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 function f in `A=R-{2/3} ` defined as `f(x)=(4x+3)/(6x-4)` is one-one and onto hence find f-1
Check the injectivity and surjectivity of the following function:
f : Z → Z given by f(x) = x3
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.
Show that the Signum Function f : R → R, given by `f(x) = {(1", if" x > 0), (0", if" x = 0), (-1", if" x < 0):}` is neither one-one nor onto.
Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but gis not injective.
(Hint: Consider f(x) = x and g(x) =|x|)
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)}
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 ∈ 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.)
Let A = {−1, 0, 1} and f = {(x, x2) : x ∈ A}. Show that f : A → A is neither one-one nor onto.
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.
If f : A → B and g : B → C are onto functions, show that gof is a onto function.
Find fog and gof if : f (x) = x2 g(x) = cos x .
If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe each of the following functions:
(1) fog
(2) gof
(3) fof
(4) f2
Also, show that fof ≠ f2
` if f : (-π/2 , π/2)` → R and g : [−1, 1]→ R be defined as f(x) = tan x and g(x) = `sqrt(1 - x^2)` respectively, describe fog and gof.
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 : C → C is defined by f(x) = x4, write f−1 (1).
Write the domain of the real function
`f (x) = sqrt([x] - x) .`
If f : R → R be defined by f(x) = (3 − x3)1/3, then find fof (x).
If f : R → R is defined by f(x) = 3x + 2, find f (f (x)).
Let f, g : R → R be defined by f(x) = 2x + l and g(x) = x2−2 for all x
∈ R, respectively. Then, find gof. [NCERT EXEMPLAR]
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
The function \[f : [0, \infty ) \to \text {R given by } f\left( x \right) = \frac{x}{x + 1} is\]
Which of the following functions form Z to itself are bijections?
The function f : [-1/2, 1/2, 1/2] → [-π /2,π/2], defined by f (x) = `sin^-1` (3x - `4x^3`), is
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
If \[f : R \to R\] is given by \[f\left( x \right) = x^3 + 3, \text{then} f^{- 1} \left( x \right)\] is equal to
Mark the correct alternative in the following question:
Let f : R \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\] R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\] Then,
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 ______.
Let f: R → R be defined by f(x) = 3x – 4. Then f–1(x) is given by ______.
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)}
Let f: R – `{3/5}` → R be defined by f(x) = `(3x + 2)/(5x - 3)`. Then ______.
Let A = {0, 1} and N be the set of natural numbers. Then the mapping f: N → A defined by f(2n – 1) = 0, f(2n) = 1, ∀ n ∈ N, is onto.
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?
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 : N → R be defined by f(x) = x2. Range of the function among the following is ____________.
`x^(log_5x) > 5` implies ______.
Find the domain of sin–1 (x2 – 4).
The trigonometric equation tan–1x = 3tan–1 a has solution for ______.
