Question

# A function whose graph is symmetrical about the origin is given by -

Hint:

## The correct answer is:

### To find the function from the given options whose graph is symmetrical about the origin.

$f(x+y)=f(x)+f(y)$

$f(x)$ is of the form $f(x)=λx$ where λ is a constant which represents a straight line passing through the origin, having a slope λ.

Hence, the function f(x+y)=f(x)+f(y) is symmetric about the origin.

### Related Questions to study

### The minimum value of is

### The minimum value of is

### is -

Hence, the given function is many one and onto.

### is -

Hence, the given function is many one and onto.

### If f(x) is a polynomial function satisfying the condition f(x). f(1/x) = f(x) + f(1/x) and f(2) = 9 then -

### If f(x) is a polynomial function satisfying the condition f(x). f(1/x) = f(x) + f(1/x) and f(2) = 9 then -

### Fill in the blank with the appropriate transition.

The movie managed to fetch decent collections ______ all the negative reviews it received.

### Fill in the blank with the appropriate transition.

The movie managed to fetch decent collections ______ all the negative reviews it received.

### If R be a relation '<' from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e. (a, b) R iff a < b, then is

Values of are {(3, 3), (3, 5), (5, 3), (5, 5)}

### If R be a relation '<' from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e. (a, b) R iff a < b, then is

Values of are {(3, 3), (3, 5), (5, 3), (5, 5)}

### Which one of the following relations on R is equivalence relation

### Which one of the following relations on R is equivalence relation

### Let R = {(x, y) : x, y A, x + y = 5} where A = {1, 2, 3, 4, 5} then

Hence, the given relation is not reflexive, symmetric and not transitive

### Let R = {(x, y) : x, y A, x + y = 5} where A = {1, 2, 3, 4, 5} then

Hence, the given relation is not reflexive, symmetric and not transitive

### Let be a relation defined by Then R is

Hence, the given relation is Reflexive, transitive but not symmetric.

### Let be a relation defined by Then R is

Hence, the given relation is Reflexive, transitive but not symmetric.

### The relation R defined in N as aRb b is divisible by a is

Hence, the given relations is reflexive but not symmetric.

### The relation R defined in N as aRb b is divisible by a is

Hence, the given relations is reflexive but not symmetric.