Advertisements
Advertisements
प्रश्न
Let A = ℝ − {3}, B = ℝ − {1}. Let f : A → B be defined by \[f\left( x \right) = \frac{x - 2}{x - 3}, \forall x \in A\] Show that f is bijective. Also, find
(i) x, if f−1(x) = 4
(ii) f−1(7)
Advertisements
उत्तर
We have,
A = R \[-\] {3} and B = R \[-\] {1}
The function f : A \[\to\] B defined by f(x) = \[\frac{x - 2}{x - 3}\]
\[\text { Let }x, y \text { in A such that }f\left( x \right) = f\left( y \right) . \text { Then }, \]
\[\frac{x - 2}{x - 3} = \frac{y - 2}{y - 3}\]
\[ \Rightarrow xy - 3x - 2y + 6 = xy - 2x - 3y + 6\]
\[ \Rightarrow - x = - y\]
\[ \Rightarrow x = y\]
\[ \therefore\text { f is one - one } .\]
\[\text { Let y } \in B . \text { Then, y } \neq 1 . \]
\[\text { The function f is onto if there exists } x \text { in A such that f }\left( x \right) = y . \]
\[\text {Now,}\]
\[f\left( x \right) = y\]
\[ \Rightarrow \frac{x - 2}{x - 3} = y\]
\[ \Rightarrow x - 2 = xy - 3y\]
\[ \Rightarrow x - xy = 2 - 3y\]
\[ \Rightarrow x\left( 1 - y \right) = 2 - 3y\]
\[ \Rightarrow x = \frac{2 - 3y}{1 - y} \in A \left[ y \neq 1 \right]\]
\[\text { Thus, for any } y \text { in B, there exists } \frac{2 - 3y}{1 - y} \text { in A such that }\]
\[f\left( \frac{2 - 3y}{1 - y} \right) = \frac{\left( \frac{2 - 3y}{1 - y} \right) - 2}{\left( \frac{2 - 3y}{1 - y} \right) - 3} = \frac{2 - 3y - 2 + 2y}{2 - 3y - 3 + 3y} = \frac{- y}{- 1} = y\]
\[ \therefore \text { f is onto } .\]
\[\text { So, f is one - one and onto fucntion } . \]
\[\text { Now }, \]
\[\text { As,} x = \left( \frac{2 - 3y}{1 - y} \right)\]
\[\text { So }, f^{- 1} \left( x \right) = \left( \frac{2 - 3x}{1 - x} \right) = \frac{3x - 2}{x - 1}\]
Therefore,f is bijective.
(i) \[f^{- 1} \left( x \right) = 4\]
\[ \Rightarrow \frac{3x - 2}{x - 1} = 4\]
\[ \Rightarrow 3x - 2 = 4x - 4\]
\[ \Rightarrow x = 2\]
\[f^{- 1} \left( x \right) = \frac{3x - 2}{x - 1}\]
\[ \Rightarrow f^{- 1} \left( 7 \right) = \frac{3\left( 7 \right) - 2}{7 - 1} = \frac{19}{6}\]
APPEARS IN
संबंधित प्रश्न
Find the number of all onto functions from the set {1, 2, 3, ..., n} to itself.
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = x3 − x
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself is one-one, onto or bijective : h(x) = x2
Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.
Give examples of two one-one functions f1 and f2 from R to R, such that f1 + f2 : R → R. defined by (f1 + f2) (x) = f1 (x) + f2 (x) is not one-one.
Find gof and fog when f : R → R and g : R → R is defined by f(x) = 2x + x2 and g(x) = x3
Find gof and fog when f : R → R and g : R → R is defined by f(x) = x and g(x) = |x| .
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, z ∈ N. Show that ho (gof) = (hog) of.
Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but gis not injective.
Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (fh).
If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x|- x, ∀x∈R" .Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).
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 A = R - {3} and B = R - {1}. Consider the function f : A → B defined by f(x) = `(x-2)/(x-3).`Show that f is one-one and onto and hence find f-1.
[CBSE 2012, 2014]
If f : R → (0, 2) defined by `f (x) =(e^x - e^(x))/(e^x +e^(-x))+1`is invertible , find f-1.
If f : A → A, g : A → A are two bijections, then prove that fog is a surjection ?
If f : C → C is defined by f(x) = x2, write f−1 (−4). Here, C denotes the set of all complex numbers.
The function
f : A → B defined by
f (x) = - x2 + 6x - 8 is a bijection if
A function f from the set of natural numbers to integers defined by
`{([n-1]/2," when n is odd" is ),(-n/2,when n is even ) :}`
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 =
If a function\[f : [2, \infty )\text{ to B defined by f}\left( x \right) = x^2 - 4x + 5\] is a bijection, then B =
The inverse of the function
\[f : R \to \left\{ x \in R : x < 1 \right\}\] given by
\[f\left( x \right) = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] is
The domain of the function f: R → R defined by f(x) = `sqrt(x^2 - 3x + 2)` is ______
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
f = {(1, 4), (1, 5), (2, 4), (3, 5)}
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
g = {(1, 4), (2, 4), (3, 4)}
A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, the highest ever
Let I be the set of all citizens of India who were eligible to exercise their voting right in the general election held in 2019. A relation ‘R’ is defined on I as follows:
R = {(V1, V2) ∶ V1, V2 ∈ I and both use their voting right in the general election - 2019}
- Three friends F1, F2, and F3 exercised their voting right in general election-2019, then which of the following is true?
Function f: R → R, defined by f(x) = `x/(x^2 + 1)` ∀ x ∈ R is not
Consider a function f: `[0, pi/2] ->` R, given by f(x) = sinx and `g[0, pi/2] ->` R given by g(x) = cosx then f and g are
The domain of the function `cos^-1((2sin^-1(1/(4x^2-1)))/π)` is ______.
Let f(n) = `[1/3 + (3n)/100]n`, where [n] denotes the greatest integer less than or equal to n. Then `sum_(n = 1)^56f(n)` is equal to ______.
