Set of multiples of 4 forms an ideal in Z the ring of intege

Set of multiples of 4 forms an ideal in Z, the ring of integers under usual addition and multiplication. This ideal is,

a prime ideal but not a maximal ideal.

a maximal ideal but not a prime ideal

both a prime ideal and a maximal ideal

neither a prime ideal nor a maximal ideal

Removable singularity

Simple pole

Branch point

Essential singularity

2

5

7

49

a and b

b and c

c and d

d and a

a = 1 and b = 0

a = 0 and b = 1

a = 1 and b not equal to 0

a = 0 and b not equal to 1

Solution

Question 1)

The correct option is:

(iv) neither a prime ideal nor a maximal ideal

Reason is, we know that every non-zero proper idea is prime in Z if and only if it is maximum because Z is a principle ideal domain. The prime ideals of Z are {0} and _pZ however maximal ideas of Z are only _pZ.

Question 2)

z=0 is an essential singularity of f(z) = sin(1/z).

Question 4)

We know that the particular solution of the differential equation will always be periodic. So, we only need to worry about complimentry solution. A complimentry solution contains periodic functions only when roots of auxiliary equation are complex. And a complimentry solution contains only periodic functions when roots of auxiliary equation are complex with real part = 0. This only happens when a=0 and b=1.

Therefore, y\'\'+y=cos(x)

Hence auxiliary equation becomes m^2+1=0

m=+-sqrt(-1) = +-i

Hence, complimentary solutions become y(x) = c₁cos(x)+c₂sin(x)

Hence, for a=0 and b=1, the solutions of differential equation contains only periodic funtions.

Therefore, correct option is the second option.