Advertisements
Advertisements
Question
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)`
Solution
t is given that `f:R -{-4/3} -> R` is defined as `f(x) = (4x)/(3x - 4)`
Let y be an arbitrary element of Range f.
Then, there exists x ∈ `R - {-4/3}` such that `y = f(x)`
`=> y = (4x)/(3x + 4)`
=>3xy + 4y = 4x Let us define g: Range `f -> R -{-4/3} " as " g(y) = (4y)/(4 -3y)`
=> x(4 - 3y) = 4y
Now `(gof)(x) = g(f(x)) = g((4x)/(3x + 4))`
`=> x = (4y)/(4 -3y)`
`= (4(4x/(3x + 4)))/(4 - 3("4x"/(3x + 4))) = "16x"/(12x + 16 - 12x) = "16x"/16 = x`
And `(fog)(y) = f(g(y)) = f((4y)/(4 - 3y))`
`= (4("4y"/(4-3y)))/(3("4y"/(4 - 3y)) + 4) = (16y)/(12y + 16 - 12y) = (16y)/16 = y`
:. `gof = I_(R - (-4/3))` and fog = I_"Range f"
Thus, g is the inverse of f i.e., f−1 = g.
Hence, the inverse of f is the map `g: Range f -> R - {-4/3}` which is given by
`g(y) = "4y"/(4 - 3y)`
The correct answer is B.
APPEARS IN
RELATED QUESTIONS
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.
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
f: {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
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] ie 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)`
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 is defined by f(x) = x2 − 3x + 2, find f(f(x)).
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.
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.
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: 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 ______.
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(x) = `(3"x" + 2)/(5"x" - 3)` then (fof)(x) is ____________.
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 ____________.
If f(x) = (ax2 – b)3, then the function g such that f{g(x)} = g{f(x)} is given by ____________.
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?
The domain of definition of f(x) = log x2 – x + 1) (2x2 – 7x + 9) 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: A → B and G B → C are one – one, then g of A → C is
If f: N → Y be a function defined as f(x) = 4x + 3, Where Y = {y ∈ N: y = 4x+ 3 for some x ∈ N} then function 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).
