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QUESTION: 1

Let f: A → B and g : B → C be one - one functions. Then, gof: A → C is

Solution:

We are given that functions f : A→B and g : B→C are both one-to-one functions.

Suppose a^{1}, a^{2} ∈ A such that (gof)(a^{1}) = (gof)(a^{2})

⇒ g(f(a1)) = g(f(a^{2})) (definition of composition) Since g is one-to-one, therefore,

f(a^{1}) = f(a^{2})

And since f is one-to-one, therefore, a^{1} = a^{2}

Thus, we have shown that if (gof)(a^{1}) = (gof)(a^{2}) then a^{1} = a^{2}

Hence, gof is one-to-one function.

QUESTION: 2

If f : A → B and g : B → C be two functions. Then, composition of f and g, gof : A → C is defined a

Solution:

A composite function is denoted by (g o f) (x) = g (f(x)). The notation g o f is read as “g of f”.

Example : Consider the functions f: A → B and g: B → C. f = {1, 2, 3, 4, 5}→ {1, 4, 9, 16, 25} and g = {1, 4, 9, 16, 25} → {2, 8, 18, 32, 50}. A = {1, 2, 3, 4, 5}, B = {16, 4, 25, 1, 9}, C = {32, 18, 8, 50, 2}.Here, g o f = {(1, 2), (2, 8), (3, 18), (4, 32), (5, 50)}.

QUESTION: 3

f: R → R is defined by f(x) = x^{2} - 2x + 1. Find f[f(x)]

Solution:

QUESTION: 4

Let f: Q→Q be a function given by f(x) = x^{2},then f^{ -1}(9) =

Solution:

If f : A → B such that y implies B then f^{-1}(y) = {x implies A : f(x) = y}

Let f^{-1}(9) = x

f(x) = 9

x^{2} = 9

x = +-3

Thus, f^{-1}(9) = {-3,3}

QUESTION: 5

Let g(x) = 1 + x – [x] and

Then f{g(x)} for all x, is equal to:

Solution:

g(x) = 1 + x - [x]

f(x) = {-1, x<0 0, x=0 1, x>0}

f(g(x)) = f[1 + x - [x]]

= f[1 + x - 1]

= f(x)

f(g(x)) is equal to f(x)

QUESTION: 6

If ƒ(x) = xsecx, then ƒ(0) =

Solution:

f(0)=0×sec0

f(0)=0×1

f(0)=0

QUESTION: 7

Let f = {(1, 3), (2, 1), (3, 2)} and g = {(1, 2), (2, 3), (3, 1)}. What is gof(2)?

Solution:

gof(2)=g[f(2)]

=g[1]=2

QUESTION: 8

If ƒ(x) = tan^{-1} x and g(x) = tan(x), then (gof)(x) =

Solution:

QUESTION: 9

If f(x) = |x| and g(x) = |5x – 2|. Then, fog =

Solution:

f(x) = |x| g(x) = |5x - 2|

fog = f(g(x) = g(x) = |5x - 2|

QUESTION: 10

If f: R → R and g: R → R defined by f(x) = 2x + 3 and g(x) = x^{2} + 7, then the value of x for which f(g(x)) = 25 is

Solution:

f : R → R g : R → R

f(g) : R → R

f(x) = 2x + 3 g(x^{2}) = x^{2} + 7

f(g(x)) = 2g(x) + 3

f(g(x)) = 2(x^{2} + 7) + 3

= 2x^{2} + 17

As it is given f(g(x)) = 25

Comparing both the eq, we get

2x^{2} + 17 = 25

2x^{2} = 8

x = __+__2

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