Relations and Functions Test

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Relations and Functions Test - 10

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Question 1
1 / -0.25

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

A
One-one

B
Does not exist

C
Not onto

D
Onto

Solution

We are given that functions f : A →B and g : B →C are both one-to-one functions.
Suppose a¹ , a² ∈A such that (gof)(a¹ ) = (gof)(a² )
⇒g(f(a1)) = g(f(a² )) (definition of composition) Since g is one-to-one, therefore,
f(a¹ ) = f(a² )
And since f is one-to-one, therefore, a¹ = a²
Thus, we have shown that if (gof)(a¹ ) = (gof)(a² ) then a¹ = a²
Hence, gof is one-to-one function.
Question 2
1 / -0.25

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

A
(gof)(x) = f(g(x)), for all x ∈A

B
(gof)(x) = g(f(x)), for all x ∈A

C
(gof)(x) = g(x), for all x ∈A

D
None of the above

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
1 / -0.25

f: R →R is defined by f(x) = x² - 2x + 1. Find f[f(x)]

A
x⁴ + 4x³ - 4x²

B
x⁴ - 4x³ + 4x²

C
x⁴ + 4x³ + 4x²

D
x⁴ - 4x³ - 4x²
Question 4
1 / -0.25

Let f: Q →Q be a function given by f(x) = x² ,then f^-1 (9) =

A
{3}

B
f

C
{-3, 3}

D
{-3}

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
x² = 9
x = +-3
Thus, f^-1 (9) = {-3,3}
Question 5
1 / -0.25

Let g(x) = 1 + x –[x] and
Then f{g(x)} for all x, is equal to:

A
g(x)

B
x

C
1

D
f(x)

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
1 / -0.25

If ƒ(x) = xsecx, then ƒ(0) =

A
−1

B
0

C
1

D
√(2)

Solution

f(0)=0 ×sec0
f(0)=0 ×1
f(0)=0
Question 7
1 / -0.25

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

A
3

B
Does not exist

C
2

D
1

Solution

gof(2)=g[f(2)]
=g[1]=2
Question 8
1 / -0.25

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

A
tan^-1 xtan(x)

B
tan^-1 xcot(x)

C
x

D
tan^-1 xsin(x)

Solution
Question 9
1 / -0.25

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

A
|g|

B
|f|

C
f

D
f²

Solution

f(x) = |x| g(x) = |5x - 2|
fog = f(g(x) = g(x) = |5x - 2|
Question 10
1 / -0.25

If f: R →R and g: R →R defined by f(x) = 2x + 3 and g(x) = x² + 7, then the value of x for which f(g(x)) = 25 is

A
±3

B
±1

C
±4

D
±2

Solution

f : R →R g : R →R
f(g) : R →R
f(x) = 2x + 3 g(x² ) = x² + 7
f(g(x)) = 2g(x) + 3
f(g(x)) = 2(x² + 7) + 3
= 2x² + 17
As it is given f(g(x)) = 25
Comparing both the eq, we get
2x² + 17 = 25
2x² = 8
x = + 2

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Relations and Functions Test - 10

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Question 1

One-one

Does not exist

Not onto

Onto

Solution

Question 2

(gof)(x) = f(g(x)), for all x ∈A

(gof)(x) = g(f(x)), for all x ∈A

(gof)(x) = g(x), for all x ∈A

None of the above

Solution

Question 3

x4 + 4x3 - 4x2

x4 - 4x3 + 4x2

x4 + 4x3 + 4x2

x4 - 4x3 - 4x2

Question 4

{3}

f

{-3, 3}

{-3}

Solution

Question 5

g(x)

x

1

f(x)

Solution

Question 6

−1

0

1

√(2)

Solution

Question 7

3

Does not exist

2

1

Solution

Question 8

tan-1 xtan(x)

tan-1 xcot(x)

x

tan-1 xsin(x)

Solution

Question 9

|g|

|f|

f

f2

Solution

Question 10

±3

±1

±4

±2

Solution

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x⁴ + 4x³ - 4x²

x⁴ - 4x³ + 4x²

x⁴ + 4x³ + 4x²

x⁴ - 4x³ - 4x²

tan^-1 xtan(x)

tan^-1 xcot(x)

tan^-1 xsin(x)

f²