Let f : N → Y be function defined as f (x) = 4 x + 3, where, 

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsShort Answer Type

361. how that f : A → B and g : B → C are onto, then g of : A → C is also onto.
110 Views

362. Let f : X → Y be an invertible function. Show that f has unique inverse.
105 Views

363.

Let f : x → Y be an invertible function. Show that the inverse of f–1 is f,

i.e.(f–1)–1 = f.

108 Views

364.  If f : R → R defined as straight f left parenthesis straight x right parenthesis equals fraction numerator 2 straight x minus 7 over denominator 4 end fraction is an invertible function, find f–1.
113 Views

Advertisement
Advertisement

365.

Let f : N → Y be function defined as f (x) = 4 x + 3, where, Y = {y ∈N : y = 4 x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.


f : N ∴ Y is defined as f(x) = 4 x + 3
where Y = {y ∈ N : y = 4 x + 3 for some x ∈ N }
Consider an arbitrary element y of Y. By the definition of Y, y = 4 x + 3, for some x in the domain N.

therefore space space space space space space space space x equals fraction numerator y minus 3 over denominator 4 end fraction

Define g : Y rightwards arrow N given by g (y) = fraction numerator straight y minus 3 over denominator 4 end fraction

Now,  g o f(x) = g(f(f))= g(4x + 3) = fraction numerator 4 straight x plus 3 minus 3 over denominator 4 end fraction equals x
and   fo g(y) = f(g(y)) = straight f open parentheses fraction numerator straight y minus 3 over denominator 4 end fraction close parentheses equals fraction numerator 4 left parenthesis straight y minus 3 right parenthesis over denominator 4 end fraction plus 3 equals straight y minus 3 plus 3 equals straight y
therefore    g o f = 1N  and f o g = 1y

rightwards double arrow   f is invertible and g is the inverse of f.

therefore           straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space fraction numerator straight x minus 3 over denominator 4 end fraction.

 

107 Views

Advertisement
366. Let Y = { n2 : n ∈ N} ⊂ N. Consider f : N → Y as f(n) = n2. Show that f is invertible. Find the inverse of f.
130 Views

367.

Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R    (B) (3, 8) ∈ R     (C) (6, 8) ∈ R    (D) (8, 7) ∈ R

155 Views

368.

Show 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.?

147 Views

Advertisement
369.

Check the injectivity and surjectivity of the following functions:
f : N → N given by f(x) = x2 

114 Views

370.

Check the injectivity and surjectivity of the following functions:
f : Z → Z given by f(x) = x2 

122 Views

Advertisement