Question

which of the following are injection or surjection or bijection?justify your answer
f:R gives rise to box brackets 0,infinity round brackets defined by f(x)=x^2​

Answer 1

Answer:

x  

1

​  

 and x  

2

​  

 are elements in domain (R) such that,

f(x  

1

​  

)=f(x  

2

​  

)

⇒  sinx  

1

​  

=sinx  

2

​  

 

Here, x  

1

​  

 may not be equal to x  

2

​  

 because sin0=sinπ.

So, 0 and π have same image 0.

∴  f is not an injective.

Surjectivetest:

y be the element in domain (R) such that,

f(x)=y

⇒  sinx=y

⇒  x=sin  

−1

(y)

Now, for y>1, x∈  

/

​  

R  ( Domain )

∴  f not surjective.

Bijectivetest:

f is not injective and surjective , then it is not bijective.

Step-by-step explanation:

Answer 2

Here is your answer mate....

Hope it helps you....