Answer
(a) $A=√(3/b)$
(b) $Ψ(x,0)=(1/(b-a))x+Ab/(b-a)$
(c) Maximum occurs at A
(d)$=a/b$
Work Step by Step
(a)
$1=∫_(-∞)^∞〖|Ψ(X,0)|^2 dx〗$
$1= ∫_0^b〖|Ψ(X,0)|^2 dx〗$
$ =∫_0^a〖|Ψ(X,0)|^2 dx〗+ ∫_a^b〖|Ψ(X,0)|^2 dx〗$
$ = ∫_0^a|(A(w/a))|^2 dx+ ∫_a^b|(A((b-x)/(b-a)))|^2 dx$
$ =∫_0^a〖(|A|^2 x^2)/a^2 dx+ ∫_a^b(|A|^2 (b-x)^2)/(b-a)^2 〗 dx$
$1=|A|^2/a^2 ∫_0^a〖x^2 dx 〗+|A|^2/(b-a)^2 ∫_a^b〖(b-x)^2 dx〗$
$1=|A|^2 {├ (1/a^2 )(x^3/3)┤|_0^a+ ├ 1/(b-a)^2 [-(b-x)^3/3┤|_a^b }$
$1=|A|^2 {a/3+(b-a)/3}$
$1=|A|^2 (b/3)$
$3=|A|^2 b$
$|A|^2=3/b $
$A=√(3/b)$
(b)
$Ψ(x,0)=A(x/a)$ , from $0≤x≤a$
$Ψ(x,0)=A((b-x)/(b-a))$, if $a≤x≤b$
$Ψ(x,0)=(1/(b-a))x+Ab/(b-a)$
See attachment for graph
(c)
From the graph on b, when x = a,
$Ψ(x,0)=A(x/a)=A(a/a)=A$
So, the maximum occurs at x=a
(d)
$P=∫_0^a〖|Ψ(x,0)|^2 dx$
$=|A|^2/a^2 ∫_0^a〖x^2 dx〗〗$
$=├ |A|^2/a^2 (x^3/3)┤|_0^a$
$=|A|^2 (a/3)$
$=(√(3/b))^2 (a/3)$
$=a/b$
Check cases
Suppose b=a
$P=a/b=a/a=1$
Suppose b = 2a
$P=a/b=a/2a=1/2$
And we see that both are consistent.
