Advertisements
Advertisements
Question
Let A = R − (2) and B = R − (1). If f: A ⟶ B is a function defined by`"f(x)"=("x"-1)/("x"-2),` how that f is one-one and onto. Hence, find f−1.
Solution
A =R − {2}, B = R − {1}
f: A → B is defined as `"f"("x") = ("x"-1)/("x"-2)`.
Let x, y ∈ A such that f (x) = f (y).
⇒ `("x"-1)/("x"-2) = ("y"-1)/("y"-2)`
⇒ ( x -1) (y - 2) = (x - 2) (y - 1)
⇒ xy - 2x - y + 2 = xy - x - 2y + 2
⇒ -2x - y = -x - 2y
⇒ 2x - x = 2y - y
⇒ x = y
∴ f is one-one.
Let y ∈B = R − {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈A such that f(x) = y.
Now,
f (x) = y
⇒`("x"-1)/("x"-2) = "y"`
⇒ x - 1 = y (x - 2)
⇒ x (1 - y) = 1 - 2y
⇒ `"x" = (1-2"y")/(1-"y")∈ "A"` .........[y ≠ 1]
Thus, for any y ∈ B, there exists `"x" = (1-2"y")/(1-"y")` ∈ A such that
`"f"((1-2"y")/(1-"y")) =((1-2"y")/(1-"y")-1)/((1-2"y")/(1-"y") - 2) = (1-2"y"-1+"y")/(1-2"y"-2+2"y") = (-"y")/-1 = "y"`
Therefore, f is onto.
Hence, function f is one-one and onto.
`"f"^-1("x") = (1-2"x")/(1-"x")`
APPEARS IN
RELATED QUESTIONS
Check the injectivity and surjectivity of the following function:
f: Z → Z given by f(x) = x2
Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.
Let S = {a, b, c} and T = {1, 2, 3}. Find F−1 of the following functions F from S to T, if it exists.
F = {(a, 2), (b, 1), (c, 1)}
Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, g: A → B be functions defined by f(x) = x2 − x, x ∈ A and g(x) = `2|x - 1/2|- 1, x in A`. Are f and g equal?
Justify your answer. (Hint: One may note that two functions f: A → B and g: A → B such that f(a) = g(a) ∀ a ∈ A are called equal functions).
Prove that the function f : N → N, defined by f(x) = x2 + x + 1, is one-one but not onto
If f : R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.
Show that the exponential function f : R → R, given by f(x) = ex, is one-one but not onto. What happens if the co-domain is replaced by`R0^+` (set of all positive real numbers)?
If A = {1, 2, 3}, show that a one-one function f : A → A must be onto.
Find gof and fog when f : R → R and g : R → R is defined by f(x) = 2x + x2 and g(x) = x3
Find fog and gof if : f (x) = x+1, g(x) = `e^x`
.
Let
f (x) =`{ (1 + x, 0≤ x ≤ 2) , (3 -x , 2 < x ≤ 3):}`
Find fof.
Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) = cat. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1and show that (gof)−1 = f −1o g−1
Let f : R `{- 4/3} `- 43 →">→ R be a function defined as f(x) = `(4x)/(3x +4)` . Show that f : R - `{-4/3}`→ Rang (f) is one-one and onto. Hence, find f -1.
Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.
If f : R → R defined by f(x) = 3x − 4 is invertible, then write f−1 (x).
If f : R → R, g : R → are given by f(x) = (x + 1)2 and g(x) = x2 + 1, then write the value of fog (−3).
If f(x) = 4 −( x - 7)3 then write f-1 (x).
Let
\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = B\] Then, the mapping\[f : A \to \text{B given by} f\left( x \right) = x\left| x \right|\] is
If the function\[f : R \to \text{A given by} f\left( x \right) = \frac{x^2}{x^2 + 1}\] is a surjection, then A =
The function
\[f : R \to R\] defined by\[f\left( x \right) = \left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)\]
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto
The function f : [-1/2, 1/2, 1/2] → [-π /2,π/2], defined by f (x) = `sin^-1` (3x - `4x^3`), is
If \[F : [1, \infty ) \to [2, \infty )\] is given by
\[f\left( x \right) = x + \frac{1}{x}, then f^{- 1} \left( x \right)\]
Let X = {1, 2, 3}and Y = {4, 5}. Find whether the following subset of X ×Y are function from X to Y or not
h = {(1,4), (2, 5), (3, 5)}
If f(x) = (4 – (x – 7)3}, then f–1(x) = ______.
The function f : A → B defined by f(x) = 4x + 7, x ∈ R is ____________.
Let f : R → R be a function defined by f(x) `= ("e"^abs"x" - "e"^-"x")/("e"^"x" + "e"^-"x")` then f(x) is
If `f : R -> R^+ U {0}` be defined by `f(x) = x^2, x ∈ R`. The mapping is
A function f : [– 4, 4] `rightarrow` [0, 4] is given by f(x) = `sqrt(16 - x^2)`. Show that f is an onto function but not a one-one function. Further, find all possible values of 'a' for which f(a) = `sqrt(7)`.
The trigonometric equation tan–1x = 3tan–1 a has solution for ______.
