Advertisements
Advertisements
प्रश्न
State with reason whether following functions have inverse h: {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
उत्तर
h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.
∴Function h is one-one.
Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y.
Thus, h is a one-one and onto function. Hence, h has an inverse.
APPEARS IN
संबंधित प्रश्न
If the function f : R → R be defined by f(x) = 2x − 3 and g : R → R by g(x) = x3 + 5, then find the value of (fog)−1 (x).
Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
Find gof and fog, if `f(x) = 8x^3` and `g(x) = x^(1/3)`
State with reason whether following functions have inverse g: {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with `f^(-1)(y) = ((sqrt(y +6) - 1)/3)`
Let f: X → Y be an invertible function. Show that f has unique inverse. (Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = IY(y) = fog2(y). Use one-one ness of f).
Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e., (f−1)−1 = f.
If f: R → R be given by `f(x) = (3 - x^3)^(1/3)` , then fof(x) is
(A) `1/(x^3)`
(B) x3
(C) x
(D) (3 − x3)
Let `f:R - {-4/3} -> R` be a function defined as `f(x) = (4x)/(3x + 4)`. The inverse of f is map g Range `f -> R -{- 4/3}`
(A) `g(y) = (3y)/(3-4y)`
(B) `g(y) = (4y)/(4 - 3y)`
(C) `g(y) = (4y)/(3 - 4y)`
(D) `g(y) = (3y)/(4 - 3y)`
Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Let f : W → W be defined as f(x) = x − 1 if x is odd and f(x) = x + 1 if x is even. Show that f is invertible. Find the inverse of f, where W is the set of all whole numbers.
If f : R → R, f(x) = x3 and g: R → R , g(x) = 2x2 + 1, and R is the set of real numbers, then find fog(x) and gof (x)
Let f: A → B and g: B → C be the bijective functions. Then (g o f)–1 is ______.
Let f: R → R be the function defined by f(x) = sin (3x+2) ∀ x ∈ R. Then f is invertible.
The composition of functions is commutative.
Every function is invertible.
If f(x) = (ax2 + b)3, then the function g such that f(g(x)) = g(f(x)) is given by ____________.
Let f : N → R : f(x) = `((2"x"−1))/2` and g : Q → R : g(x) = x + 2 be two functions. Then, (gof) `(3/2)` is ____________.
If f : R → R, g : R → R and h : R → R are such that f(x) = x2, g(x) = tan x and h(x) = log x, then the value of (go(foh)) (x), if x = 1 will be ____________.
If f(x) = `(3"x" + 2)/(5"x" - 3)` then (fof)(x) is ____________.
Let f : R – `{3/5}`→ R be defined by f(x) = `(3"x" + 2)/(5"x" - 3)` Then ____________.
If f : R → R defind by f(x) = `(2"x" - 7)/4` is an invertible function, then find f-1.
If f : R → R defined by f(x) `= (3"x" + 5)/2` is an invertible function, then find f-1.
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}
- Two neighbors X and Y ∈ I. X exercised his voting right while Y did not cast her vote in a general election - 2019. Which of the following is true?
`f : x -> sqrt((3x^2 - 1)` and `g : x -> sin (x)` then `gof : x ->`?
Domain of the function defined by `f(x) = 1/sqrt(sin^2 - x) log_10 (cos^-1 x)` is:-
Let 'D' be the domain of the real value function on Ir defined by f(x) = `sqrt(25 - x^2)` the D is :-
If f(x) = [4 – (x – 7)3]1/5 is a real invertible function, then find f–1(x).
Let `f : R {(-1)/3} → R - {0}` be defined as `f(x) = 5/(3x + 1)` is invertible. Find f–1(x).
