If the function f : - 1 , 1 → 0 , 2 , defined by f x = a x + b is a surjection, then find a and b .

Given, f : - 1 , 1 → 0 , 2 and f x = a x + b is a surjection. Since, f x = a x + b is a surjection which means the range of the f is equal to its co-domain. ⇒ Range of f = Co-domain of f = 0 , 2 Now, f - 1 = a - 1 + b = b - a = 0 . . . i and f 1 = a 1 + b = a + b = 2 . . . ii From i and ii , we have b - a + a + b = 0 + 2 ⇒ 2 b = 2 ⇒ b = 1 By substituting the value of b in i , we have 1 - a = 0 ⇒ a = 1 ∴ a = 1 and b = 1

If the function f : - 1 , 1 → 0 , 2 , defined by f x = a x + b is a surjection, then find a and b .

Given, f : - 1 , 1 → 0 , 2 and f x = a x + b is a surjection. Since, f x = a x + b is a surjection which means the range of the f is equal to its co-domain. ⇒ Range of f = Co-domain of f = 0 , 2 Now, f - 1 = a - 1 + b = b - a = 0 . . . i and f 1 = a 1 + b = a + b = 2 . . . ii From i and ii , we have b - a + a + b = 0 + 2 ⇒ 2 b = 2 ⇒ b = 1 By substituting the value of b in i , we have 1 - a = 0 ⇒ a = 1 ∴ a = 1 and b = 1

If f x = cos log x , then show that f 1 x · f 1 y - 1 2 f x y + f x y = 0 .

Given, f x = cos log x and we need to show f 1 x · f 1 y - 1 2 f x y + f x y = 0 . Since, f x = cos log x then f 1 x = cos log 1 x = cos - log x , f 1 y = cos log 1 y = cos - log y f x y = cos log x y = cos log x - log y , f x y = cos log x y = cos log x + log y At LHS , f 1 x · f 1 y - 1 2 f x y + f x y = cos log x · cos log y - 1 2 cos log x - log y + cos log x + log y = cos log x · cos log y - cos log x · cos log y = 0 ∴ f 1 x · f 1 y - 1 2 f x y + f x y = 0

If f y = y 1 - y 2 and g y = y 1 + y 2 , then show that f o g y = y .

Given, f y = y 1 - y 2 and g y = y 1 + y 2 Here, f o g y = f g y = f y 1 + y 2 ⇒ y 1 + y 2 1 - y 1 + y 2 2 = y 1 + y 2 1 + y 2 - y 2 1 + y 2 ⇒ y 1 + y 2 × 1 + y 2 1 = y Hence, f o g ( y ) = y proved.