Let * be binary operation defined on R by a * b = 1 + a b , ∀ a , b ∈ R . Then the operation * is.

Commutative but not associative.

Associative but not commutative.

Neither associative nor commutative.

Both commutative and associative.

Let R be a relation on the set N of all natural numbers, defined by a R b ⇔ a is a factor of b . Then, R is

reflexive and transitive but not symmetric

reflexive and symmetric but not transitive

symmetric and transitive but not reflexive

Show that the function f : R → R defined by f x = x x 2 + 1 , ∀ x ∈ ℝ is neither one-one nor onto. Also, if g : R → R is defined as g ( x ) = 2 x - 1 , find fog ( x ) .

Given, y = x x 2 + 1 ⇒ y x 2 − x + y = 0 Here a = y , b = - 1 and c = y ∴ x = − ( − 1 ) ± 1 − 4 y 2 2 y Clearly, for every value of y , x will have two different values . So the function is many-one not one - one. Since 1 − 4 y 2 ≥ 0 ⇒ ( 1 + 2 y ) ( 1 − 2 y ) ≥ 0 ⇒ − 1 2 ≤ y ≤ 1 2 That means no matter what is x , y belongs to the interval − 1 2 , 1 2 . Hence, the function is not onto. Now, f o g x = f g x = f 2 x - 1 = 2 x - 1 2 x - 1 2 + 1 = 2 x + 1 4 x 2 - 4 x + 1 + 1 = 2 x + 1 2 2 x 2 - 2 x + 1