Question

Let the domain of x be the set {1}. Which of the following functions are equal to 1

(a) f(x)=(x)^2,g(x)=x
(b) f(a)=x,g(x)=1–x
(c) f(x)=(x)^2+x+2,g(x)=(x+1)^2
(d) none of these

(a) f(x) = x^2,g(x)=x is Correct Answer

Please Give Full Explaination & Solution​

Answer 1

It gives three cases

Case I:

When f(x)=1 is true.

In this case remaining two are false i.e. f(y)=1 and f(z)=2

This means x and y have the same image, so f(x) is not an injective, which is a contradiction as the function is injective.

Case II:

When f(y)



=1 is true then the remaining statements are false.

∴f(x)



=1 and f(z)=2

⟹ Either of the x,y mapped to 2

Or both x and y are not mapped to 1.

So, either both associate to 2 or 3,

Thus, it is not injective which is a contradiction.

Thus, both the cases I and II are not true.

Case III:

When f(z)



=2 is true then remaining statements are false

∴ If f(x)



=1 and f(y)=1

But f is injective

Thus, we have f(x)=2,f(y)=1 and f(z)=3

Hence, f

−1

(1)=y