(i)

Note that f(-x) is a reflection in the y-axis, corresponding to figure (B)

f(x - 1) is a shift right by 1 unit, corresponding to figure ©

-f(x) is a reflection in the x-axis, corresponding to figure (A) (or process of elimination)

(ii)

Note that if y = 2^{-x^2} then as x \to \pm \infty, y \to 0, and that y is increasing for x < 0, (for instance, by noting the behaviour of y = 2^x or by differentiating) decreasing for x > 0. So y has a maximum at x = 0, y = 1. This gives the graph:

In the second case we have y = 2^{2x - x^2} = 2^{-(x^2 - 2x)} = 2^{-(x^2 - 2x + 1) + 1} = 2^{1 - (x - 1)^2} = 2 \times 2^{-(x - 1)^2}, which is the first graph shifted to the right 1 unit then scaled by 2 parallel to the y-axis. The single stationary point (0,1) will be translated to (1,2). Plotting both gives the figure:

(iii)

We have:

\displaystyle \int_0^1 2^{-(x - c)^2} \mathrm dx = \int_0^1 2^{-(c - x)^2} \mathrm dx = \int_{c - 1}^c 2^{-u^2} \mathrm du

(translating the curve appropriately)

Informally, we want to pick c to make 2^{-u^2} as large as possible for as long as possible on [c - 1, c]. Since 2^{-u^2} decreases away from 0 and is even, this is done by taking [c - 1, c] to be centred around the maximum, 0. That is, 1 - c = c, so that c = \dfrac 1 2, the c that maximises I(c).

**Alternatively**: By the fundamental theorem of calculus we have:

I'(c) = 2^{-c^2} - 2^{-(c - 1)^2}

Note that if I'(c) = 0 we have c^2 = (c - 1)^2, so that c = \dfrac 1 2. Note that for c < \dfrac 1 2, we have I'(c) > 0 and for c > \dfrac 1 2 we have I'(c) < 0. So c = \dfrac 1 2 is indeed the maximum.