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Python
Observe the following code for a function bisection: def bisection(f, a, b, N): for i in range(N): m = (a+b)/2 if f(a)*f(m)<0: b = m elif f(m)*f(b)<0: a = m return (a+b)/2 Below are five pieces of code, each of which calls this bisection function, and some of which depend on some variable y, that has already been assigned to some fixed integer. Match each piece of code with the value that it produces given that each answer can only be used once.
bisection(lambda x: x**2-14*x+y, -11, 7, 100)
(lambda y: bisection(lambda x: x**2-14*x+y, 7, 25, 100))(32)
bisection(lambda x: 10000*x-111231, 7, 25, 100)
bisection(lambda x: x**2-14*x+y, 7, 25, 5)
bisection(lambda x: x**2-14*x+y, 7, 25, 100).
The following options are: 11.242640687119284, 11.1231, 11.21875, 11.123105625617661 and 2.7573593128807152
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