Test: Introduction To Functions - JEE MCQ

Here, f(1) = 8, f(2) = 7, f(3) = 6.

Since, different points of domain have the different f-image in the range, therefore f is a one-one function.

Clearly, mappings given in options (a), (b), (c) and (d) satisfy the given conditions and are both one-one and onto.

F(n) = n+2 if n is odd
so if n=1, then f(1)= 3
if n=3 , f(3)=5
On the other hand
F(n)= n+3 if n is even
if n= 2,f(2)=5
if n=4 ,f(4)= 7
Clearly f(3) = f(2) = 5
but 3 does not equals to 3, so it is not one one i.e. injective also 1€N
 i.e. codomain
but here 1€ range does not equals to codomain so it is not surjective.

 f: R → R is defined as f(x) = x⁴
Let x, y ∈ R such that f(x) = f(y).
=> x⁴ = y⁴
=> x =+-y
Therefore, f(x¹) = f(x²) does not imply that x¹=x².
For instance,f(1) = f(-1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.

Suppose x1 and x2 are real numbers such that f(x1) = f(x2). (We need to show x1 = x2 .)
5x1  = 5x2 
Dividing by 5 on both sides gives
x1 = x2  (function is one - one)
Let y ∈R. (We need to show that  x in R such that f(x) = y.)
If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. It follows that
f(x) = 5((y + 2)/5)          
by the substitution and the definition of f
       = y + 2 
       = y                
by basic algebra
Hence, f is onto.

Let x₁, x₂ ∈ A and let f(x₁) = f(x₂)

or x₁ = x₂ So,f is one-one.
To find whether f is onto or not, first let us find the range of f.
Let 
x is defined if y≠1, i.e., the range of f is R−{1} which is also the co-domain of f. Also, for no value of y, x can be 3, i.e., if we put 3
= x = then 3y − 3 = 3y − 2 or −3 = −2
which is not possible. Hence, f is onto.

Since, the given function has minimum value 75 which is attained at x = 2 and maximum value 89 which is attained at x = 3. Hence, the range of f is [75, 89].

f(x)=f(−x). So, f is many-one
Also 

So, f is into

The function f(x) = x² + 2, x ∈ (−∞,∞) is not injective as f(1) = f(−1) but 1 ≠ −1.
The function f(x) = (x − 4)(x − 5), x ∈(−∞,∞) is not one-one as f(4) = f(5), but 4 ≠ 5
The function,

is also not injective as f(1) = f(−1), but 1 ≠ −1.
For the function, f(x) = |x + 2|, x ∈[−2,∞).
Let f(x) = f(y), x , y ∈ [−2,∞) ⇒|x + 2| = |y + 2| ⇒ x + 2 = y + 2 ⇒ x = y
So, fis an injection.

The function f(x) = 2x - 1 is surjective.

Explanation:

A function is surjective (also called onto) if for every element y in the codomain, there exists an element x in the domain such that f(x) = y.

In this case, the function f(x) = 2x - 1 is a linear function. For any given y value, we can find an x value such that f(x) = y:

y = 2x - 1
x = (y + 1) / 2

So, for any y value, there exists an x value that satisfies the equation. Therefore, the function is surjective.

Graphically, y = f(x) = (x − 1)(x − 2)(x − 3), which is clearly many-one and onto.

(i) f(1)=1,f(2)=2,f(3)=3
(ii) f(1)=1,f(2)=3,f(3)=2
(iii) f(1)=2,f(2)=3,f(3)=1
(iv) f(1)=2,f(2)=1,f(3)=3
(v) f(1)=3,f(2)=1,f(3)=2
(vi) f(1)=3,f(2)=2,f(3)=1
Since, f is onto, all elements of {1,2,3} have unique pre-image.
The above cases are possible.
Since, every element 1,2,3 has either of image 1,2,3 and that image is unique.
∴  f is one-one.
∴ Function f:A→A is one-one.

The total number of bijective functions from a set A containing 106 elements to itself is (106)! .

We are tasked with analyzing the function f defined as f(x) = x³, where Z is the set of integers. We need to determine whether the function is one-one (injective), onto (surjective), or neither.

Step 1: Check if the function is one-one (injective)
A function is one-one if f(a) = f(b) implies a = b.

For f(x) = x³, suppose f(a) = f(b). Then:
a³ = b³.
Taking the cube root on both sides (which is valid for integers), we get:
a = b.

Thus, f(x) = x³ is one-one.

Step 2: Check if the function is onto (surjective)
A function is onto if every element in the codomain (Z) has a pre-image in the domain (Z).

For f(x) = x³, we need to check if for every y ∈ Z, there exists an x ∈ Z such that x³ = y.
However, not all integers y can be expressed as the cube of an integer x. For example:

• If y = 2, there's no integer x such that x³ = 2.

• Similarly, if y = -3, there's no integer x such that x³ = -3.

Thus, f(x) = x³ is not onto because not every integer in the codomain has a pre-image in the domain.

Step 3: Conclusion
From the above analysis:
The function f(x) = x³ is one-one but not onto.

Final Answer:
d) One-one but not onto.

Hence, f(x) is constant function.