Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
f: `R_+ -> [-5, oo]` given by `f(x) = 9x^2 + 6x - 5`
To show: f is one-one and onto.
Let us assume that f is not one-one.
Therefore there exist two or more numbers for which images are same.
For x1, x2 ∈ R+ and x1 ≠ x2
Since x1 and x2 are positive,
9(x1 + x2) + 6 > 0
∴ x1 − x2 = 0⇒ x1 = x2
Therefore, it contradicts our assumption.
Hence the function f is one-one.
Now, let is prove that f is onto.
A function f : X → Y is onto if for every y ∈ Y, there exist a pre-image in X.
f(x) = 9x2 + 6x −5
= 9x2 + 6x + 1 - 6
=(3x + 1)2 - 6
Now, for all x ∈ R+ or [0,∞), f(x) ∈ [−5, ∞)
∴ Range = co-domain.
Hence, f is onto.
Therefore, function f is invertible.
Now, let y = 9x2 + 6x − 5
1) `f^(-1) (10) = (sqrt(10+6)-1)/3 = (4-1)/3 = 1`
2) `f^(-1) (y) = 4/3`
`:.(sqrt(y + 6) -1)/3 = 4/3`
`=> sqrt(y + 6)-1 = 4`
`=> sqrt(y+6) = 5`
`=> y + 6 = 25`
=> y = 19
APPEARS IN
संबंधित प्रश्न
Show that f: [–1, 1] → R, given by f(x) = `x/(x + 2)` is one-one. Find the inverse of the function f: [–1, 1] → Range f.
(Hint: For y in Range f, y = `f(x) = x/(x + 2)` for some x in [–1, 1] i.e., `x = (2y)/(1 - y)`)
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)`.
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.
If f: R → R be given by `f(x) = (3 - x^3)^(1/3)`, then fof(x) is ______.
If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
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 = {(1, 2), (3, 5), (4, 1) and g = {(2, 3), (5, 1), (1, 3)}. Then g o f = ______ and f o g = ______.
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 ____________.
Which one of the following functions is not invertible?
The inverse of the function `"y" = (10^"x" - 10^-"x")/(10^"x" + 10^-"x")` is ____________.
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
Which statement correctly describes an onto (surjective) function?
Which property of inverse functions is represented by \((f^{-1})^{-1}=f\)?
In the graphical example, what are the domain and range of the given function?
Which statement expresses that an inverse is unique whenever it exists?
