Show that f is invertible and find the inverse of f. Here, W is

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 Multiple Choice QuestionsShort Answer Type

381.

Let * be a binary operation on the set Q of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ?

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382. Consider f : R+ → [4, ⊂) given by f(x) = x+ 4. Show that f is invertible with the inverse f–1 of f given by straight f to the power of negative 1 end exponent left parenthesis straight y right parenthesis equals square root of straight y minus 4 end root where R+ is the set of all non-negative real numbers.
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383.

Let f : R → R be defined as f (x) = 3 x. Choose the correct answer. (A) f is one-one onto   (B) f is many-one onto  (C) f is one-one but not onto (D) f is neither onc-one nor onto.

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384.

how that the function f : R* → R* defined by

straight f left parenthesis straight x right parenthesis equals 1 over straight x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?

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385. Show that the function f : R →R , defined as f (x) = x2 , is neither one-to-one nor onto.
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386.

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 negative.

 

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387. Find f(A), if f(x) = x2–5x –14 and straight A equals space open square brackets table row 3 cell negative 5 end cell row cell negative 4 end cell 2 end table close square brackets
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 Multiple Choice QuestionsLong Answer Type

388.

Let A= R × R and * be a binary operation on A defined by

(a, b) * (c, d) = (a+c, b+d)

Show that * is commutative and associative. Find the identity element for *

on A. Also find the inverse of every element (a, b) ∈ A.

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389.

Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b+ad) for (a, b), (c, d) element of A. Then find
(i) The identify element of * in A.
(ii) Invertible elements of A, and write the inverse of elements (5, 3) and open parentheses 1 half comma space 4 close parentheses.

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390. Let space straight f colon straight W rightwards arrow straight W space be space defined space as
straight f left parenthesis straight n right parenthesis space equals space open curly brackets table row cell straight n minus 1 comma space space if space straight n space is space odd end cell row cell straight n plus 1 comma space if space straight n space is space even end cell end table close curly brackets
Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers. 


Let f: W→W be defined as
straight f left parenthesis straight n right parenthesis space equals space open curly brackets table row cell straight n minus 1 comma space if space straight n space is space odd end cell row cell straight n plus 1 comma space space if space straight n space is space even end cell end table close curly brackets
We need to prove that 'f' is invertible.
In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection.
A function f: A→B is a one-one function or an injection, if
f(x) = f(y) ⇒ x = y for all x, y ∈ A.
Case i:
If x and y are odd.
Let f(x) = f(y)
⇒x − 1 = y − 1
⇒x = y
Case ii:
If x and y are even,
Let f(x) = f(y)
⇒x + 1 = y + 1
⇒x = y
Thus, in both the cases, we have,
f(x) = f(y) ⇒ x = y for all x, y ∈ W.
Hence f is an injection. 

Let n be an arbitrary element of W.
If n is an odd whole number, there exists an even whole number n − 1 ∈ W such that
f(n − 1) = n − 1 + 1 = n.
If n is an even whole number, then there exists an odd whole number n + 1 ∈ W such that f(n + 1) = n + 1 − 1 = n. Also, f(1) = 0 and f(0) = 1
Thus, every element of W (co-domain) has its pre-image in W (domain).
So f is an onto function.

Thus, it is proved that f is an invertible function.
Thus, a function g: B→A which associates each element y ∈ B to a unique element x ∈ A such that f(x) = y is called the inverse of f.
That is, f(x) = y ⇔ g(y) = x.
The inverse of f is generally denoted by f -1.

Now let us find the inverse of f.
Let x, y ∈ W such that f(x) = y
⇒x + 1 = y, if x is even
And
straight x minus 1 space equals space straight y comma space if space straight x space is space odd
rightwards double arrow space space straight x space equals space open curly brackets table row cell straight y minus 1 comma space if space straight y space is space odd end cell row cell straight y plus 1 comma if space straight y space is space even end cell end table close curly brackets
rightwards double arrow space straight f to the power of negative 1 end exponent left parenthesis straight y right parenthesis space equals open curly brackets table row cell straight y minus 1 comma space if space straight y space is space odd end cell row cell straight y plus 1 comma space if space straight y space straight i space even end cell end table close curly brackets
Interchange comma space straight x space and space straight y comma space we space have comma space
rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals open curly brackets table row cell straight x minus 1 comma space if space straight x space is space odd end cell row cell straight x plus 1 comma space if space straight x space is space even end cell end table close curly brackets
Re space writing space the space above space we space have comma
space space rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space open curly brackets table row cell straight x plus 1 comma space if space straight x space is space even space end cell row cell straight x minus 1 comma space if space straight x space is space odd end cell end table close curly brackets
Thus comma space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis

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