Advertisements
Advertisements
प्रश्न
State with reason whether following functions have inverse
f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
उत्तर
f: {1, 2, 3, 4} → {10}defined as:
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10
∴f is not one-one.
Hence, function f does not have an inverse.
APPEARS IN
संबंधित प्रश्न
Let f : W → W be defined as
`f(n)={(n-1, " if n is odd"),(n+1, "if n is even") :}`
Show that f is invertible a nd find the inverse of f. Here, W is the set of all whole
numbers.
Let f, g and h be functions from R to R. Show that
`(f + g)oh = foh + goh`
`(f.g)oh = (foh).(goh)`
Find gof and fog, if f(x) = |x| and g(x) = |5x - 2|
if f(x) = `(4x + 3)/(6x - 4), x ≠ 2/3` show that fof(x) = x, for all x ≠ 2/3 . What is the inverse of f?
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+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f−1 of given f by `f^(-1) (y) = sqrt(y - 4)` where R+ is the set of all non-negative real numbers.
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).
Consider f: {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c. Find f−1 and show that (f−1)−1 = f.
Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e., (f−1)−1 = f.
Consider f: `R_+ -> [-5, oo]` given by `f(x) = 9x^2 + 6x - 5`. Show that f is invertible with `f^(-1) (y) ((sqrt(y + 6)-1)/3)`
Hence Find
1) `f^(-1)(10)`
2) y if `f^(-1) (y) = 4/3`
where R+ is the set of all non-negative real numbers.
Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by g (x) = αx + β, then what value should be assigned to α and β
Let f: R → R be defined by f(x) = 3x 2 – 5 and g: R → R by g(x) = `x/(x^2 + 1)` Then gof is ______.
Let f: [0, 1] → [0, 1] be defined by f(x) = `{{:(x",", "if" x "is rational"),(1 - x",", "if" x "is irrational"):}`. Then (f o f) x is ______.
Let f: N → R be the function defined by f(x) = `(2x - 1)/2` and g: Q → R be another function defined by g(x) = x + 2. Then (g o f) `3/2` is ______.
Let f = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then g o f = ______ and f o g = ______.
The composition of functions is commutative.
If f(x) = (ax2 + b)3, then the function g such that f(g(x)) = g(f(x)) is given by ____________.
Let f : R – `{3/5}`→ R be defined by f(x) = `(3"x" + 2)/(5"x" - 3)` Then ____________.
If f(x) = (ax2 – b)3, then the function g such that f{g(x)} = g{f(x)} is given by ____________.
Which one of the following functions is not invertible?
The inverse of the function `"y" = (10^"x" - 10^-"x")/(10^"x" + 10^-"x")` is ____________.
Consider the function f in `"A = R" - {2/3}` defiend as `"f"("x") = (4"x" + 3)/(6"x" - 4)` 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 ->`?
The domain of definition of f(x) = log x2 – x + 1) (2x2 – 7x + 9) is:-
If `f(x) = 1/(x - 1)`, `g(x) = 1/((x + 1)(x - 1))`, then the number of integers which are not in domian of gof(x) are
Let A = `{3/5}` and B = `{7/5}` Let f: A → B: f(x) = `(7x + 4)/(5x - 3)` and g:B → A: g(y) = `(3y + 4)/(5y - 7)` then (gof) is equal to
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).
