Advertisements
Advertisements
प्रश्न
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)`.
Advertisements
उत्तर
f: R+ → [–5, ∞) is given as f(x) = 9x2 + 6x – 5.
Let y be an arbitrary element of [–5, ∞).
Let y = 9x2 + 6x – 5.
⇒ y = (3x + 1)2 – 1 – 5
⇒ y = (3x + 1)2 – 6
⇒ (3x + 1)2 = y + 6
⇒ `3x + 1 = sqrt(y + 6)` ...[As y ≥ 5 ⇒ y + 6 > 0]
⇒ `x = (sqrt(y + 6) - 1)/3`
∴ f is onto, there by range f = [–5, ∞).
Let us define g: [–5, ∞) → R+ as g(y) = `(sqrt(y + 6) - 1)/3`.
We now have:
(gof)(x) = g(f(x))
= g(9x2 + 6x – 5)
= g((3x + 1)2 – 6)
= `sqrt((3x + 1)^2 - 6 + 6 - 1)/3`
= `(3x + 1 - 1)/3`
= x
And (fog)(y) = f(g(y))
= `f((sqrt(y + 6) - 1)/3)`
= `[3((sqrt(y + 6) - 1)/3) + 1]^2 - 6`
= `(sqrt(y + 6))^2 - 6`
= y + 6 – 6
= y
∴ `gof = I_R` and `fog = I_([-5, ∞)`
Hence, f is invertible and the inverse of f is given by`f^(-1)(y) = g(y) = (sqrt(y + 6) - 1)/3`.
संबंधित प्रश्न
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|.
Find gof and fog, if f(x) = 8x3 and `g(x) = x^(1/3)`.
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)}
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)}
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.
If f: R → R be given by `f(x) = (3 - x^3)^(1/3)`, then fof(x) is ______.
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
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.
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)
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: A → B and g: B → C be the bijective functions. Then (g o f)–1 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 = ______.
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 associative.
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(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?
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 ->`?
The domain of definition of f(x) = log x2 – x + 1) (2x2 – 7x + 9) is:-
Domain of the function defined by `f(x) = 1/sqrt(sin^2 - x) log_10 (cos^-1 x)` 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
