Answer:
{ c ∣ c ≠ 2
, 12, -1, 0, c ∈ R }
Step-by-step explanation:
Considering the set
![\left\{\left(2,\:8\right),\left(12,\:3\right),\:\left(c,\:4\right),\:\left(-1,\:8\right),\:\left(0,\:3\right)\right\}](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cleft%282%2C%5C%3A8%5Cright%29%2C%5Cleft%2812%2C%5C%3A3%5Cright%29%2C%5C%3A%5Cleft%28c%2C%5C%3A4%5Cright%29%2C%5C%3A%5Cleft%28-1%2C%5C%3A8%5Cright%29%2C%5C%3A%5Cleft%280%2C%5C%3A3%5Cright%29%5Cright%5C%7D)
As we know that
- A function is said to be a relation if every x-value has one and only one y-value.
So, the value of c must not be equal to 2, 12, -1, 0 i.e. c ≠ 2, 12, -1, 0
Therefore,
{ c ∣ c ≠ 2
, 12, -1, 0, c ∈ R }