Advertisements
Advertisements
प्रश्न
Let A = R – {2} and B = R – {1}. If f: A `→` B is a function defined by f(x) = `(x - 1)/(x - 2)` then show that f is a one-one and an onto function.
Advertisements
उत्तर
Let A = R – {2}, B = R – {1}
f: A `→` B s.t. f(x) = `(x - 1)/(x - 2)`
For x1, x2 ∈ A
f(x1) = f(x2)
`\implies (x_1 - 1)/(x_1 - 2) = (x_2 - 1)/(x_2 - 2)`
`\implies (x_1 - 1)/(x_1 - 2) - 1 = (x_2 - 1)/(x_2 - 2) - 1`
`\implies 1/(x_1 - 2) = 1/(x_2 - 2)`
`\implies` x1 – 2 = x2 – 2
`\implies` x1 = x2
∴ f(x) is one-one function.
Also if f(x) = y, where y ∈ B.
`\implies (x - 1)/(x - 2)` = y
`\implies` x – 1 = xy – 2y
`\implies` 2y – 1 = xy – x
`\implies` 2y – 1 = x(y – 1)
x = `(2y - 1)/(y - 1) ∈ A`

Clearly every element y ∈ B is associated to x = `(2y - 1)/(y - 1)` of set A.
So Range of f = B `\implies` f is into
Hence f is one-one and onto function.
APPEARS IN
संबंधित प्रश्न
Show that the modulus function f : R → R given by f(x) = |x| is neither one-one nor onto, where |x| is x if x is positive or 0 and |x| is − x if x is negative.
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is a bijective function.
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = x3 + 1
Set of ordered pair of a function? If so, examine whether the mapping is injective or surjective :{(x, y) : x is a person, y is the mother of x}
Show that f : R→ R, given by f(x) = x — [x], is neither one-one nor onto.
Verify associativity for the following three mappings : f : N → Z0 (the set of non-zero integers), g : Z0 → Q and h : Q → R given by f(x) = 2x, g(x) = 1/x and h(x) = ex.
If f : A → B and g : B → C are one-one functions, show that gof is a one-one function.
Find fog and gof if : f(x)= x + 1, g (x) = 2x + 3 .
` if f : (-π/2 , π/2)` → R and g : [−1, 1]→ R be defined as f(x) = tan x and g(x) = `sqrt(1 - x^2)` respectively, describe fog and gof.
If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x|- x, ∀x∈R" .Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).
Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1
If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)−1 = f−1 og −1.
Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by f(x) = x2 and g(x) = sin (π x/2). Show that g−1 exists but f−1 does not exist. Also, find g−1.
Let A and B be two sets, each with a finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.
Which one of the following graphs represents a function?

Let C denote the set of all complex numbers. A function f : C → C is defined by f(x) = x3. Write f−1(1).
Let f be a function from C (set of all complex numbers) to itself given by f(x) = x3. Write f−1 (−1).
Let \[f : \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \to R\] be a function defined by f(x) = cos [x]. Write range (f).
Let A = {x ∈ R : −4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by \[f\left( x \right) = \frac{\left| x \right|}{x}\]Write the range of f.
Let \[f : \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \to\] A be defined by f(x) = sin x. If f is a bijection, write set A.
Write the domain of the real function
`f (x) = 1/(sqrt([x] - x)`.
The inverse of the function
\[f : R \to \left\{ x \in R : x < 1 \right\}\] given by
\[f\left( x \right) = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] is
Let the function f: R → R be defined by f(x) = cosx, ∀ x ∈ R. Show that f is neither one-one nor onto
Let A = {1, 2, 3, ...n} and B = {a, b}. Then the number of surjections from A into B is ______.
The function f : R → R defined by f(x) = 3 – 4x is ____________.
Let f : R `->` R be a function defined by f(x) = x3 + 4, then f is ______.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:
A function f: x → y is said to be one – one (or injective) if:
