Advertisements
Advertisements
प्रश्न
Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Advertisements
उत्तर
f: R → R is given by,
f(x) = 4x + 3
One-one:
Let f(x) = f(y).
⇒ 4x + 3 = 4y + 3
⇒ 4x = 4y
⇒ x = y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 4x + 3.
⇒ `x = (y - 3)/4 ∈ R`
Therefore, for any y ∈ R, there exists `x = (y - 3)/4 ∈ R` such that
`f(x) = f((y - 3)/4)`
= `4((y - 3)/4) + 3`
= y
∴ f is onto.
Thus, f is one-one and onto and therefore, f−1 exists.
Let us define g: R→ R by `g(x) = (y - 3)/4`.
Now, (gof)(x) = g(f(x))
= g(4x + 3)
= `((4x + 3) - 3)/4 `
= x
(fog)(y) = f(g(y))
= `f((y - 3)/4)`
= `4((y - 3)/4) + 3`
= y – 3 + 3
= y
∴ gof = fog = IR
Hence, f is invertible and the inverse of f is given by `f^(-1) = g(y) = (y - 3)/4`.
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 : 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: {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.
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)}
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}` given by
If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
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: 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: R → R be the function defined by f(x) = sin (3x+2) ∀ x ∈ R. Then f is invertible.
The composition of functions is commutative.
The composition of functions is associative.
Every function is invertible.
If f : R → R, g : R → R and h : R → R is such that f(x) = x2, g(x) = tanx and h(x) = logx, then the value of [ho(gof)](x), if x = `sqrtpi/2` will be ____________.
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 ____________.
Let f : R → R be the functions defined by f(x) = x3 + 5. Then f-1(x) is ____________.
Let f : R – `{3/5}`→ R be defined by f(x) = `(3"x" + 2)/(5"x" - 3)` Then ____________.
Which one of the following functions is not invertible?
If f is an invertible function defined as f(x) `= (3"x" - 4)/5,` then f-1(x) is ____________.
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?
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 'D' be the domain of the real value function on Ir defined by f(x) = `sqrt(25 - x^2)` the D is :-
If f: A → B and G B → C are one – one, then g of A → C is
Let `f : R {(-1)/3} → R - {0}` be defined as `f(x) = 5/(3x + 1)` is invertible. Find f–1(x).
