(a) \( a = -0.917 \), \( b = 0 \) (M1A1, N2).

Working:
Attempts to find an intersection point (M1)
Set \( f(x) = g(x) \): \( 1 – x^2 = e^{2x} \).
Define \( h(x) = 1 – x^2 – e^{2x} \).
At \( x = -1 \): \( h(-1) = 0 – e^{-2} \approx -0.135 < 0 \).
At \( x = 0 \): \( h(0) = 1 – 1 = 0 \), so \( b = 0 \).
Test \( x = -0.95 \): \( h(-0.95) \approx -0.0525 < 0 \).
Test \( x = -0.9 \): \( h(-0.9) \approx 0.025 > 0 \).
Root between \( -0.95 \) and \( -0.9 \), approximated as \( a \approx -0.917 \) (A1).

[2 marks]

(b) \( A = 0.240 \) (M1A1, N2).

Working:
Attempts to form the required integral (M1)
\( A = \int_a^b [f(x) – g(x)] \, dx = \int_{-0.917}^{0} (1 – x^2 – e^{2x}) \, dx \).
Antiderivative \( F(x) = x – \frac{x^3}{3} – \frac{1}{2} e^{2x} \).
At \( x = 0 \): \( F(0) = 0 – 0 – \frac{1}{2} = -0.5 \).
At \( x = -0.917 \): \( F(-0.917) \approx -0.922 + 0.2609 – 0.0791 \approx -0.7403 \).
\( A = F(0) – F(-0.917) \approx -0.5 – (-0.7403) \approx 0.240 \) (A1).

[2 marks]

Total [4 marks]