Advertisements
Advertisements
प्रश्न
If f(x) = [4 – (x – 7)3]1/5 is a real invertible function, then find f–1(x).
Advertisements
उत्तर
Given f(x) = [4 – (x – 7)3]1/5 is a real invertible function.
Let f(x) = y
`\implies` y = [4 – (x – 7)3]1/5
`\implies` y5 = 4 – (x – 7)3
`\implies` (x – 7)3 = 4 – y5
`\implies` x – 7 = [4 – y5]1/3
`\implies` x = 7 + (4 – y5)1/3
`\implies` f1(y) = 7 + (4 – y5)1/3 ` {{:(∵ f(x) "is invertible"), (∴ f(x) = "y"),(\implies x = f^-1("y")):}`
Hence f1(x) = 7 + (4 – x5)1/3
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: {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)`
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 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: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)`
If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
Let f: A → B and g: B → C be the bijective functions. Then (g o f)–1 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.
Every function is invertible.
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?
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:-
Domain of the function defined by `f(x) = 1/sqrt(sin^2 - x) log_10 (cos^-1 x)` 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
