (M1) \).

Use integration by parts, let \( u = x \), \( dv = e^{-x} \, dx \):

\[ du = dx, \quad v = -e^{-x} \quad (A1) \]

\[ \int x e^{-x} \, dx = x (-e^{-x}) – \int (-e^{-x}) \, dx \]

\[ = -x e^{-x} + \int e^{-x} \, dx \quad (A1) \]

\[ = -x e^{-x} – e^{-x} + c \quad (A1) \]

Evaluate from 0 to \( b \):

\[ \left[ -x e^{-x} – e^{-x} \right]_0^b \quad (M1) \]

\[ = \left( -b e^{-b} – e^{-b} \right) – \left( 0 – e^0 \right) \]

\[ = -b e^{-b} – e^{-b} + 1 \]

\[ = \frac{-b – 1}{e^b} + 1 \]

\[ = \frac{e^b – b – 1}{e^b} \quad (A1)(AG) \]

[5 marks]

(c) (i) Use l’Hôpital’s rule for \( \lim_{b \to \infty} \frac{e^b – b – 1}{e^b} \):

Form is \( \frac{\infty}{\infty} \), apply l’Hôpital’s rule (M1):

\[ \lim_{b \to \infty} \frac{e^b – b – 1}{e^b} = \lim_{b \to \infty} \frac{e^b – 1}{e^b} \quad (A1) \]

Still \( \frac{\infty}{\infty} \), apply again:

\[ \lim_{b \to \infty} \frac{e^b}{e^b} = 1 \quad (A1) \]

[3 marks]

(c) (ii) Value of \( A_1 \):

\[ A_1 = \lim_{b \to \infty} \int_0^b x e^{-x} \, dx = \lim_{b \to \infty} \frac{e^b – b – 1}{e^b} = 1 \quad (A1) \]

[1 mark]

(d) (i) Value of \( A_4 \):

Correct integral \( \int_0^b x^4 e^{-x} \, dx \) with large \( b \quad (M1) \).

\[ A_4 = 24 \quad (A1) \]

[2 marks]

(d) (ii) Value of \( A_5 \):

Correct integral \( \int_0^b x^5 e^{-x} \, dx \) with large \( b \quad (M1) \).

\[ A_5 = 120 \quad (A1) \]

[2 marks]

(e) Suggest expression for \( A_n \):

\[ A_n = n! \quad (A1) \]

[1 mark]

(f) Prove \( A_n = n! \) by induction:

Base case (\( n = 1 \)):

\[ A_1 = 1 = 1! \quad (A1) \]

So true for \( n = 1 \).

Assume true for \( n = k \): \( A_k = \int_0^\infty x^k e^{-x} \, dx = k! \quad (M1) \).

Prove for \( n = k + 1 \):

\[ A_{k+1} = \int_0^\infty x^{k+1} e^{-x} \, dx \]

Integrate by parts, let \( u = x^{k+1} \), \( dv = e^{-x} \, dx \):

\[ du = (k + 1) x^k \, dx, \quad v = -e^{-x} \quad (A1) \]

\[ \int x^{k+1} e^{-x} \, dx = x^{k+1} (-e^{-x}) – \int (-e^{-x}) (k + 1) x^k \, dx \quad (M1) \]

\[ = -x^{k+1} e^{-x} + (k + 1) \int x^k e^{-x} \, dx \]

Evaluate from 0 to \( \infty \):

\[ \left[ -x^{k+1} e^{-x} \right]_0^\infty + (k + 1) \int_0^\infty x^k e^{-x} \, dx \quad (A1) \]

Using \( \lim_{x \to \infty} x^{k+1} e^{-x} = 0 \):

\[ \left[ 0 – 0 \right] + (k + 1) A_k \]

By assumption, \( A_k = k! \):

\[ A_{k+1} = (k + 1) k! = (k + 1)! \quad (A1) \]

Since true for \( n = 1 \) and if true for \( n = k \), then true for \( n = k + 1 \), by induction, \( A_n = n! \) for all \( n \in \mathbb{Z}^+ \quad (R1) \).

[6 marks]