In a $k$-way set associative cache,main memory block mapping in range?
Number of sets in cache = v. So, main memory block j will be mapped to
  set (j mod v), which will be any one of the cache lines from (j mod v)
  * k to (j mod v) * k + (k-1). (Associativity plays no role in mapping- k-way associativity means there are k spaces for a block and hence
  reduces the chances of replacement.)


This is a simple concept of k-way set associate mapping.

To understand it better I would like to take an example:

Example:


2-way set associative 
4 blocks
Lines of a set are placed in sequence one after another (as in the
question)
The main memory blocks are numbered 0 onwards (as in the
question)


Empty 2-way set associative Cache Memory :

╔═══╤═══════════════╗
║   │     Cache     ║
╠═══╪═══════╤═══════╣
║   │ Set 1 │ Set 2 ║
╟───┼───────┼───────╢
║ 0 │   -   │   -   ║
╟───┼───────┼───────╢
║ 1 │   -   │   -   ║
╟───┼───────┼───────╢
║ 2 │   -   │   -   ║
╟───┼───────┼───────╢
║ 3 │   -   │   -   ║
╚═══╧═══════╧═══════╝


Memory reference (data asked for processing) in the order:

4, 5, 9, 7 

Reference 4:

4 % 4 = 0 (goes to Block 0 Set 1) Cache miss as it doesn't already present in the cache

As it had 4 blocks, we mod it the reference address to know which location of cache should I put it.

╔═══╤═══════════════╗
║   │     Cache     ║
╠═══╪═══════╤═══════╣
║   │ Set 1 │ Set 2 ║
╟───┼───────┼───────╢
║ 0 │   4   │   -   ║
╟───┼───────┼───────╢
║ 1 │   -   │   -   ║
╟───┼───────┼───────╢
║ 2 │   -   │   -   ║
╟───┼───────┼───────╢
║ 3 │   -   │   -   ║
╚═══╧═══════╧═══════╝


Reference 5:

4 % 5 = 1 (goes to Block 1 Set 1) Cache miss

╔═══╤═══════════════╗
║   │     Cache     ║
╠═══╪═══════╤═══════╣
║   │ Set 1 │ Set 2 ║
╟───┼───────┼───────╢
║ 0 │   4   │   -   ║
╟───┼───────┼───────╢
║ 1 │   5   │   -   ║
╟───┼───────┼───────╢
║ 2 │   -   │   -   ║
╟───┼───────┼───────╢
║ 3 │   -   │   -   ║
╚═══╧═══════╧═══════╝


Reference 9:

4 % 9 = 1 (goes to Block 1 Set 2) Cache miss

╔═══╤═══════════════╗
║   │     Cache     ║
╠═══╪═══════╤═══════╣
║   │ Set 1 │ Set 2 ║
╟───┼───────┼───────╢
║ 0 │   4   │   -   ║
╟───┼───────┼───────╢
║ 1 │   5   │   9   ║
╟───┼───────┼───────╢
║ 2 │   -   │   -   ║
╟───┼───────┼───────╢
║ 3 │   -   │   -   ║
╚═══╧═══════╧═══════╝


Reference 4:

4 % 7 = 3 (goes to Block 3 Set 1) Cache miss

╔═══╤═══════════════╗
║   │     Cache     ║
╠═══╪═══════╤═══════╣
║   │ Set 1 │ Set 2 ║
╟───┼───────┼───────╢
║ 0 │   4   │   -   ║
╟───┼───────┼───────╢
║ 1 │   5   │   9   ║
╟───┼───────┼───────╢
║ 2 │   -   │   -   ║
╟───┼───────┼───────╢
║ 3 │   7   │   -   ║
╚═══╧═══════╧═══════╝


Visualization of Cache memory for you undersatnding:

╔═════════════════════════════════════════════════════╗
║                    Cache Memory                     ║
╠════════════════╤════════════╤═══════════════════════╣
║ Memory address │ References │                       ║
╟────────────────┼────────────┼───────────────────────╢
║       0        │     4      │                       ║
╟────────────────┼────────────┤ Cache line 0 elements ║
║       1        │     -      │                       ║
╟────────────────┼────────────┼───────────────────────╢
║       2        │     5      │                       ║
╟────────────────┼────────────┤ Cache line 1 elements ║
║       3        │     9      │                       ║
╟────────────────┼────────────┼───────────────────────╢
║       4        │     -      │                       ║
╟────────────────┼────────────┤ Cache line 2 elements ║
║       5        │     -      │                       ║
╟────────────────┼────────────┼───────────────────────╢
║       6        │     7      │                       ║
╟────────────────┼────────────┤ Cache line 3 elements ║
║       7        │     -      │                       ║
╚════════════════╧════════════╧═══════════════════════╝


Both of the two set of cache line 1 is full, therefore calculation the location of first and last element of line gives us the answer.

Lets now check if the answer is option (1) as you mentioned


  
  (j mod v) ∗ k  to  (j mod v) ∗ k + (k − 1)
  


Given:

v = 4 (four blocks)

k = 2 (two-way)

First element of cache line 1 = (j mod v) ∗ k = (5 mod 4) ∗ 2 = 2 yes, its located at memory location 2.

Last element of cache line 1 = (j mod v) ∗ k + (k − 1) = (9 mod 4) ∗ 2 + (2 - 1) = 3 yes, its located at memory location 3.

Done.

Answer: (j mod v) ∗ k to (j mod v) ∗ k + (k − 1)