Advertisements
Advertisements
Question
Let \[f\left( x \right) = \frac{1}{1 - x} . \text{Then}, \left\{ f o \left( fof \right) \right\} \left( x \right)\]
Options
\[\text{x for all x} \in R\]
\[\text{x for all x} \in R - \left\{ 1 \right\}\]
\[\text{x for all x} \in R - \left\{ 0, 1 \right\}\]
none of these
Advertisements
Solution
\[\text{Domain of f}:\]
\[1 - x \neq 0\]
\[ \Rightarrow x \neq 1\]
\[\text{Domain of f} = R - \left\{ 1 \right\}\]
\[\text{Range of f}: \]
\[y = \frac{1}{1 - x}\]
\[ \Rightarrow 1 - x = \frac{1}{y}\]
\[ \Rightarrow x = 1 - \frac{1}{y}\]
\[ \Rightarrow x = \frac{y - 1}{y}\]
\[ \Rightarrow y \neq 0\]
\[\text{Range of f} = R - \left\{ 0 \right\}\]
\[So,f: R - \left\{ 1 \right\} \to R - \left\{ 0 \right\} andf: R - \left\{ 1 \right\} \to R - \left\{ 0 \right\} \]
\[\text{Range of f is not a subset of the domain of f}.\] \[\text{Domain}\left( fof \right)=\left\{ x: x\text{in domain of f and f}\left( x \right) \ \text{in domain of f} \right\}\]
\[\text{Domain}\left( fof \right)=\left\{ x: x \ in R - \left\{ 1 \right\}\text{and}\frac{1}{1 - x} \in R - \left\{ 1 \right\} \right\}\]
\[ \text{Domain}\left( fof \right)=\left\{ x: x \neq 1 \text{and}\frac{1}{1 - x} \neq 1 \right\}\]
\[\text{Domain}\left( fof \right)=\left\{ x: x \neq 1 \text{and}1 - x \neq 1 \right\}\]
\[\text{Domain}\left( fof \right)=\left\{ x: x \neq 1 \text{and}x \neq 0 \right\}\]
\[\text{Domain}\left( fof \right)=R - \left\{ 0, 1 \right\}\]
\[\left( \text{f of} \right)\left( x \right) = f\left( f\left( x \right) \right) = f\left( \frac{1}{1 - x} \right) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{1 - x}{1 - x - 1} = \frac{1 - x}{- x} = \frac{x - 1}{x}\]
\[\text{For range of f of}, x \neq 0\]
\[\text{Now,f of} : R - \left\{ 0, 1 \right\} \to R - \left\{ 0 \right\} \text{and}f: R - \left\{ 1 \right\} \to R - \left\{ 0 \right\}\]
\[\text{Range of f of is not a subset of domain of f}.\]
\[\text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \text{in domain of f of and}\left( fof \right)\left( x \right) \text{in domain of f} \right\}\]
\[\text{Domain}\left( f o\left( \text{f of} \right) \right)=\left\{ x: x \in R - \left\{ 0, 1 \right\}\text{and}\frac{x - 1}{x} \in R - \left\{ 1 \right\} \right\}\]
\[ \text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \neq 0, 1 \text{ and }\frac{x - 1}{x} \neq 1 \right\}\]
\[\text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \neq 0, 1 \text{ and }x-1 \neq x \right\}\]
\[\text{Domain}\left( f o\left( fof \right) \right)=\left\{ x: x \neq 0, 1 \text{ and }x \in R \right\}\]
\[\text{Domain}\left( f o\left( f of \right) \right)=R - \left\{ 0, 1 \right\}\]
\[\left( fo\left( fof \right) \right)\left( x \right) = f\left( \left( fof \right)\left( x \right) \right)\]
\[ = f\left( \frac{x - 1}{x} \right)\]
\[ = \frac{1}{1 - \frac{x - 1}{x}}\]
\[ = \frac{x}{x - x + 1}\]
\[ = x\]
\[\text{So},\left( fo\left( fof \right) \right)\left( x \right) = x, \text{where}x \neq 0, 1\]
So, the answer is (c).
APPEARS IN
RELATED QUESTIONS
Show that the modulus function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is –x, if x is negative.
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, 3), (b, 2), (c, 1)}
Give an example of a function which is one-one but not onto ?
Classify the following function as injection, surjection or bijection :
f : Q → Q, defined by f(x) = x3 + 1
Show that the function f : R − {3} → R − {2} given by f(x) = `(x-2)/(x-3)` is a bijection.
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself is one-one, onto or bijective : h(x) = x2
If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.
Let f : R → R and g : R → R be defined by f(x) = x2 and g(x) = x + 1. Show that fog ≠ gof.
Find fog and gof if : f (x) = x+1, g(x) = `e^x`
.
If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
Find f −1 if it exists : f : A → B, where A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {a, b, c}.
If f : C → C is defined by f(x) = x2, write f−1 (−4). Here, C denotes the set of all complex numbers.
Let C denote the set of all complex numbers. A function f : C → C is defined by f(x) = x3. Write f−1(1).
If f : R → R is defined by f(x) = 3x + 2, find f (f (x)).
If the mapping f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, then write fog. [NCERT EXEMPLAR]
Let
f : R → R be given by
\[f\left( x \right) = \left[ x^2 \right] + \left[ x + 1 \right] - 3\]
where [x] denotes the greatest integer less than or equal to x. Then, f(x) is
(d) one-one and onto
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 ) :}`
A function f from the set of natural numbers to the set of integers defined by
\[f\left( n \right)\begin{cases}\frac{n - 1}{2}, & \text{when n is odd} \\ - \frac{n}{2}, & \text{when n is even}\end{cases}\]
If \[g \left( f \left( x \right) \right) = \left| \sin x \right| \text{and} f \left( g \left( x \right) \right) = \left( \sin \sqrt{x} \right)^2 , \text{then}\]
Let
\[A = \left\{ x \in R : x \geq 1 \right\}\] The inverse of the function,
\[f : A \to A\] given by
\[f\left( x \right) = 2^{x \left( x - 1 \right)} , is\]
Let
\[f : [2, \infty ) \to X\] be defined by
\[f\left( x \right) = 4x - x^2\] Then, f is invertible if X =
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)
Write about strcmp() function.
Let f: R → R be the function defined by f(x) = 4x – 3 ∀ x ∈ R. Then write f–1
Let N be the set of natural numbers and the function f: N → N be defined by f(n) = 2n + 3 ∀ n ∈ N. Then f is ______.
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
{(a, b): a is a person, b is an ancestor of a}
Let C be the set of complex numbers. Prove that the mapping f: C → R given by f(z) = |z|, ∀ z ∈ C, is neither one-one nor onto.
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)}
The function f: R → R defined as f(x) = x3 is:
Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible outcomes of the throw every time belongs to set {1,2,3,4,5,6}. Let A be the set of players while B be the set of all possible outcomes.
A = {S, D}, B = {1,2,3,4,5,6}
- Raji wants to know the number of functions from A to B. How many number of functions are possible?
Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by y = x2.
Answer the following questions using the above information.
- Let f: {1,2,3,....} → {1,4,9,....} be defined by f(x) = x2 is ____________.
Function f: R → R, defined by f(x) = `x/(x^2 + 1)` ∀ x ∈ R is not
If f; R → R f(x) = 10x + 3 then f–1(x) is:
The domain of the function `cos^-1((2sin^-1(1/(4x^2-1)))/π)` is ______.
If log102 = 0.3010.log103 = 0.4771 then the number of ciphers after decimal before a significant figure comes in `(5/3)^-100` is ______.
If A = {x ∈ R: |x – 2| > 1}, B = `{x ∈ R : sqrt(x^2 - 3) > 1}`, C = {x ∈ R : |x – 4| ≥ 2} and Z is the set of all integers, then the number of subsets of the set (A ∩ B ∩ C) C ∩ Z is ______.
Let A = {1, 2, 3, ..., 10} and f : A `rightarrow` A be defined as
f(k) = `{{:(k + 1, if k "is odd"),( k, if k "is even"):}`.
Then the number of possible functions g : A `rightarrow` A such that gof = f is ______.
The trigonometric equation tan–1x = 3tan–1 a has solution for ______.
