RD Sharma XII Vol 1 2021 Solutions for Class 12 Humanities Maths Chapter 2 Functions are provided here with simple step-by-step explanations. These solutions for Functions are extremely popular among class 12 Humanities students for Maths Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2021 Book of class 12 Humanities Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2021 Solutions. All RD Sharma XII Vol 1 2021 Solutions for class 12 Humanities Maths are prepared by experts and are 100% accurate.

Page No 2.31:

Question 1:

Give an example of a function
(i) which is one-one but not onto
(ii) which is not one-one but onto
(iii) which is neither one-one nor onto

Answer:

(i) which is one-one but not onto.

f: ZZ given by f(x)=3x+2

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
 f(x)=f(y)
3x + 2 =3y + 2
3x = 3y
x = y
f(x) = f(y) x = y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
3x + 2 = y
3x = y - 2

x = y-23. It may not be in the domain (Zbecause if we take y = 3,x = y-23 = 3-23 = 13∉ domain Z.
So, for every element in the co domain there need not be any element in the domain such that f(x) = y.
Thus, f is not onto.

(ii) which is not one-one but onto.
f: ZN {0} given by f(x) = |x|

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
|x| = |y|
x= ±y
So, different elements of domain f may give the same image.
So, f is not one-one.

Surjectivity:
Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
|x| = y
x±y, which is an element in Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.

(iii) which is neither one-one nor onto.

f: ZZ given by f(x) = 2x2 + 1

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
fx = fy2x2+1 = 2y2+12x2 = 2y2x2 = y2x = ±y
So, different elements of domain f may give the same image.
Thus, f is not one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
2x2+1=y2x2=y-1x2=y-12x=±y-12, ∉ Z always.For example, if we take, y = 4,x=±y-12=±4-12=±32, ∉ ZSo, x may not be in Z (domain).

Thus, f is not onto.

Page No 2.31:

Question 2:

Which of the following functions from A to B are one-one and onto?
(i) f1 = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}
(ii) f2 = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}
(iii) f3 = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Answer:

(i) f1 = {(1, 3), (2, 5), (3, 7)} ; A = {1, 2, 3}, B = {3, 5, 7}

Injectivity:
f1 (1) = 3
f1(2) = 5
f1 (3) = 7
Every element of A has different images in B.
So, f1 is one-one.

Surjectivity:
Co-domain of f1 = {3, 5, 7}
Range of f1 =set of images  =  {3, 5, 7}
Co-domain = range
So, f1 is onto.

(ii) f2 = {(2, a), (3, b), (4, c)} ; A = {2, 3, 4}, B = {a, b, c}

Injectivity:
f2 (2) = a
f2 (3) = b
f2 (4) = c
Every element of A has different images in B.
So, f2 is one-one.

Surjectivity:
Co-domain of f2 = {a, b, c}
Range of f2 = set of images = {a, b, c}
Co-domain = range
So, f2 is onto.

(iii) f3 = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}

Injectivity:
f3 (a) = x
f3(b) = x
f3 (c) = z
f3 (d) = z

a and b have the same image x. (Also c and d have the same image z)
So, f3 is not one-one.

Surjectivity:
Co-domain of f1 ={x, y, z}
Range of f1 =set of images  =  {x, z}
So, the co-domain  is not same as the range.
So, f3 is not onto.

Page No 2.31:

Question 3:

Prove that the function f : NN, defined by f(x) = x2 + x + 1, is one-one but not onto.

Answer:

f : NN, defined by f(x) = x2 + x + 1

Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
x2+x+1=y2+y+1x2-y2+x-y=0x+yx-y+x-y=0x-yx+y+1=0x-y=0       x+y+1 cannot be zero because x and y are natural numbersx=y
 So, f is one-one.

Surjectivity:
The minimum number in N is 1.When x=1,x2+x+1=1+1+1=3x2+x+13, for every x in N.fx will not assume the values 1 and 2.So, f is not onto.

Page No 2.31:

Question 4:

Let A = {−1, 0, 1} and f = {(x, x2) : xA}. Show that f : AA is neither one-one nor onto.

Answer:

A = {−1, 0, 1} and f = {(x, x2) : xA}
Given, f(x) = x2

Injectivity:
f(1) = 12=1 and
f(-1)=(-1)2=1

1 and -1 have the same images.
So, is not one-one.

Surjectivity:
Co-domain of  f  = {-1, 0, 1}

f(1) = 12 = 1,
f(-1) = (-1)2 = 1 and
f(0) = 0
Range of = {0, 1}
So, both are not same.
Hence,is not onto.

Page No 2.31:

Question 5:

Classify the following functions as injection, surjection or bijection :
(i) f : NN given by f(x) = x2
(ii) f : ZZ given by f(x) = x2
(iii) f : NN given by f(x) = x3
(iv) f : ZZ given by f(x) = x3
(v) f : RR, defined by f(x) = |x|
(vi) f : ZZ, defined by f(x) = x2 + x
(vii) f : ZZ, defined by f(x) = x − 5
(viii) f : RR, defined by f(x) = sinx
(ix) f : RR, defined by f(x) = x3 + 1
(x) f : RR, defined by f(x) = x3x
(xi) f : RR, defined by f(x) = sin2x + cos2x
(xii) f : Q − {3} → Q, defined by fx=2x+3x-3
(xiii) f : QQ, defined by f(x) = x3 + 1
(xiv) f : RR, defined by f(x) = 5x3 + 4
(xv) f : RR, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x2
(xvii) f : R → R, defined by f(x) = xx2+1                                                                                                                    [NCERT EXEMPLAR]

Answer:

(i) f : NN, given by f(x) = x2

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x)=f(y)

x2=y2x=y       (We do not get ± because x and y are in N)

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x2=yx=y, which may not be in N.For example, if y=3,x=3 is not in N.

So, f is not a surjection.

So, f is not a bijection.

(ii) f : ZZ, given by f(x) = x2

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x= f(y)

x2=y2x=±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x2=yx=±y which may not be in Z.For example, if y=3,x=±3 is not in Z.

So, f is not a surjection.

So, f is not a bijection.

(iii) f : NN, given by f(x) = x3

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

f(x= f(y)

x3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

x3=yx=y3which may not be in N.For example, if y=3,x=33 is not in N.

So, f is not a surjection and  f is not a bijection.

(iv) f : ZZ, given by f(x) = x3

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

f(x= f(y)

x3=y3x=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x3=yx=y3 which may not be in Z.For example, if y=3,x=33 is not in Z.

So, f is not a surjection and f is not a bijection.

(v) f : RR, defined by f(x) = |x|

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

f(x= f(y)

x=yx=±y

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x=yx=±yZ

So, f is a surjection and  f is not a bijection.

(vi) f : ZZ, defined by f(x) = x2 + x

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x= f(y)

x2+x=y2+yHere, we cannot say that x = y.For example, x = 2 and y = - 3Then, x2+x=22+2= 6y2+y=-32-3= 6So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x2+x=yHere, we cannot say xZ.For example, y =-4.x2+x=-4x2+x+4=0x=-1±-152=-1±i152 which is not in Z.

So, f is not a surjection and  f is not a bijection.

(vii) f : ZZ, defined by f(x) = x − 5

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f(x= f(y)

x - 5 = y - 5

x = y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

x - 5 = y

x = y + 5, which is in Z.

So, f is a surjection and f is a bijection.

(viii) f : RR, defined by f(x) = sinx

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

sinx=sinyHere, x may not be equal to y because sin0=sinπ.So, 0 and π have the same image 0.

So, f is not an injection .

Surjection test:

Range of f = [-1, 1]

Co-domain of f = R

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) f : RR, defined by f(x) = x3 + 1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

x3+1=y3+1x3=y3x=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (R)such that f(x) = y for some element x in R (domain).

f(x) = y

x3+1=yx=y-13R

So, f is a surjection.

So, f is a bijection.

(x) f : RR, defined by f(x) = x3x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

x3-x=y3-yHere, we cannot say x=y.For example, x=1 and y=-1x3-x=1-1=0y3-y=-13--1-1+1=0So, 1 and -1 have the same image 0.

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

x3-x=yBy observation we can say that there exist some x in R, such that x3-x=y.

So, f is a surjection and  f is not a bijection.

(xi) f : RR, defined by f(x) = sin2x + cos2x

f(x) = sin2x + cos2= 1

So, f(x) = 1 for every x in R.

So, for all elements in the domain, the image is 1.

So, f is not an injection.

Range of f = {1}

Co-domain of fR

Both are not same.

So, f is not a surjection and  f is not a bijection.

(xii) f : Q − {3} → Q, defined by fx=2x+3x-3

Injection test:

Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).

f(x) = f(y)

2x+3x-3=2y+3y-32x+3y-3=2y+3x-32xy-6x+3y-9=2xy-6y+3x-99x=9yx=y

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).

f(x) = y

2x+3x-3=y2x+3=xy-3y2x-xy=-3y-3x2-y=-3y+1x=3y+1y-2, which is not defined at y=2.

So, f is not a surjection and f is not a bijection.

(xiii) f : QQ, defined by f(x) = x3 + 1

Injection test:

Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

f(x= f(y)

x3+1=y3+1x3=y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).

f(x) = y

x3+1=yx=y-1,3 which may not be in Q.For example, if y= 8,x3+1= 8x3=7x=73, which is not in Q.

So, f is not a surjection and f is not a bijection.

So, f is a surjection and f is a bijection.

(xiv) f : RR, defined by f(x) = 5x3 + 4

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

5x3+4 = 5y3+45x3= 5y3x3= y3x=y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

5x3+4=y5x3=y-4x3=y-45x=y-453R

So, f is a surjection and f is a bijection.

(xv) f : RR, defined by f(x) = 3 − 4x

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

3-4x=3-4y-4x=-4yx= y

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

3-4x=y4x=3-yx=3-y4R

So, f is a surjection and f is a bijection.

(xvi) f : RR, defined by f(x) = 1 + x2

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

1+x2=1+y2x2=y2x= ±y

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

f(x) = y

1+x2=yx2=y-1x=±y-1 which may not be in RFor example, if y=0,x=±-1=±i is not in R.

So, f is not a surjection and f is not a bijection.

(xvii)  f : R → R, defined by f(x) = xx2+1

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

f(x= f(y)

xx2+1=yy2+1xy2+x=x2y+yxy2-x2y+x-y=0-xy-y+x+1x-y=0x-y1-xy=0x=y or x=1y

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in (domain).

f(x) = y

xx2+1=yyx2-x+y=0x=--1±1-4y22y, if y0=1±1-4y22y, which may not be in RFor example, if y=1, thenx=1±1-42=1±i32, which is not in RSo, f is not surjection and f is not bijection.

So, is not a surjection and f is not a bijection.

Page No 2.31:

Question 6:

If f : AB is an injection, such that range of f = {a}, determine the number of elements in A.

Answer:

Range of f = {a}
So, the number of images of  f = 1
Since, is an injection, there will be exactly one image for each element of f .
So, number of elements in A = 1.

Page No 2.31:

Question 7:

Show that the function f : R − {3} → R − {2} given by fx=x-2x-3 is a bijection.

Answer:

f : R − {3} → R − {2} given by
fx=x-2x-3
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
 f(x) = f(y)
x-2x-3=y-2y-3x-2y-3=y-2x-3xy-3x-2y+6=xy-3y-2x+6x=y

So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).
f(x) = y
x-2x-3=yx-2=xy-3yxy-x=3y-2xy-1=3y-2x=3y-2y-1, which is in R-{3}
So, for every element in the co-domain, there exists some pre-image in the domain.
is onto.
Since, is both one-one and onto, it is a bijection.



Page No 2.32:

Question 8:

Let A = [-1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:

(i) f(x) = x2                                (ii) g(x) = |x|                                (iii) h(x) = x2                                                               [NCERT EXEMPLAR]

Answer:

(i) f : A  A, given by f(x) = x2

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

x2y2

x = y

So, f is one-one.

Surjection test:

Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain)

f(x) = y

x2 = y

x = 2y, which may not be in A.

For example, if y = 1, then

x = 2, which is not in A.

So, f is not onto.

So, f is not bijective.

(ii) g(x) = |x|

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

|x| = |y|

x = ±y

So, f is not one-one.

Surjection test:

For y = -1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

(iii) h(x) = x2

Injection test:

Let x and y be any two elements in the domain (A), such that f(x) = f(y).

f(x) = f(y)

x2y2

x = ±y

So, f is not one-one.

Surjection test:

For y = -1, there is no value of x in A.

So, f is not onto.

So, f is not bijective.

 

Page No 2.32:

Question 9:

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:

(i) {(x, y) : x is a person, y is the mother of x}
(ii) {(a, b) : a is a person, b is an ancestor of a}                                                                                                               [NCERT EXEMPLAR]

Answer:

(i) f = {(xy) : x is a person, y is the mother of x}

As, for each element x in domain set, there is a unique related element y in co-domain set.

So, f is the function.

Injection test:

As, y can be mother of two or more persons

So, f is not injective.

Surjection test:

For every mother y defined by (xy), there exists a person x for whom y is mother.

So, f is surjective.

Therefore, f is surjective function.

(ii) g = {(ab) : a is a person, b is an ancestor of a}

Since, the ordered map (a, b) does not map 'a' - a person to a living person.
So, g is not a function.

Page No 2.32:

Question 10:

Let A = {1, 2, 3}. Write all one-one from A to itself.

Answer:

A ={1, 2, 3}
Number of elements in  A = 3
Number of one-one functions = number of ways of arranging 3 elements = 3! = 6

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}

Page No 2.32:

Question 11:

If f : RR be the function defined by f(x) = 4x3 + 7, show that f is a bijection.

Answer:

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
4x3+7=4y3+74x3=4y3x3=y3x=y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
4x3+7=y4x3=y-7x3=y-74x=y-743R
So, for every element in the co-domain, there exists some pre-image in the domain.
f is onto.
Since, f is both one-to-one and onto, it is a bijection.

Page No 2.32:

Question 12:

Show that the exponential function f : RR, given by f(x) = ex, is one-one but not onto. What happens if the co-domain is replaced by R0+ (set of all positive real numbers)?

Answer:

f : RR, given by f(x) = ex

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
 f(x)=f(y)
ex=eyx=y

So, f is one-one.

Surjectivity:
We know that range of ex is (0, ∞) = R+
Co-domain = R
Both are not same.
So, f is not onto.

If the co-domain is replaced by R+, then the co-domain and range become the same and in that case, f is onto and hence, it is a bijection.

Page No 2.32:

Question 13:

Show that the logarithmic function f : R0+R given by fx=loga x, a>0 is a bijection.

Answer:

f:R+R given by fx= loga x, a>0
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
 f(x) = f(y)
loga x=loga yx=y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R+(domain).
f(x) = y
loga x=yx=ay R+
So, for every element in the co-domain, there exists some pre-image in the domain.
f is onto.
Since f is one-one and onto, it is a bijection.

Page No 2.32:

Question 14:

If A = {1, 2, 3}, show that a one-one function f : AA must be onto.

Answer:

A ={1, 2, 3}
Number of elements in  A = 3
Number of one - one functions = number of ways of arranging 3 elements = 3! = 6
So, the possible one -one functions can be the following:

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}
Here, in each function, range = {1, 2, 3}, which is same as the co-domain.
So, all the functions are onto.

Page No 2.32:

Question 15:

If A = {1, 2, 3}, show that a onto function f : AA must be one-one.

Answer:

A ={1, 2, 3}
Possible onto functions from A to A can be the following:

(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}

Here, in each function, different elements of the domain have different images.
So, all the functions are one-one.

Page No 2.32:

Question 16:

Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.

Answer:

We know that every onto function from A to itself is one-one.
So, the number of one-one functions = number of bijections = n!

Page No 2.32:

Question 17:

Give examples of two one-one functions f1 and f2 from R to R, such that f1 + f2 : RR. defined by (f1 + f2) (x) = f1 (x) + f2 (x) is not one-one.

Answer:

We know that f1: RR, given by f1(x)=x, and f2(x)=-x are one-one.
Proving f1is one-one:
Let f1x=f1yx=y
So, f1 is one-one.

Proving f2is one-one:
Let f2x=f2y-x=-yx=y
So, f2 is one-one.

Proving (f1 + f2) is not one-one:
Given:
(f1 + f2) (x) = f1 (x) + f2 (x)= x + (-x) =0
So, for every real number x, (f1 + f2) (x)=0
So, the image of ever number in the domain is same as 0.
Thus, (f1 + f2) is not one-one.

Page No 2.32:

Question 18:

Give examples of two surjective functions f1 and f2 from Z to Z such that f1 + f2 is not surjective.

Answer:

We know that f1: RR, given by f1(x) = x, and f2(x) = -x are surjective functions.
Proving f1is surjective :
Let y be an element in the co-domain (R), such that f1(x) = y.
f1(x) = y
x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f1is surjective .

Proving f2 is surjective :Let f2(x)=f2(y)x=yx=y
Let y be an element in the co domain (R) such that f2(x) = y.
 f2(x) = y
x = y, which is in R.
So, for every element in the co-domain, there exists some pre-image in the domain.1(x)=f1(y)x=y
So, f2is surjective .

Proving (f1 + f2) is not surjective :
Given:
(f1 + f2) (x) = f1 (x) + f2 (x)= x + (-x) =0
So, for every real number x, (f1 + f2) (x)=0
So, the image of every number in the domain is same as 0.
Range = {0}
Co-domain = R
So, both are not same.
So, f1 + f2is not surjective.

Page No 2.32:

Question 19:

Show that if f1 and f2 are one-one maps from R to R, then the product f1×f2 : RR defined by f1×f2 x=f1 x f2 x need not be one-one.

Answer:

We know that f1: RR, given by f1(x) = x, and f2(x) = x are one-one.
Proving f1is one-one:
Let x and y be two elements in the domain R, such that
f1(x) = f1(y)
x = y
et f1(x)=f1(y)x=y
So, f1is one-one.

Proving f2is one-one:
Let x and y be two elements in the domain R, such that
f2(x) = f2(y)
x = y
et f1(x)=f1(y)x=y
So, f2is one-one.

Proving f1 × f2 is not one-one:
Given:
f1 × f2x=f1 x × f2 x=x × x=x2Let x and y be two elements in the domain R, such thatf1 × f2x=f1 × f2yx2 = y2x=± ySo, f1 × f2 is not one-one.f1×f2

Page No 2.32:

Question 20:

Suppose f1 and f2 are non-zero one-one functions from R to R. Is f1f2 necessarily one-one? Justify your answer. Here, f1f2:RR is given by f1f2 x=f1 xf2 x for all xR.

Answer:

We know that f1: RR, given by f1(x)=x3and f2(x)=x are one-one.
Injectivity of f1:
Let x and y be two elements in the domain R, such that
f1x=f2yx3=yx=y3RLet f1(x)=f1(y)x=y

So, f1is one-one.

Injectivity of f2:
Let x and y be two elements in the domain R, such that
f2x=f2yx=y xR.Let f2(x)=f2(y)x=yx=y
So, f2 is one-one.

Proving f1f2is not one-one:
Given that f1f2x=f1xf2x=x3x=x2
Let x and y be two elements in the domain R, such that

f1f2x=f1f2yx2=y2x=±y
So, f1f2 is not one-one.

Page No 2.32:

Question 21:

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(i) an injective map from A to B
(ii) a mapping from A to B which is not injective
(iii) a mapping from A to B.

Answer:

(i) {(2, 7), (3, 6), (4, 5)}

(ii) {(2, 2), (3, 2), (4, 5)}

(iii) {(2, 5), (3, 6), (4, 7)}

Disclaimer: There are many more possibilities of each case.

Page No 2.32:

Question 22:

Show that f : R R, given by f(x) = x - [x], is neither one-one nor onto.

Answer:

We have, f(x) = x - [x]

Injection test:

f(x) = 0 for all x Z

So, f is a many-one function.

Surjection test:

Range (f) = [0, 1) R.

So, f is an into function.

Therefore, f is neither one-one nor onto.

Page No 2.32:

Question 23:

Let f : N N be defined by

fn=n+1, if n is oddn-1, if n is even

Show that f is a bijection.                                                                                                                                                   [CBSE 2012, NCERT]

Answer:

We have,fn=n+1, if n is oddn-1, if n is evenInjection test:Case I: If n is odd,Let x, yN such that fx=fyAs, fx=fyx+1=y+1x=yCase II: If n is even,Let x, yN such that fx=fyAs, fx=fyx-1=y-1x=ySo, f is injective.Surjection test:Case I: If n is odd,As, for every nN, there exists y=n-1 in N such thatfy=fn-1=n-1+1=nCase II: If n is even,As, for every nN, there exists y=n+1 in N such thatfy=fn+1=n+1-1=nSo, f is surjective.So, f is a bijection.



Page No 2.46:

Question 1:

Find gof and fog when f : RR and g : RR are defined by
(i) f(x) = 2x + 3                and       g(x) = x2 + 5
(ii) f(x) = 2x + x2             and       g(x) = x3
(iii) f(x) = x2 + 8              and       g(x) = 3x3 + 1
(iv) f(x) = x                       and       g(x) = |x|
(v) f(x) = x2 + 2x − 3       and       g(x) = 3x − 4
(vi) f(x) = 8x3                   and       g(x) = x1/3

Answer:

Given, f : RR and g : RR
So, gof : RR  and fog : RR

(i) f(x) = 2x + 3  and g(x) = x2 + 5
Now, (gof) (x)
= g (f (x))
= g
(2x +3)
= (2x + 3)2 + 5
= 4x2+ 9 + 12x +5
=4x2+  12x + 14

(fog) (x)
=f (g (x))
= f (x2
+ 5)
= 2 (x2+ 5) +3
= 2 x2+ 10 + 3
= 2x2 + 13

(ii) f(x) = 2x + x2 and g(x) = x3
gof x=g f x=g 2x+x2=2x+x23fog x=f g x=f x3=2 x3+x32=2x3+x6

(iii) f(x) = x2 + 8  and g(x) = 3x3 + 1
gof x=g fx=g x2+8=3 x2+83+1fog x=f g x=f 3x3+1=3x3+12+8=9x6+6x3+1+8=9x6+6x3+9

(iv) f(x) = x and g(x) = |x|
gof x=g fx=g x=xfog x=f g x=f x=x

(v) f(x) = x2 + 2x − 3 and g(x) = 3x − 4
gof x=g fx=g x2+2x-3=3 x2+2x-3-4=3x2+6x-9-4=3x2+6x-13fog x=f g x=f 3x-4=3x-42+2 3x-4-3=9x2+16-24x+6x-8-3=9x2-18x+5

(vi) f(x) = 8x3 and g(x) = x1/3
gof x=g f x=g 8x3=8x313=2x313=2xfog x=f g x=f x13=8 x133=8x

Page No 2.46:

Question 2:

Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.

Answer:

f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}

f : {3, 9, 12} → {1, 3,4} and g : {1, 3, 4, 5} → {3, 9}

Co-domain of f is a subset of the domain of g.
So, gof exists and gof : {3, 9, 12} → {3, 9}
gof 3=g f 3=g 1=3gof 9=g f 9=g 3=3gof 12=g f 12=g 4=9gof =3, 3, 9, 3, 12, 9
Co-domain of g is a subset of the domain of f.
So, fog exists and fog : {1, 3, 4, 5} → {3, 9, 12}
fog 1=f g 1=f 3=1fog 3=f g 3=f 3=1fog 4=f g 4=f 9=3fog 5=f g 5=f 9=3fog=1, 1, 3, 1, 4, 3, 5, 3

Page No 2.46:

Question 3:

Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.

Answer:

f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f : {1, 4, 9, 16} → {-1, -2, -3, 4} and g : {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof : {1, 4, 9, 16} → {-2, -4, -6, 8}
gof 1=g f 1=g -1=-2gof 4=g f 4=g -2=-4gof 9=g f 9=g -3=-6gof 16=g f 16=g 4=8So, gof=1, -2, 4, -4, 9, -6, 16, 8

But the co-domain of g is not same as the domain of f.
So, fog does not exist.

Page No 2.46:

Question 4:

Let A = {a, b, c}, B = {u v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as :

    f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.

Show that f and g both are bijections and find fog and gof.

Answer:

Proving f is a bijection:
f = {(a, v), (b, u), (c, w)} and f : A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u v, w}
Range of f = {u v, w}
Both are same.
So,  f is onto.
Hence, f is a bijection.

Proving g is a bijection:
g = {(u, b), (v, a), (w, c)} and g : B → A
Injectivity of g: No two elements of have the same image in A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.

Finding  fog:
Co-domain of g is same as the domain of f.
So, fog exists and fog : {u v, w} {u v, w}
fog u=f g u=f b=ufog v=f g v=f a=vfog w=f g w=f c=wSo, fog =u, u, v, v, w, w

Finding gof:
Co-domain of f is same as the domain of g.
So, fog exists and gof : {a, b, c} {a, b, c}
gof a=g f a=g v=agof b=g f b=g u=bgof c=g f c=g w=cSo, gof=a, a, b, b, c, c


Page No 2.46:

Question 5:

Find  fog (2) and gof (1) when : f : R → R ; f(x) = x2 + 8 and g : R → R; g(x) = 3x3 + 1.

Answer:

fog 2=f g 2=f3×23+1=f25=252+8=633gof 1=g f 1=g 12+8=g 9=3×93+1=2188

Page No 2.46:

Question 6:

Let R+ be the set of all non-negative real numbers. If f : R+R+ and g : R+R+ are defined as fx=x2 and gx=+x, find fog and gof. Are they equal functions?

Answer:

Given,  f : R+R+ and g : R+R+
So,  fog : R+R+  and gof : R+R+
Domains of fog  and gof  are the same.
fog x=f g x=f x=x2=xgof x=g f x=g x2=x2=xSo, fog x=gof x,xR+
Hence, fog = gof

Page No 2.46:

Question 7:

Let f : RR and g : RR be defined by f(x) = x2 and g(x) = x + 1. Show that foggof.

Answer:

Given,  f : RR and g : R → R.
So, the domains of f and g are the same.

fog x=f g x=f x+1=x+12=x2+1+2xgof x=g f x=g x2=x2+1
So,  fog ≠ gof

Page No 2.46:

Question 8:

Let f : RR and g : RR be defined by f(x) = x + 1 and g(x) = x − 1. Show that fog = gof = IR.

Answer:

Given,  f : RR and g : R → R
fog R → R and gof : R → R (Also, we know that IR : R → R)
So, the domains of all fog, gof and IRare the same.
fog x=f g x=f x-1=x-1+1=x=IR x      ... 1gof x=g f x=g x+1=x+1-1=x=IR x      ... 2From 1 and 2fog x=gof x=IR x, xRHence, fog=gof=IR

Page No 2.46:

Question 9:

Verify associativity for the following three mappings : f : N → Z0 (the set of non-zero integers), g : Z0Q and h : QR given by f(x) = 2x, g(x) = 1/x and h(x) = ex.

Answer:

Given that f : N → Z0 , g : Z0Q and h : QR .
gof : N Q  and hog : Z0 → R
h o (gof ) : N R and (hog) o f: N → R
So, both have the same domains.
gof x=g f x=g 2x=12x         ...1hog x=h g x=h 1x=e1x       ...2Now,h ogof x=hgof x=h 12x=e12x           [from 1]hog o fx=hog f x= hog 2x=e12x    [from 2]h ogof x=hog o fx, xNSo, h ogof=hog o f
Hence, the associative property has been verified.

Page No 2.46:

Question 10:

Consider f : NN, g : NN and h : NR defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z for all x, y, zN. Show that ho (gof) = (hog) of.

Answer:

Given, f : NN, g : NN and h : NR
gof : NN and hog : NR
ho (gof) : NR and (hog) of : NR
So, both have the same domains.
gof x=g f x=g 2x=3 2x+4=6x+4  ...1hog x=hg x=h 3x+4=sin 3x+4        ... 2Now,h o gof x=h gof x=h6x+4 = sin 6x+4    [from 1]hog o f x=hog f x=hog 2x=sin 6x+4    [from 2]So, h o gof x=hog o f x, xNHence, h o gof=hog o f

Page No 2.46:

Question 11:

Give examples of two functions f : NN and g : NN, such that gof is onto but f is not onto.

Answer:

Let us consider a function f : NN given by f(x) = x +1 , which is not onto.
[This not onto because if we take 0 in N (co-domain), then,
0=x+1
x=-1N]

Let us consider g : Ngiven by
g x=x-1, if x>11, if x=1Now, let us find gof xCase 1: x>1gof x=g f x=g x+1=x+1-1=xCase 2: x=1gof x=g f x=g x+1=1From case-1 and case-2, gof x=x, xN, which is an identity function and, hence, it is onto.

Page No 2.46:

Question 12:

Give examples of two functions f : NZ and g : Z → Z, such that gof is injective but g is not injective.

Answer:

Let f : NZ be given by f (x) = x, which is injective.
(If we take f(x) = f(y), then it gives x = y)

Let g : Z → Z be given by g (x) = |x|, which is not injective.
If we take f(x) = f(y), we get:
|x| = |y|
x = ±y

Now, gof : N → Z.
gof x=g f x=g x=x
Let us take two elements x and y in the domain of gof , such that
gof x=gof yx=yx=y We don't get ± here because x, y ∈N
So, gof is injective.
 

Page No 2.46:

Question 13:

If f(x)=1-x1+x, then find fof(x).

Answer:

Given: f(x)=1-x1+x,find fof(x).= ?
Here,
fofx=ffxfofx=f1-x1+x=1-1-x1+x1+1-x1+x fx=1-x1+x=1+x-1-x1+x+1-x=2x2=x
Hence, the value of fofx will be x.

Page No 2.46:

Question 14:

If f : AB and g : BC are one-one functions, show that gof is a one-one function.

Answer:

Given,  f : AB and g : B → C are one - one.
Then, gof : AB
Let us take two elements x and y from A, such that
gof x=gof yg f x=g f yf x=f y As, g is one-onex=y As, f is one-one
Hence, gof is one-one.

Page No 2.46:

Question 15:

If f : AB and g : BC are onto functions, show that gof is a onto function.

Answer:

Given,  f : AB and g : B → C are onto.
Then, gof : AC
Let us take an element z in the co-domain (C).
Now, z is in C and g : B → C is onto.
So, there exists some element y in B, such that g (y) = z ... (1)
Now, y is in B and f : AB is onto.
So, there exists some x in A, such that f (x) = y ... (2)
From (1) and (2),
z = g (y) = g (f (x)) = (gof) (x)
So, z = (gof) (x), where x is in A.
Hence, gof is onto.



Page No 2.54:

Question 1:

Find fog and gof  if
(i) fx=ex, gx=loge x
(ii) fx=x2, gx=cos x
(iii) fx=|x|, g (x)=sin x
(iv) fx=x+1, gx=ex
(v) fx=sin-1 x, gx=x2
(vi) fx=x+1, gx=sin x
(vii) fx=x+1, gx=2x+3
(viii) fx=c, c  R, gx=sin x2
(ix) fx=x2+2, gx=1-11-x

Answer:

i f x=ex, gx=loge xf:R0,; g:0,RComputing fog:Clearly, the range of g is a subset of the domain of f.fog : 0,Rfog x=f g x=f loge x=loge ex=xComputing gof:Clearly, the range of f is a subset of the domain of g.fog : RRgof x=g f x=g ex=loge ex=x

ii f x=x2, gx=cos xf:R[0, ) ; g:R-1, 1Computing fog:Clearly, the range of g is not a subset of the domain of f.Domain fog=x: xdomain of g and gxdomain of fDomain fog=x: xR and cos x R}Domain of fog=Rfog: RRfog x=f g x=f cos x=cos2xComputing gof:Clearly, the range of f is a subset of the domain of g.fog : RRgof x=g f x=g x2=cos x2


iii f x=x, gx=sin xf:R0, ; g:R-1, 1Computing fog:Clearly, the range of g is a subset of the domain of f.fog : RRfog x=f g x=f sin x=sin xComputing gof:Clearly, the range of f is a subset of the domain of g.fog : RRgof x=g f x=g x=sin x

iv f x=x+1, gx=exf:RR; g:R[1, )Computing fog:Clearly, range of g is a subset of domain of f.fog : RRfog x=f g x=f ex=ex+1Computing gof:Clearly, range of f is a subset of domain of g.fog : RRgof x=g f x=g x+1=ex+1

v f x=sin-1x, gx=x2f:-1,1-π2,π2 ; g:R[0, )Computing fog:Clearly, the range of g is not a subset of the domain of f.Domain fog=x: xdomain of g and gxdomain of fDomain fog=x: xR and x2-1,1Domain fog=x: xR and x-1,1Domain of fog=-1,1fog: -1,1Rfog x=f g x=f x2=sin-1 x2Computing gof:Clearly, the range of f is a subset of the domain of g.fog : -1,1Rgof x=g f x=g sin-1x=sin-1 x2

vi fx=x+1, gx=sin xf:RR ; g:R-1, 1Computing fog:Clearly, the range of g is a subset of the domain of f.fog: RRfog x=f g x=f sin x=sin x+1Computing gof:Clearly, the range of f is a subset of the domain of g.fog : RRgof x=g f x=g x+1=sin x+1

vii f x=x+1, gx=2x+3f:RR ; g:RRComputing fog:Clearly, the range of g is a subset of the domain of f.fog: RRfog x=f g x=f 2x+3=2x+3+1=2x+4Computing gof:Clearly, the range of f is a subset of the domain of g.fog : RRgof x=g f x=g x+1=2 x+1+3=2x+5

viii f x=c, gx=sin x2f:Rc ; g:R0, 1Computing fog:Clearly, the range of g is a subset of the domain of f.fog: RRfog x=f g x=f sin x2=cComputing gof:Clearly, the range of f is a subset of the domain of g.fog : RRgof x=g f x=g c=sin c2

ix fx=x2+2f:R[2,)  gx=1-11-xFor domain of g: 1-x0 x1Domain of g=R-1gx=1-11-x=1-x-11-x=-x1-xFor range of g:y=-x1-xy-xy=-xy=xy-xy=xy-1x=yy-1Range of g =R-1So, g: R-1R-1Computing fog:Clearly, the range of g is a subset of the domain of f.fog: R-1Rfog x=f g x=f -xx-1=-xx-12+2=x2+2x2+2-4x1-x2=3x2-4x+21-x2Computing gof:Clearly, the range of f is a subset of the domain of g.gof : RRgof x=g f x=g x2+2=1-11-x2+2=1-1-x2+1=x2+2x2+1

Page No 2.54:

Question 2:

Let f(x) = x2 + x + 1 and g(x) = sin x. Show that foggof.

Answer:

fog x=f g x=fsin x=sin2x+sin x+1and gof x=g f x=g x2+x+1= sin x2+x+1So, foggof.

Page No 2.54:

Question 3:

If f(x) = |x|, prove that fof = f.

Answer:

Domains of  f and fof are same as R.
fof x=f f x=f x= x =x=f xSo,fof x=f x, xRHence, fof=f

Page No 2.54:

Question 4:

If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f2

Also, show that fof f2

Answer:

f(x) and g(x) are polynomials.
f : R
R and g : R R.
So, fog : R and gof : R R.
i fog x=f g x=f x2+1=2 x2+1+5=2x2+2+5=2x2+7

ii gof x=g f x=g 2x+5=2x+52+1=4x2+20x+26

iii fof x=f f x=f 2x+5=2 2x+5+5=4x+10+5=4x+15

iv f2 x=fx×fx=2x+52x+5=2x+52=4x2+20x+25



→→  →

Page No 2.54:

Question 5:

If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
 

Answer:

We know thatf:R-1, 1 and g: RRClearly, the range of f is a subset of the domain of g.gof:RRgof x=g f x=g sin x=2sin x

Clearly, the range of g is a subset of the domain of f.fog:RRSo, fog x=f g x=f 2x=sin 2x

Clearly, foggof

Page No 2.54:

Question 6:

Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (fh).

Answer:

We know that f:R-1, 1 and g:RRClearly, the range of g is a subset of the domain of f.fog:RRNow, fh x=fxhx=sin x cos x=12 sin 2xDomain of fh is R.Since range of sin x is [-1,1],-1sin 2x1-12sin x212Range of fh  =-12, 12So, fh:R-12, 12Clearly, range of fh is a subset of g.gofh:RR⇒domains of fog and gofh are the same.So, fog x=f g x=f 2x=sin 2xand gofhx= g fh x=g sinx cos x=2sin x cos x=sin 2xfog x= gofhx, xRHence, fog = gofh

Page No 2.54:

Question 7:

Let  f  be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.

Answer:

Given, f:RRSince gx=2x is a polynomial, g:RRClearly, gof:RR and f+f:RRSo, domains of gof and f+f are the same.gof x=g f x=2 fxf+f x=fx+fx=2 fxgof x=f+f x, xRHence, gof =f+f

Page No 2.54:

Question 8:

If fx=1-x and gx=loge x are two real functions, then describe functions fog and gof.

Answer:

 fx=1-xFor domain, 1-x≥0x1domain of f =(-, 1]f:(-, 1]0, gx=loge xClearly, g : 0, RComputation of fog:Clearly, the range of g is not a subset of the domain of f.So,we need to compute the domain of fog.Domain fog=x : xDomain g and gxDomain of fDomain fog=x: x0,  and loge x ∈ (-, 1]Domain fog=x:x0,  and x (0, e]Domain fog=x: x (0, e]Domain fog=(0, e]fog: 0, eRSo, fog x=f g x=f loge x=1-loge x Computation of gof:Clearly, the range of f is  a subset of the domain of g.gof:(-,1]Rgof x=g f x=g 1-x=loge1-x=loge 1-x12=12loge 1-x

Page No 2.54:

Question 9:

If f:-π2,π2R and g:-1, 1R be defined as
fx=tan x and gx=1-x2 respectively, describe fog and gof.

Answer:

g x=1-x2x20, x-1, 1-x20, x-1, 11-x21, x-1, 1We know that 1-x2≥0⇒0≤1-x2≤1⇒Range of gx=0, 1So, f:-π2, π2R and g:-1, 1 0, 1Computation of fog:Clearly, the range of g is a subset of the domain of f.So, fog: -1, 1Rfog x=f g x=f 1-x2=tan 1-x2Computation of gof:Clearly, the range of f is not a subset of the domain of g.Domain gof=xdomain of f and fxdomain of gDomain gof=x-π2, π2 and tan x -1,1Domain gof=x-π2, π2 and x -π4, π4Domain gof=x-π4, π4Now, gof:-π4, π4RSo, gof x=g f x=g tan x=1-tan2x 

Page No 2.54:

Question 10:

If fx=x+3 and gx=x2+1 be two real functions, then find fog and gof.

Answer:

fx=x+3For domain,x+30x-3Domain of f =[-3, ∞)Since f is a square root function, range of f =[0, ∞)f: [-3, ∞)[0, ∞)g(x)=x2+1 is a polynomial.g:RRComputation of fog:Range of g  is not a subset of the domain of f.and domain fog=x: xdomain of g and gxdomain of fxDomain fog=x:xR and  x2+1[-3, ∞)Domain fog=x:xR and  x2+1-3Domain fog=x:xR and  x2+40Domain fog=x:xR and xRDomain fog=Rfog:RRfog x=fg x=f x2+1=x2+1+3=x2+4Computation of gof:Range of f  is a subset of the domain of g.gof: [-3, ∞)Rgof x=g f x=g x+3=x+32+1=x+3+1=x+4

Page No 2.54:

Question 11:

Let f be a real function given by fx=x-2.
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f2

Also, show that foff2 .

Answer:

fx=x-2For domain,x-20x2Domain of f=[2,)Since fis a square-root function, range of f=0,So, f:[2,)0,i fofRange of f is not a subset of the domain of f.Domainfof=x: x domain of fand fxdomain of fDomainfof=x: x [2,) and x-2[2,)Domainfof=x: x [2,) and x-22Domainfof=x: x [2,) and  x-24Domainfof=x: x [2,) and  x6Domainfof=x: x6Domainfof=[6, ∞)fof :[6, ∞)Rfof x=f f x=f x-2=x-2-2

ii fofof= (fof) ofWe have, f:[2,)0, and fof : [6, )RRange of f is not a subset of the domain of fof.Then, domainfofof=x: x domain of fand fxdomain of fofDomainfofof=x: x [2,) and x-2[6,)Domainfofof=x: x [2,) and x-26Domainfofof=x: x [2,) and  x-236Domainfofof=x: x [2,) and  x38Domainfofof=x: x38Domainfofof=[38, ∞)fof :[38,)RSo, fofof x=fof f x=fof x-2=x-2-2-2

iii We have, fofof x=x-2-2-2So, fofof 38=38-2-2-2=36-2-2=6-2-2=2-2=0

iv We have, fof=x-2-2f2x=fx×fx=x-2×x-2=x-2So, fof  f2



Page No 2.55:

Question 12:

Let fx=1+x,0x23-x,2<x3. Find fof.

Answer:

fx=1+x,0x23-x,2<x3It can be written as,fx= 1+x,0x11+x,1<x23-x,2<x3  When,0x1Then, f(x)=1+xNow when ,0x1 then ,1x+12Then, f(f(x))=1+1+x=2+x   1f(x)<2When ,1<x2Then, f(x)=1+xNow when ,1<x2 then,2<x+13Then, f(f(x))=3-1+x=2-x  2f(x)<3When ,2<x3Then, f(x)=3-xNow when ,2<x3 then ,03-x<1Then, f(f(x))=1+3-x=4-x     0f(x)<1ffx= 2+x,0x12-x,1<x24-x,2<x3  

Page No 2.55:

Question 13:

If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| –x,xR. Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).

Answer:

Given: f(x) = |x| + x 
and g(x) = |x| – x,xR

fog = f(g(x)) = g(x) + g(x)                           =x - x+x - x

Therefore,
 f(g(x)) = 0                x04x             x<0fog = 4x       x<00          x0

gof = g(f(x))=f(x)-f(x)                          =x+x-x+xg(f(x))=0         x00         x<0
Therefore, g(f(x)) = gof = 0

Now, fog(−3) =(4)(−3) = −12                 (since, fog = 4x for x < 0)

fog(5) = 0                                                (since, fog = 0 for x  0)

gof(−2) = 0                                               (since, gof = 0 for x < 0)



Page No 2.68:

Question 1:

State with reasons whether the following functions have inverse:
(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Answer:

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have:
f (1) = f (2) = f (3) = f (4) = 10
f is not one-one.
f is not a bijection.
So, f does not have an inverse.

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
g (5) = g (7) = 4
f is not one-one.
f is not a bijection.
So, f does not have an inverse.

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, different elements of the domain have different images in the co-domain.
h is one-one.
Also, each element in the co-domain has a pre-image in the domain.
h is onto.
h is a bijection.
h has an inverse and it is given by
h-1={(7, 2), (9, 3), (11, 4), (13, 5)}

Page No 2.68:

Question 2:

Find f −1 if it exists : f : AB, where
(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

Answer:

(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
Given: f(x) = 3 x
So,  f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Clearly, this is one-one.
Range of f = Range of f =B
So, f is a bijection and, thus, f -1exists.  
Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
Given: f(x) = x2
So, f = {(1, 1), (3, 9), (5, 25), (7,49), (9, 81)}
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A) .
f is not a bijection.
So, f-1does not exist.

Page No 2.68:

Question 3:

Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat} defined as f (1) = a, f (2) = b, f (3) = c, g (a) = apple, g (b) = ball and g (c) =  cat. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1 and show that (gof)−1 = f 1o g−1.

Answer:

f=1, a, 2, b, 3, c and g=a, apple, b, ball, c, catClearly, f and g are bijections.So, f and g are invertible.Now,f-1=a, 1, b, 2, c, 3 and g-1=apple, a, ball, b, cat, cSo, f-1o g-1=apple, 1, ball, 2, cat, 3         ...1f:1, 2, 3a, b, c and g:a, b, capple, ball, catSo, gof:1, 2, 3apple, ball, catgof 1=g f 1=g a=applegof 2=g f 2=g b=ball,and gof 3=g f 3=g c=catgof =1, apple, 2, ball, 3, catClearly, gofis a bijection.So, gof is invertible.gof-1=apple, 1, ball, 2, cat, 3          ...2From 1 and 2, we get:gof-1=f-1o g-1

Page No 2.68:

Question 4:

Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f : AB, g : BC be defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1og−1.

Answer:

 fx=2x+1f=1, 21+1, 2, 22+1, 3, 23+1, 4, 24+1=1, 3, 2, 5, 3, 7, 4, 9gx=x2-2g=3, 32-2, 5, 52-2, 7, 72-2, 9, 92-2=3, 7, 5, 23, 7, 47, 9, 79Clearly and g are bijections and, hence, f-1:BA and g-1: CB exist.So, f-1=3, 1, 5, 2, 7, 3, 9, 4 and g-1=7, 3, 23, 5, 47, 7, 79, 9Now, f-1 o g-1:CAf-1 o g-1=7, 1, 23, 2, 47, 3, 79, 4        ...1Also, f:AB and g:BC,gof:AC, gof-1:CASo, f-1 o g-1and gof-1 have same domains.gofx=g f x=g 2x+1=2x+12-2 gofx=4x2+4x+1-2 gofx=4x2+4x-1Then, gof1=g f 1=4+4-1=7,gof2=g f 2=4+4-1=23,gof3=g f 3=4+4-1=47 and gof4=g f 4=4+4-1=79So, gof=1, 7, 2, 23, 3, 47, 4, 79gof-1=7, 1, 23, 2, 47, 3, 79, 4     ...2From 1 and 2, we get: gof-1=f-1 o g-1

Page No 2.68:

Question 5:

Show that the function f : QQ, defined by f(x) = 3x + 5, is invertible. Also, find f−1

Answer:

Injectivity of f:
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)
3x + 5 =3y + 5
3x = 3y
x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y

3x+5=y3x=y-5x=y-53Q domain


f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f  -1:
Let f-1x=y                    ...1x=fyx=3y+5x-5=3yy=x-53So, f-1x=x-53       [from 1]

Page No 2.68:

Question 6:

Consider f : RR given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer:

Injectivity of f :
Let x and y be two elements of domain (R), such that
f(x) = f(y)
4x + 3 = 4y + 3
4x = 4y
x = y
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (R), such that f(x) = y.

4x+3=y4x=y- 3x=y- 34RDomain


f is onto.
So, f is a bijection and, hence, is invertible.

Finding f  -1:
Let f-1x=y                   ...1x=fyx=4y+3x-3=4yy=x-34So, f-1x=x-34       [from 1]

Page No 2.68:

Question 7:

Consider f : RR+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1x=x-4, where R+ is the set of all non-negative real numbers.

Answer:

Injectivity of f :
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)
x2+4=y2+4x2=y2x=y   as co-domain as R+
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (Q), such that f(x) = y

x2+4=yx2=y-4x=y-4R


f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f  -1:
Let f-1x=y                    ...1x=fyx=y2+4x-4=y2y=x-4So, f-1x=x-4          [from 1]

Page No 2.68:

Question 8:

If fx=4x+36x-4, x23, show that fof(x) = x for all x23. What is the inverse of f?

Answer:

fofx=ffx=f 4x+36x-4=44x+36x-4+364x+36x-4-4=16x+12+18x-1224x+18-24x+16=34x34=xfofx=x=IX, where I is an identity function.So, f=f-1 Hence, f-1=4x+36x-4

Page No 2.68:

Question 9:

Consider f : R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f-1x=x+6-13.

Answer:

Injectivity of f :
Let x and y be two elements of domain (R+), such that
f(x)=f(y)
9x2+6x-5=9y2+6y-59x2+6x=9y2+6yx=y As, x, yR+
So, f is one-one.

Surjectivity of f:
Let y is in the co domain (Q) such that f(x) = y

9x2+6x-5=y9x2+6x=y+59x2+6x+1=y+6 Adding 1 on both sides3x+12=y+63x+1=y+63x=y+6-1x=y+6-13R+domain


f is onto.
So, f is a bijection and hence, it is invertible.

Finding f  -1:
Let f-1x=y                          ...1x=fyx=9y2+6y-5x+5=9y2+6yx+6=9y2+6y+1       adding 1 on both sidesx+6=3y+123y+1=x+63y=x+6-1y=x+6-13So, f-1x=x+6-13     [from 1]

Page No 2.68:

Question 10:

If f : RR be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).

Answer:

Injectivity of f :
Let x and y be two elements in domain (R),
 
such that,  x3-3=y3-3            x3=y3            x=y
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (R) such that f(x) = y

x3-3=yx3=y+3x=y+33R


f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f  -1:
Let f-1x=y                         ...1x=fyx=y3-3x+3=y3y=x+33 = f-1x       [from 1]So, f-1x=x+33 Now, f-124=24+33=273=333=3 and f-15=5+33=83=233=2 



Page No 2.69:

Question 11:

A function f : RR is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).

Answer:

Injectivity of f:
Let x and y be two elements of domain (R), such that
fx=fyx3+4=y3+4x3=y3x=y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (R), such that f(x) = y.

x3+4=yx3=y-4x=y-43R domain


f is onto.
So, f is a bijection and, hence, is invertible.

Finding f  -1:
Let f-1x=y              ...1x=fyx=y3+4x-4=y3y=x-43So, f-1x=x-43       [from 1]f-13=3-43 =-13=-1

Page No 2.69:

Question 12:

If f : QQ, g : QQ 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−1og −1.

Answer:

Injectivity of f:
Let x and y be two elements of domain (Q), such that
f(x) = f(y)
2x = 2y
x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y.

2x= yx= y2Q   domain


f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f  -1:
Let f-1x=y             ...1x=fyx=2yy=x2  So, f-1x=x2        from 1

Injectivity of g:
Let x and y be two elements of domain (Q), such that
g(x) = g(y)
x + 2 = y + 2
x = y
So, g is one-one.

Surjectivity of g:
Let y be in the co domain (Q), such that g(x) = y.

x+2=yx= 2-yQ   domain


g is onto.
So, g is a bijection and, hence, it is invertible.

Finding g -1:
Let g-1x=y             ...2x=gyx=y+2y=x-2So, g-1x=x-2     From 2

Verification of (gof)−1 = f−1og −1:

fx=2x; gx=x+2and f-1x=x2; g-1x=x-2Now, f-1o g-1x=f-1g-1xf-1o g-1x=f-1x-2f-1o g-1x=x-22      ...3gofx=g f x=g 2x=2x+2Let gof-1x=y  ....   4x=gofyx=2y+22y=x-2y=x-22 gof-1x=x-22   [from 4   ... 5]From 3 and 5gof-1=f-1o g-1

Page No 2.69:

Question 13:

Let A = R - {3} and B = R - {1}. Consider the function f : A B defined by f(x) = x-2x-3. Show that f is one-one and onto and
hence find f-1.                                                                                                                                                                        [CBSE 2012, 2014]

Answer:

We have,

A = R - {3} and B = R - {1}
The function f : A  B defined by f(x) = x-2x-3

Let x,yA such that fx=fy. Then,x-2x-3=y-2y-3xy-3x-2y+6=xy-2x-3y+6-x=-yx=y f is one-one.Let yB. Then, y1.The function f is onto if there exists xA such that fx=y.Now,fx=yx-2x-3=yx-2=xy-3yx-xy=2-3yx1-y=2-3yx=2-3y1-yA    y1Thus, for any yB, there exists 2-3y1-yA such thatf2-3y1-y=2-3y1-y-22-3y1-y-3=2-3y-2+2y2-3y-3+3y=-y-1=y f is onto.So, f is one-one and onto fucntion.Now,As, x=2-3y1-ySo, f-1x=2-3x1-x=3x-2x-1

Page No 2.69:

Question 14:

Consider the function f : R+-9, given by f(x) = 5x2 + 6x - 9. Prove that f is invertible with f-1(y) = 54+5y-35.      [CBSE 2015]

Answer:

We have,fx=5x2+6x-9Let y=5x2+6x-9=5x2+65x-95=5x2+2×x×35+925-925-95=5x+352-925-95=5x+352-95-9=5x+352-545y+545=5x+3525y+5425=x+3525y+5425=x+35x=5y+545-35x=5y+54-35Let gy=5y+54-35Now,fogy=fgy=f5y+54-35=55y+54-352+65y+54-35-9=55y+54+9-65y+5425+65y+54-185-9=5y+63-65y+545+65y+54-185-9=5y+63-18-455=y=IY, Identity functionAlso, gofx=gfx=g5x2+6x-9=55x2+6x-9+54-35=25x2+30x-45+54-35=25x2+30x+9-35=5x+32-35=5x+3-35=x=IX, Identity functionSo, f is invertible.Also, f-1y=gy=5y+54-35

Page No 2.69:

Question 15:

 Let f : Nbe a function defined as (x)=9x2+6x-5. Show that : NS, where  is the range of f, is invertible. find the inverse of and hence find-1(43) and -1(163).

Answer:

We have,

f : Nis a function defined as f (x) = 9x2 + 6- 5.

Let y(x) = 9x2 + 6- 5

y=9x2+6x-5y=9x2+6x+1-1-5y=9x2+6x+1-6y=3x+12-6y+6=3x+12

y+6=3x+1             yNy+6-1=3xx=y+6-13gy=y+6-13         Let x=gy

Now,

fogy=fgy=fy+6-13=9y+6-132+6y+6-13-5=9y+6-2y+6+19+2y+6-1-5=y+6-2y+6+1+2y+6-2-5=y=IY, Identity function

gofx=gfx=g9x2+6x-5=9x2+6x-5+6-13=9x2+6x+1-13=3x+12-13=3x+1-13=3x3=x=IX, Identity function

Since, fog(y) and gof(x) are identity function.

Thus, f is invertible.

So, f-1x=gx=x+6-13.

Now,

-1(43) = 43+6-13=49-13=7-13=63=2

And -1(163) = 163+6-13=169-13=13-13=123=4

Page No 2.69:

Question 16:

Let f  --43 R be a function defined as f(x) =4x3x+4 . Show that
     f : R --43  Rang (f) is one-one and onto. Hence, find -1.

Answer:

The function f:R--43R-43 is given by fx=4x3x+4.
Injectivity: Let x, yR--43 be such that
fx=fy4x3x+4=4y3y+44x3y+4=4y3x+412xy+16x=12xy+16y16x=16yx=y
Hence, f is one-one function.
Surjectivity: Let y be an arbitrary element of R-43. Then,
f(x) = y
4x3x+4=y4x=3xy+4y4x-3xy=4yx=4y4-3y
As yR-43, 4y4-3yR.
Also, 4y4-3y-43 because 4y4-3y=-4312y=-16+12y0=-16, which is not possible.
Thus,
x=4y4-3yR--43 such that
fx=f4x3x+4=44y4-3y34y4-3y+4=16y12y+16-12y=16y16=y, so every element in R-43 has pre-image in R--43.
Hence, f is onto.
Now,
x=4y4-3y
Replacing x by f-1x and y by x, we have
 f-1x=4x4-3x   

Page No 2.69:

Question 17:

Let A = R – {2} and B = R – {1}. If f : AB is a function defined by f(x)=x-1x-2, show that f is one-one and onto. Find f1.

Answer:

Given: f(x)=x-1x-2

To show f is one-one:

Let fx1=fx2x1-1x1-2=x2-1x2-2x1-1x2-2=x2-1x1-2x1x2-2x1-x2+2=x1x2-2x2-x1+2-2x1-x2=-2x2-x1-2x1+x1=-2x2+x2-x1=-x2x1=x2Hence, f is one-one.To show f is onto: Let yB y=fxy=x-1x-2yx-2=x-1xy-2y=x-1xy-x=2y-1xy-1=2y-1x=2y-1y-1Thus, for every value of y in R-1, there exists a pre-image x=2y-1y-1 in R-2.Hence, f is onto.

Since, is one-one and onto
Therefore, is invertible with f-1y=2y-1y-1.

Hence, f-1x=2x-1x-1.

 

Page No 2.69:

Question 18:

Show that the function f : NN defined by f(x) = x2 + x + 1 is one-one but not onto. Find the inverse of f : NS, where S is range of f.

Answer:

Given: The function N → N defined by f(x) = x2 + x + 1

To show f is one-one:

Let fx1=fx2x12+x1+1=x22+x2+1x12+x1=x22+x2x12+x1-x22-x2=0x12-x22+x1-x2=0x1-x2x1+x2+x1-x2=0x1-x2x1+x2+1=0x1-x2=0 or x1+x2+1=0x1=x2 or x1=-x2+1x1=x2          x1, x2NHence, f is one-one.To show f is not onto: Since fx=x2+x+1 f1=3f2=7f3=13and so onThus, Range of f=3, 7, 13, ...NHence, f is not onto.Now, Let f:NRange of fy=x2+x+1x2+x+1-y=0x2+x+1-y=0x=-1±12-411-y21x=-1±1-4+4y2x=-1±4y-32x=-1+4y-32 or x=-1-4y-32x=-1+4y-32         xNHence, f-1x=-1+4x-32.


 

Page No 2.69:

Question 19:

Consider the bijective function f: R+(7, ) given by f(x) = 16x2 + 24x + 7, where R+
is the set of positive real numbers. Find the inverse function of f.

Answer:

Given: fx=16x2+24x+7, f:R+7, 
Consider, y=fx...............(1)
y=16x2+24x+9-2    y=4x+32-24x+32=y+24x+3=y+2x=y+2-34f-1y=x=y+2-34
Hence, the inverse of f is x+2-34

Page No 2.69:

Question 20:

Let f : [−1, ∞) → [−1, ∞) be given by f(x) = (x + 1)2 − 1, x ≥ −1. Show that f is invertible. Also, find the set S = {x : f(x) = f−1 (x)}.

Answer:

Injectivity: Let x and y [-1, ), such that fx=fyx+12-1=y+12-1x+12=y+12x+1=y+1x=ySo, f is a injection.Surjectivity: Let y [-1, ). Then, fx=yx+12-1=yx+1=y+1x=y+1-1Clearly, x=y+1-1 is real for all y-1.Thus, every element y [-1, ) has its pre-image x[-1, ) given by x=y+1-1.f is a surjection.So, f is a bijection.Hence, f is invertible.Let f-1x=y                          ...(1)fy=xy+12-1=xy+12=x+1y+1=x+1y=±x+1-1f-1x=±x+1-1   [from 1]fx=f-1xx+12-1=±x+1-1x+12=±x+1x+14=x+1x+1x+13-1=0x+1=0 or x+13-=0x=-1 or x+13=1x=-1 or x+1=1x=-1 or x=0S=0, -1

Page No 2.69:

Question 21:

Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : AA, g : AA 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.

Answer:

 f is not one-one because
f-1=-12=1and f1=12=1
-1 and 1 have the same image under f.
f is not a bijection.
So, f -1 does not exist.

Injectivity of g:
Let x and y be any two elements in the domain (A), such that
gx=gysin πx2=sin πy2πx2=πy2x=y
So, g is one-one.

Surjectivity of g:
Range of g sin π-12, sin π12 = sin -π2, sin π2 = -1, 1 = (co-domain of g)
g is onto.
g is a bijection.
So, g-1 exists.

Also,
let g-1x=y                      ...1gy=xsinπy2=xπy2=sin-1 xy=2πsin-1 xg-1x=2πsin-1 x        [from 1]

Page No 2.69:

Question 22:

Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.

Answer:

Injectivity:
Let x and y be two elements in the domain (R), such that
fx=fycosx+2=cosy+2x+2=y+2 or x+2=2π-y+2x=or x+2=2π-y-2x=or x=2π-y-4So, we cannot say that x=yFor example,cosπ2=cos 3π2=0So,π2 and 3π2  have the same image 0.
f is not one-one.
f is not a bijection.
Thus, f  is not invertible.

Page No 2.69:

Question 23:

If A = {1, 2, 3, 4} and B = {a, b, c, d}, define any four bijections from A to B. Also give their inverse functions.

Answer:

f1=1, a, 2, b, 3, c, 4, df1-1=a, 1, b, 2, c, 3, d, 4f2=1, b, 2, a, 3, c, 4, df2-1=b, 1, a, 2, c, 3, d, 4f3=1, a, 2, b, 4, c, 3, df3-1=a, 1, b, 2, c, 4, d, 3f4=1, b, 2, a, 4, c, 3, df4-1=b, 1, a, 2, c, 4, d, 3
Clearly, all these are bijections because they are one-one and onto.

Page No 2.69:

Question 24:

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.

Answer:

 A and B are two non empty sets. Let f be a function from A to B .It is given that there is injective map from A to B. That means f is one-one function .It is also given that there is injective map from  B to A .That means every element of set B has its image in set A.f is onto function or surjective. f is bijective.If a function is both injective and surjective, then the function is bijective.     

Page No 2.69:

Question 25:

If f : AA, g : AA are two bijections, then prove that
(i) fog is an injection
(ii) fog is a surjection

Answer:

Given: AA, g : AA are two bijections.
Then,  fog : AA

(i) Injectivity of fog:
Let x and y be two elements of the domain (A), such that
fogx=fogyfgx=fgygx=gy As, f is one-onex=y          As, g is one-one
So,  fog is an injection.

(ii) Surjectivity of fog:
Let z be an element in the co-domain of fog (A).
Now, z∈A co-domain of f and f is a surjection.So, z=fy, where y∈A domain of f ...1Now, y∈A co-domain of g and g is a surjection.Soy=gx, where x∈A domain of g ...2From 1 and 2,z=fy=fgx=fogx, where xAdomain of fog
So,  fog is a surjection.



Page No 2.72:

Question 1:

Let A=x  R : -1 x1=B and C=x  R : x0 and
let S=x, y  A×B : x2+y2=1 and S0=x, y  A×C : x2+y2=1. Then,
(a) S defines a function from A to B
(b) S0 defines a function from A to C
(c) S0 defines a function from A to B
(d) S defines a function from A to C

Answer:

(a) S defines a function from A to B

Let xA-1x1Now, x2+y2=1y2=1-x2y=±1-x2-1y1 yBThus, S defines a function from A to B.

Page No 2.72:

Question 2:

f : RR given by fx=x+x2 is
(a) injective
(b) surjective
(c) bijective
(d) None of these

Answer:

fx=x+x2=x±x=0 or 2x⇒ Each element of the domain has 2 images.
f is not a function.
So, the answer is (d).

Page No 2.72:

Question 3:

If f : AB given by 3fx+2-x=4 is a bijection, then
(a) A=x  R : -1 < x < , B=x  R : 2 < x < 4
(b) A=x  R : -3 < x < , B=x  R : 2 < x < 4
(c) A=x  R : -2 < x < , B=x  R : 2 < x < 4
(d) None of these

Answer:

(d) None of these

f:AB3f(x)+2-x=4 3f(x)=4-2-xTaking log on both the sides , f(x) log 3=log 4-2-xf(x)=log 4-2-xlog 3Logaritmic function will only be defined if 4-2-x>04>2-x22>2-x2>-x-2<xx -2,That means  A=xR:-2<x<As we know that, f(x)=log 4-2-xlog 3We take x=0 -2,f(x)=1 which does not belong to any of the options .

Page No 2.72:

Question 4:

The function f : RR defined by fx=2x+2x is
(a) one-one and onto
(b) many-one and onto
(c) one-one and into
(d) many-one and into

Answer:

(d) many-one and into

Graph for the given function is as follows.


A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y.
That means it is many one function.

From the given graph we can see that the range is [2,) 
and R is the co-domain of the given function.
Hence, Co-domainRange
Therefore, the given function is into.



 

Page No 2.72:

Question 5:

Let the function f : R--bR-1 be defined by

fx=x+ax+b, ab. Then,

(a) f is one-one but not onto
(b) f is onto but not one-one
(c) f is both one-one and onto
(d) None of these

Answer:

(c) f is both one-one and onto

Injectivity:
Let x and y be two elements in the domain R- {-b}, such that
fx=fyx+ax+b=y+ay+bx+ay+b=x+by+axy+bx+ay+ab=xy+ax+by+abbx+ay=ax+bya-bx=a-byx=y
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain of f, i.e. R-{1}, such that f (x)=y

fx=yx+ax+b=yx+a=yx+ybx-yx=yb-ax1-y=yb-ax=yb-a1-yR--b
So,  f is onto.



Page No 2.73:

Question 6:

The function f : AB defined by fx=-x2+6x-8 is a bijection if
(a) A=(-, 3] and B=(-, 1]
(b) A=[-3, ) and B=(-, 1]
(c) A=(-, 3] and B=[1, )
(d) A=[3, ) and B=[1, )

Answer:

(a) A=(-, 3] and B=(-, 1]


fx=-x2+6x-8 , is a polynomial functionAnd the domain of polynomial function is real number.xR

f(x) =-x2+6x-8       =-x2-6x+8       =-x2-6x+9-1       =-x-32+1Maximum value of -x-32 woud be 0Maximum value of -x-32+1 woud be 1 f(x) (-,1]



We can see from the given graph that function is symmetrical about x=3& the given function is bijective .So, x would be either (-,3] or [3,)The correct option which satisfy A and B both is: 
A=(-, 3] and B=(-, 1]

Page No 2.73:

Question 7:

Let A=x  R : -1x1=B. Then, the mapping f : AB given by fx=xx is
(a) injective but not surjective
(b) surjective but not injective
(c) bijective
(d) none of these

Answer:

Injectivity:
Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that
fx=fyxx=yyxx=yyx2=y2x=y

Case-2: Let x and y be two negative numbers, such that
fx=fyxx=yyx-x=y-y-x2=-y2x2=y2x=y

Case-3: Let x be positive and y be negative.
Then, xyfx=xx is positive and fy=yy is negativefxfySo, xyfxfy
From the 3 cases, we can conclude that  f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
Case-1: Let y>0. Then, 0<y≤1y=fx=xx>0x>0x=xfx=yxx=yxx=yx2=yx=y  A We do not get ± because x>0Case-2: Let y<0. Then, -1≤y<0y=fx=xx<0x<0x=-xfx=yxx=yx-x=y-x2=yx2=-yx=--y  A We do not get ± because x>0
f is onto.
f is a bijection.
So, the answer is (c).

Page No 2.73:

Question 8:

Let f : RR be given by fx=x2+x+1-3 where [x] denotes the greatest integer less than or equal to x. Then, f(x) is
(a) many-one and onto
(b) many-one and into
(c) one-one and into
(d) one-one and onto

Answer:

(b) many-one and into

f : RR

fx=x2+x+1-3

It is many one function because in this case for two different values of x
we would get the same value of f(x) .
For x=1.1, 1.2 Rf(1.1)=1.12 +1.1+1-3           =1.21+2.1-3           =1+2-3           =0f(1.1)=1.22 +1.2+1-3           =1.44+2.2-3           =1+2-3           =0


It is into function because for the given domain we would only get the integral values of
f(x).
but R is the codomain of the given function.
That means , CodomainRange
Hence, the given function is into function.

Therefore, f(x) is many one and into

Page No 2.73:

Question 9:

Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : MR defined by f(A) = |A| for every AM, is
(a) one-one and onto
(b) neither one-one nor onto
(c) one-one but-not onto
(d) onto but not one-one

Answer:


M=A=abcd: a, b, c, dR f: MR is given by fA=A

Injectivity:
f0000=0000=0and f1000=1000=0f0000=f1000=0
So, f is not one-one.

Surjectivity:
Let y be an element of the co-domain, such that
fA=-y, A=abcdabcd=yad-bc=ya, b, c, dR A=abcdM
f is onto.
So, the answer is (d).

Page No 2.73:

Question 10:

The function f : [0, )R given by fx=xx+1 is
(a) one-one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) onto but not one-one

Answer:

Injectivity:
Let x and y be two elements in the domain, such that
fx=fyxx+1=yy+1xy+x=xy+yx=y
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that
y=fxy=xx+1xy+y=xxy-1=-yx=-yy-1Range of f=R-1co domain (R)
f is not onto.
So, the answer is (b).

Page No 2.73:

Question 11:

The range of the function fx=7-xPx-3 is
(a) {1, 2, 3, 4, 5}
(b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3, 4}
(d) {1, 2, 3}

Answer:

We know that
7-x>0; x-3 0 and 7-xx-3x<7; x3 and 2x10x<7; x3 and x5So, x=3, 4, 5Range of f=P3-37-3, P4-37-4 , P7-55-3=4P0, 3P1, 2P2=1, 3, 2=1, 2, 3
So, the answer is (d).

Page No 2.73:

Question 12:

A function f  from the set of natural numbers to integers defined by

fn=n-12, when n is odd-n2, when n is evenis

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto both

Answer:

(d) one-one and onto both

Injectivity:

Let x and y be any two elements in the domain (N).
Case-1: Both x and y are even.Let fx=fy-x2=-y2-x=-yx=yCase-2: Both x and y are odd.Let fx=fyx-12=y-12x-1=y-1x=yCase-3Let x be even and y be odd. Then, fx=-x2and fy=y-12Then, clearly xy fxfyFrom all the cases, f is one-one.

Surjectivity:

Co-domain of f=Z=...,-3, -2, -1, 0, 1, 2, 3, ....Range of f=..., -3-12, --22,-1-12, 02, 1-12, -22, 3-12, ...Range of f=...,-2, 1, -1, 0, 0, -1, 1,...Range of f=..., -2, -1, 0, 1, 2, ....Co-domain of f=Range of f
f is onto.

Page No 2.73:

Question 13:

Let f be an injective map with domain {x, y, z} and range {1, 2, 3}, such that exactly one of the following statements is correct and the remaining are false.

fx=1, fy1, fz2.

The value of f-1 1 is
(a) x
(b) y
(c) z
(d) none of these

Answer:

Case-1: Let fx=1 be true.Then, fy≠1 and fz2 are false.So, f(y)=1 and fz=2⇒ fx=1, fy=1x and y have the same images.This contradicts the fact that fis one-one.Case-2: Let fy≠1 be true.Then, fx=1 and fz2 are false.So,  fx≠1 and fz=2⇒ fx1, fy1 and fz=2⇒There is no pre-image for 1.This contradicts the fact that range is 1, 2, 3.Case-3: Let fz2 be true.Then, fx=1 and fy1 are false.So,  fx≠1 and fy=1fx=2, fy=1 and fz=3fy=1f-11=y
So, the answer is (b).

Page No 2.73:

Question 14:

Which of the following functions form Z to itself are bijections?
(a) fx=x3
(b) fx=x+2
(c) fx=2x+1
(d) fx=x2+x

Answer:

a f is not onto because for = 3∈Co-domain(Z), there is no value of x∈Domain(Z)x3=3x=33Zf is not onto.So, fis not a bijection.

(b) Injectivity:
Let x and y be two elements of the domain (Z), such that
 x+2=y+2x=y
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that
y=fxy=x+2x=y-2Z Domain
f is onto.
So, f is a bijection.

c fx=2x+1 is not onto because if we take 4 ∈ Zco domain, then 4=fx4=2x+12x=3x=32ZSo, f  is not a bijection.d f0=02+0=0and f-1=-12+-1=1-1=0⇒0 and -1 have the same image.f is not one-one.So, f is not a bijection.

So, the answer is (b).

Page No 2.73:

Question 15:

Which of the following functions from A=x : -1x1 to itself are bijections?
(a) fx=x2
(b) gx=sinπ x2
(c) hx=|x|
(d) kx=x2

Answer:

a Range of f =-12, 12 ≠ ASo, f is not a bijection.b Range =sin-π2, sinπ2=-1,1=ASo, g is a bijection.c h-1=-1=1and h1=1=1⇒-1 and 1 have the same imagesSo, h is not a bijection.d k-1=-12=1and k1=12=1⇒-1 and 1 have the same imagesSo, k is not a bijection.
So, the answer is (b).

Page No 2.73:

Question 16:

Let A=x : -1x1 and f : AA such that fx=x|x|, then f is
(a) a bijection
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective

Answer:

Injectivity:
Let x and y be any two elements in the domain A.

Case-1: Let x and y be two positive numbers, such that
fx=fyxx=yyxx=yyx2=y2x=y

Case-2: Let x and y be two negative numbers, such that
fx=fyxx=yyx-x=y-y-x2=-y2x2=y2x=y

Case-3: Let x be positive and y be negative.
Then, xyfx=xx is positive and fy=yy is negativefxfySo, xyfxfy
So, f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)
Case-1: Let y>0. Then, 0<y≤1y=fx=xx>0x>0x=xfx=yxx=yxx=yx2=yx=y  A We do not get ±, as x>0Case-2: Let y<0Then-1≤y<0y=fx=xx<0x<0x=-xfx=yxx=yx-x=y-x2=yx2=-yx=--y  A We do not get ±, as x>0
f is onto
f is a bijection.
So, the answer is (a).

Page No 2.73:

Question 17:

If the function f : RA given by fx=x2x2+1 is a surjection, then A =
(a) R
(b) [0, 1]
(c) [0, 1)
(d) [0, 1)

Answer:


As f is surjective, range of f=co-domain of fA= range of f fx= x2x2+1,  y=x2x2+1yx2+1= x2y-1x2+y= 0x2= -yy-1x=y1-yy1-y0y[0, 1)Range of f= [0, 1)A= [0, 1)
So, the answer is (d).

Page No 2.73:

Question 18:

If a function f : [2, )  B defined by fx=x2-4x+5 is a bijection, then B =
(a) R
(b) [1, ∞)
(c) [4, ∞)
(d) [5, ∞)

Answer:

Since f is a bijection, co-domain of f = range of f
B = range of f
Given: fx=x2-4x+5Let fx=yy=x2-4x+5x2-4x+5-y=0Discrimant, D=b2-4ac0,-42-4×1×5-y016-20+4y04y4y1y[1, )Range of f=[1, )B=[1, )
So, the answer is (b).



Page No 2.74:

Question 19:

The function f : RR defined by
fx=x-1 x-2 x-3 is
(a) one-one but not onto
(b) onto but not one-one
(c) both one and onto
(d) neither one-one nor onto

Answer:

fx=x-1x-2x-3

Injectivity:
f1=1-11-21-3=0f2=2-12-22-3=0f3=3-13-23-3=0⇒ f1=f2=f3=0
So, f is not one-one.

Surjectivity:
Let y be an element in the co domain R, such that
y=fxy=x-1x-2x-3Since yR and xR, f is onto.

So, the answer is (b).

Page No 2.74:

Question 20:

The function f : -1/2, 1/2, 1/2-π/2, π/2 , defined by fx=sin-1 3x-4x3, is
(a) bijection
(b) injection but not a surjection
(c) surjection but not an injection
(d) neither an injection nor a surjection

Answer:

fx=sin-13x-4x3fx=3sin-1x

Injectivity:
Let x and y be two elements in the domain -12, 12 , such that
fx=fy3sin-1x=3sin-1ysin-1x=sin-1yx=y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain, such that
fx=y3sin-1x=ysin-1x=y3x=siny3-12, 12
f is onto.
f is a bijection.
So, the answer is (a).

Page No 2.74:

Question 21:

Let f : RR be a function defined by fx=e|x|-e-xex+e-x. Then,
(a) f is a bijection
(b) f is an injection only
(c) f is surjection on only
(d) f is neither an injection nor a surjection

Answer:

(d) f is neither an injection nor a surjection

f : RR

fx=e|x|-e-xex+e-xFor x=-2 and -3 R f(-2) =e-2-e2e-2+e2           =e2-e2e-2+e2           = 0& f(-3) =e-3-e3e-3+e3              =e3-e3e-3+e3              = 0Hence, for different values of x we are getting same values of f(x)That means , the given function is many one .
Therefore, this function is not injective.

For x<0f(x) =0For x>0f(x) =ex-e-xex+e-x        =ex+e-xex+e-x-2e-xex+e-x        =1-2e-xex+e-xThe value of 2e-xex+e-x is always  positive.Therefore, the value of f(x) is always less than 1Numbers more than 1 are not included in the range but they are included in codomain.As the codomain is R. CodomainRangeHence, the given function is not onto .
Therefore, this function is not surjective .

Page No 2.74:

Question 22:

Let f : R-nR be a function defined by

fx=x-mx-n, where mn. Then,

(a) f is one-one onto
(b) f is one-one into
(c) f is many one onto
(d) f is many one into

Answer:

Injectivity:
Let x and y be two elements in the domain R-{n}, such that
fx=fyx-mx-n=y-my-nx-my-n=x-ny-mxy-nx-my+mn=xy-mx-ny+mnm-nx=m-nyx=y
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that
fx=yx-mx-n=yx-m=xy-nyny-m=xy-xny-m=xy-1x=ny-my-1, which is not defined for y =1So, 1 ∈ R co domain has no pre image in R-n
f is not onto.
Thus, the answer is (b).

Page No 2.74:

Question 23:

Let f : RR be a function defined by fx=x2-8x2+2. Then,  f is
(a) one-one but not onto
(b) one-one and onto
(c) onto but not one-one
(d) neither one-one nor onto

Answer:

Injectivity:
Let x and y be two elements in the domain (R), such that
fx=fyx2-8x2+2=y2-8y2+2x2-8y2+2=x2+2y2-8x2y2+2x2-8y2-16=x2y2-8x2+2y2-1610x2=10y2x2=y2x=±y
So, f is not one-one.

Surjectivity:
f-1=-12-8-12+2=1-81+2=-73 and f1=12-812+2=1-81+2=-73f-1=f1=-73
f is not onto.
The correct answer is (d).

Page No 2.74:

Question 24:

f : RR is defined by fx=ex2-e-x2ex2+e-x2 is
(a) one-one but not onto
(b) many-one but onto
(c) one-one and onto
(d) neither one-one nor onto

Answer:

(d) neither one-one nor onto

We have,fx=ex2-e-x2ex2+e-x2Here, -2, 2RNow, 2-2But, f2=f-2Therefore, function is not one-one.And,The minimum value of the function is 0 and maximum value is 1That is range of the function is 0, 1 but the co-domain of the function is given R.Therefore, function is not onto.function is neither one-one nor onto.

Page No 2.74:

Question 25:

The function f : RR, fx=x2 is
(a) injective but not surjective
(b) surjective but not injective
(c) injective as well as surjective
(d) neither injective nor surjective

Answer:

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y). Then,

x2=y2x=±y


So, f is not one-one.

Surjectivity:
As f-1=-12=1and f1=12=1, f-1=f1
So, both -1 and 1 have the same images.
f is not onto.
So, the answer is (d).

x2+x+1=y2+y+1(x2y2)+(xy)=0(x+y)(xy)+(xy)=0(xy)(x+y+1)=0xy=0 (x+y+1) cannot be zero because x and y are natural numbers)x=y

Page No 2.74:

Question 26:

A function f from the set of natural numbers to the set of integers defined by

fnn-12,when n is odd-n2,when n is evenis

(a) neither one-one nor onto
(b) one-one but not onto
(c) onto but not one-one
(d) one-one and onto

Answer:

Injectivity:
Let x and y be any two elements in the domain (N).
Case-1: Both x and y are even.Let fx=fy-x2=-y2-x=-yx=yCase-2: Both x and y are odd.Let fx=fyx-12=y-12x-1=y-1x=yCase-3Let x be even and y be odd. Then, fx=-x2and fy=y-12Then, clearly xy fxfyFrom all the cases, f is one-one.

Surjectivity:
Co-domain of f=Z=...,-3, -2, -1, 0, 1, 2, 3, ....Range of f=..., -3-12, --22,-1-12, 02, 1-12, -22, 3-12, ...Range of f=...,-2, 1, -1, 0, 0, -1, 1,...Range of f=..., -2, -1, 0, 1, 2, ....Co-domain of f=Range of f
f is onto.
So, the answer is (d).

Page No 2.74:

Question 27:

Which of the following functions from A=x  R : -1x1 to itself are bijections?

(a) fx=|x|
(b) fx=sinπ x2
(c) fx=sinπ x4
(d) None of these

Answer:

(b) fx=sinπ x2

It is clear that  f(x) is one-one.

Range of f=sinπ-12, sinπ12=sin -π2, sinπ2=-1,1= A=Co domain of f
⇒ f is onto.
So, f is a bijection.

Page No 2.74:

Question 28:

Let f : ZZ be given by fx=x2, if x is even0, if x is odd
Then,  f is
(a) onto but not one-one
(b) one-one but not onto
(c) one-one and onto
(d) neither one-one nor onto

Answer:

Injectivity:
Let x and y be two elements in the domain (Z), such that
fx=fyCase-1: Let both x and y be even.Then,fx=fyx2=y2x=yCase-2: Let both x and y be odd.Then,fx=fy0=0Here, we cannot determine whether x=y.
So, f is not one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that
Co-domain of f =Z={0, ±1, ±2, ±3, ±4, ...}Range of f=0, 0, ±22, 0, ±42 ,...=0, ±1, ±2, ...Co-domain of f=Range of f
f is onto.
So, the answer is (a).

Page No 2.74:

Question 29:

The function f : RR defined by fx=6x+6|x| is
(a) one-one and onto
(b) many one and onto
(c) one-one and into
(d) many one and into

Answer:

(d) many one and into

Graph of the given function is as follows :



A line parallel to X axis is cutting the graph at two different values.

Therefore, for two different values of x we are getting the same value of y .
That means it is many one function .

From the given graph we can see that the range is [2,) 
and R is the codomain of the given function .
Hence, CodomainRange
Therefore, the given function is into .
 



Page No 2.75:

Question 30:

Let fx=x2 and gx=2x. Then, the solution set of the equation fog x=gof x is
(a) R
(b) {0}
(c) {0, 2}
(d) none of these

Answer:


Since fogx=gofx, fgx=gfxf2x=gx22x2=2x222x=2x2x2=2xx2-2x=0xx-2=0x=0, 2x0, 2
So, the answer is (c).

Page No 2.75:

Question 31:

If f : RR is given by fx=3x-5, then f-1x
(a) is given by 13x-5
(b) is given by x+53
(c) does not exist because f is not one-one
(d) does not exist because f is not onto

Answer:

Clearly, f is a bijection.
So, f -1 exists.
Let f-1x=y                   ...1fy=x3y-5=x3y=x+5y=x+53f-1x=x+53          [from 1]
So, the answer is (b).

Page No 2.75:

Question 32:

If g f x=sin x and f g x=sin x2, then
(a) fx=sin2 x, gx=x
(b) fx=sin x, gx=|x|
(c) fx=x2, gx=sin x
(d) f and g cannot be determined.

Answer:

If we solve it  by the trial-and-error method, we can see that (a) satisfies the given condition.
From (a):
fx=sin2x and gx=xfgx=fx=sin2 x=sin x2
So, the answer is (a).

Page No 2.75:

Question 33:

The inverse of the function f : Rx  R : x < 1 given by fx=ex-e-xex+e-x is
(a) 12 log 1+x1-x

(b) 12 log 2+x2-x

(c) 12 log 1-x1+x

(d) none of these

Answer:

Let f-1x=y         ...1fy=xey-e-yey+e-y=xe-ye2y-1e-ye2y+1=xe2y-1=xe2y+1e2y-1=xe2y+xe2y1-x=x+1e2y=1+x1-x2y=loge 1+x1-xy=12loge 1+x1-xf-1x=12loge 1+x1-x                [from 1]
So, the answer is (a).

Page No 2.75:

Question 34:

Let A=x  R : x  1. The inverse of the function, f : AA given by fx=2x x-1, is

(a) 12x x-1

(b) 12 1+1+4 log2 x

(c) 12 1-1+4 log2 x

(d) not defined

Answer:

Let f-1x=y...1fy=x2yy-1=x2y2-y=xy2-y=log2 xy2-y+14=log2 x+14y-122=4log2 x+14y-12=±4log2 x+12y=12±4log2 x+12y=12+4log2 x+12                      ∵ y ≥1So, f-1x=12(1+1+4log2 x )       [from 1]
So, the answer is (b).

Page No 2.75:

Question 35:

Let A=x  R : x 1 and f : AA  be defined as fx=x 2-x. Then, f-1 x is
(a) 1+1-x
(b) 1-1-x
(c) 1-x
(d) 1±1-x

Answer:

Let y be the element in the codomain R such that        f-1x=y ...1fy=x and y ≤1y2-y=x2y-y2=xy2-2y+x=0y2-2y=-xy2-2y+1=1-xy-12=1-xy-1=±1-xy=1±1-xy=1-1-x        ∵ y ≤1
The correct answer is (b).

Page No 2.75:

Question 36:

Let fx=11-x. Then, f o fof x
(a) x for all x  R
(b) x for all x  R-1
(c) x for all x  R-0, 1
(d) none of these

Answer:

Domain of f:1-x0x1Domain of f=R-1Range of f:y=11-x1-x=1yx=1-1yx=y-1yy0Range of f=R-0So, f:R-1R-0 and f:R-1R-0 Range of f is not a subset of the domain of f.Domain fof=x: xdomain of f and fxdomain of fDomain fof=x: xR-1 and 11-xR-1 Domain fof=x: x 1 and 11-x1Domain fof=x: x 1 and 1-x1Domain fof=x: x 1 and x0Domain fof=R-0, 1fofx=ffx=f11-x=11-11-x=1-x1-x-1=1-x-x=x-1xFor range of fof, x0Now, fof:R-0, 1R -0 and f:R-1R-0Range of fof is not a subset of domain of f.Domainf o fof=x: xdomain of fof and fofxdomain of fDomain f o fof=x: xR-0, 1 and x-1xR-1 Domainf o fof=x: x 0, 1 and x-1x1Domain f o fof=x: x 0, 1 and x-1xDomain f o fof=x: x 0, 1 and xRDomain f o fof=R-0, 1fofofx=ffofx=fx-1x=11-x-1x=xx-x+1=xSo, fofofx=x, where x0,1
So, the answer is (c).

Page No 2.75:

Question 37:

If the function f : RR be such that fx=x-x, where [x] denotes the greatest integer less than or equal to x, then f-1 x is

(a) 1x-x

(b) [x] − x

(c) not defined

(d) none of these

Answer:

f(x) = x - [x]
We know that the range of f is [0, 1).
Co-domain of f = R
As range of f Co-domain of ff is not onto.
f is not a bijective function.
So,  f -1 does not exist.
Thus, the answer is (c).

Page No 2.75:

Question 38:

If F : [1, )[2, ) is given by fx=x+1x, then f-1 x equals

(a) x+x2-42

(b) x1+x2

(c) x-x2-42

(d) 1+x2-4

Answer:

Let f-1x = yfy=xy+1y=xy2+1=xyy2-xy+1=0y2-2×y×x2+x22-x22+1=0y2-2×y×x2+x22=x2-14y-x22=x2-14y-x2=x2-42y=x2+x2-42y=x+x2-42f-1x=x+x2-42

So, the answer is (a).

Page No 2.75:

Question 39:

Let gx=1+x-x and fx=-1,x<00,x=0, 1,x>0, where [x] denotes the greatest integer less than or equal to x. Then for all x, f g x is equal to
(a) x
(b) 1
(c) f(x)
(d) g(x)

Answer:

(b) 1

When, -1<x<0Then, g(x)=1+x-x               =1+x--1=2+xfg(x)=1 When, x=0Then, g(x)=1+x-x               =1+x-0=1+xfg(x)=1When, x>1Then, g(x)=1+x-x               =1+x-1=xfg(x)=1

Therefore, for each interval f(g(x))=1

Page No 2.75:

Question 40:

Let fx=α xx+1, x-1. Then, for what value of α is f fx=x?
(a) 2
(b) -2
(c) 1
(d) −1

Answer:

(d) −1

  ffx=x fαxx+1=xααxx+1αxx+1+1=xα2xαx+x+1=xα2x=αx2+x2+xα2x-αx2-x2-x=0α2x-αx2-x2+x=0Solving for the α we get,α=--x2±-x22-4×x×-x2+x2x  =x2±x4+4x3+4x22x  =x+1, -1Here, -1 is independent of x,for, α=-1, ffx=x   



Page No 2.76:

Question 41:

The distinct linear functions that map [−1, 1] onto [0, 2] are
(a) fx=x+1, gx=-x+1
(b) fx=x-1, gx=x+1
(c) fx=-x-1, gx=x-1
(d) None of these

Answer:

Let us substitute the end-points of the intervals in the given functions. Here, domain = [-1, 1] and range =[0, 2]
By substituting -1 or 1 in each option, we get:

Option (a):
f-1=-1+1=0 and f1=1+1=2g-1=1+1=2 and g1=-1+1=0
So, option (a) is correct.

Option (b):
f-1=-1-1=-2 and f1=1-1=0g-1=-1+1=0 and g1=1+1=2
Here, f(-1) gives -20, 2
So, (b) is not correct.

Similarly, we can see that (c) is also not correct.

Page No 2.76:

Question 42:

Let f : [2, )X be defined by fx=4x-x2. Then, f is invertible if X =
(a) [2, )
(b) (-, 2]
(c) (-, 4]
(d) [4, )

Answer:

Since f is invertible, range of f = co domain of f = X
So, we need to find the range of f to find X.
For finding the range, let
fx=y4x-x2=yx2-4x=-yx2-4x+4=4-yx-22=4-yx-2=±4-yx=2±4-yThis is defined only when 4-y0y4X=Range of f=(-,4]
So, the answer is (c).

Page No 2.76:

Question 43:

If f : R -1, 1 is defined by fx=-x|x|1+x2, then f-1 x equals
(a) x1-x

(b) Sgn x x1-x

(c) -x1-x

(d) None of these

Answer:

(b) -Sgn x x1-x

We have, fx=-x|x|1+x2    x-1, 1Case-IWhen, x<0,Then, x=-xAnd fx>0Now,fx=-x-x1+x2y=x21+x2y1=x21+x2y+1y-1=x2+1+x2x2-1-x2           Using Componendo and dividendoy+1y-1=2x2+1-1-y+1y-1=2x2+12y1-y=2x2y1-y=x2x=-y1-y                         As x<0x=-y1-y                        As y>0To find the inverse interchanging x and y we get,f-1x=-x1-x            ...iCase-IIWhen, x>0,Then, x=xAnd fx<0Now,fx=-xx1+x2y=-x21+x2y1=-x21+x2y+1y-1=-x2+1+x2-x2-1-x2           Using Componendo and dividendoy+1y-1=1-2x2-11+y1-y=12x2+11-y1+y=2x2+1-2y1+y=2x2x2=-y1+yx=-y1+y                         As x>0x=y1-y                        As y<0To find the inverse interchanging x and y we get,f-1x=x1-x            ...iiCase-IIIWhen, x=0,Then, fx=0Hence, f-1x=0     ...iiiCombinig equation i , ii and iii we get,f-1x=-Sgnxx1-x  

Page No 2.76:

Question 44:

Let [x] denote the greatest integer less than or equal to x. If fx=sin-1x, gx=x2 and hx=2x,12x12, then
(a) fogohx=π2
(b) fogohx=π
(c) hofog=hogof
(d) hofoghogof

Answer:

(c) hofog=hogof

We have,gx=x2       =0           As 12x 12 14x2 12fogx=fgx=sin-10                       =0hofogx=hfgx=2×0=0

Andfx=sin-1xNow,for, x12, 12fxπ6, π4fx0.52, 0.78gofx=0    As, fx0.52, 0.78                       =0hogofx=hgfx=2×0=0

  hofog=hogof=0
 

Page No 2.76:

Question 45:

If gx=x2+x-2 and 12 gofx=2x2-5x+2, then f(x) is equal to
(a) 2 x-3
(b) 2 x+3
(c) 2 x2+3x+1
(d) 2 x2-3x-1

Answer:

We will solve this problem by the trial-and-error method.
Let us check option (a) first.
If fx = 2x-312gofx = gfx= 12g2x-3= 122x-32+2x-3-2= 124x2+9-12x+2x-3-2= 124x2-10x+4= 2x2-5x+2
The given condition is satisfied by (a).
So, the answer is (a).

Page No 2.76:

Question 46:

If fx=sin2 x and the composite function gfx=sin x, then g(x) is equal to
(a) x-1
(b) x
(c) x+1
(d) -x

Answer:

(b)

 If we take gx = x, thengfx = gsin2x = sin2x = ±sin x = sin x

So, the answer is (b).

Page No 2.76:

Question 47:

If f : RR is given by fx=x3+3, then f-1x is equal to
(a) x1/3-3
(b) x1/3+3
(c) x-31/3
(d) x+31/3

Answer:

(c)

Let f-1x = yfy = xy3+3 = xy3 = x-3y = x-33 y = x-313
So, the answer is (c).

Page No 2.76:

Question 48:

Let fx=x3 be a function with domain {0, 1, 2, 3}. Then domain of f-1 is
(a) {3, 2, 1, 0}
(b) {0, −1, −2, −3}
(c) {0, 1, 8, 27}
(d) {0, −1, −8, −27}

Answer:

(c) {0, 1, 8, 27}


fx=x3Domain = 0, 1, 2, 3Range = 03, 13, 23, 33 = 0, 1, 8, 27So, f 0, 0, 1, 1, 2, 8, 3, 27f-1 = 0, 0, 1, 1, 8, 2, 27, 3Domain of f-1 = 0, 1, 8, 27

Page No 2.76:

Question 49:

Let f : RR be given by fx=x2-3. Then, f-1 is given by
(a) x+3
(b) x+3
(c) x+3
(d) None of these

Answer:

(d)

Let f-1x=yfy=xy2-3=xy2=x+3y=±x+3
So, the answer is (d).

Page No 2.76:

Question 50:

Mark the correct alternative in the following question:

Let f : R R be given by f(x) = tanx. Then, f -1(1) is

(a) π4                              (b) nπ+π4:nZ                              (c) does not exist                              (d) none of these

Answer:

We have,f:RR is given byfx=tanxf-1x=tan-1x f-11=tan-11=nπ+π4:nZ

Hence, the correct alternative is option (b).

Page No 2.76:

Question 51:

Mark the correct alternative in the following question:

Let f : R R be defined as f(x) = 2x, if x>3     x2, if 1<x33x, if x1      

Then, find f(-1) + f(2) + f(4)

(a) 9                              (b) 14                              (c) 5                              (d) none of these

Answer:

We have,fx=2x, if x>3     x2, if 1<x33x, if x1      Now,f-1+f2+f4=3-1+22+24=-3+4+8=9

Hence, the correct alternative is option (a).

Page No 2.76:

Question 52:

Mark the correct alternative in the following question:

Let A = {1, 2, ... , n} and B = {a, b}. Then the number of subjections from A into B is

(a) nP2                               (b) 2n- 2                               (c) 2n- 1                               (d) nC2

Answer:

As, the number of surjections from A to B is equal to the number of functions from A to B minus the number of functions from A to B whose images are proper subsets of B.

And, the number of functions from a set with n number of elements into a set with m number of elements = mn

So, the number of subjections from A into B where A = {1, 2, ... , n} and B = {ab} is 2n - 2.           (As, two functions can be many-one into functions)

Hence, the correct alternative is option (b).

Page No 2.76:

Question 53:

Mark the correct alternative in the following question:

If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is

(a) 720                                                 (b) 120                                                 (c) 0                                                 (d) none of these

Answer:

As, the number of bijection from A into B can only be possible when provided nAnBBut here nA<nBSo, the number of bijection i.e. one-one and onto mappings from A to B=0

Hence, the correct alternative is option (c).

Page No 2.76:

Question 54:

Mark the correct alternative in the following question:

If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is

(a) 10C7                                            (b) 10C7× 7!                                              (c) 710                                              (d)107

Answer:

As, the number of one-one functions from A to B with m and n elements, respectively = nPm = nCm× m!

So, the number of one-one functions from A to B with 7 and 10 elements, respectively = 10P7 = 10C7 × 7!

Hence, the correct alternative is option (b).

Page No 2.76:

Question 55:

Mark the correct alternative in the following question:

Let f : R - 35 R be defined by f(x) = 3x+25x-3. Then,

(a) f -l(x) = f(x)                         (b) f -1(x) = -f(x)                         (c) fof(x) = -x                         (d) -1(x) = 119f(x)

Answer:

We have,

f : R - 35  R is defined by f(x) = 3x+25x-3

fofx=ffx=f3x+25x-3=33x+25x-3+253x+25x-3-3=9x+65x-3+215x+105x-3-3=9x+6+10x-65x-315x+10-15x+95x-3=19x19=xLet y=3x+25x-35xy-3y=3x+25xy-3x=3y+2x5y-3=3y+2x=3y+25y-3f-1y=3y+25y-3So, f-1x=3x+25x-3=fx

Hence, the correct alternative is option (a).



Page No 2.77:

Question 56:

Let f : RR be defined by fx=1x. Then,  f is
(a) one-one
(b) onto
(e) bijective
(d) not defined

Answer:

Given: The function f : R → R be defined by fx=1x.

To check f is one-one:

Let fx1=fx21x1=1x2x1=x2Hence, f is one-one.To check f is onto: Since, y=1xx=1yyR-0RThere is no pre-image of y=0.Hence, f is not onto.

Hence, the correct option is (a).

​

Page No 2.77:

Question 57:

Let f : RR be defined by f(x) = 3x2 – 5 and g : RR by gx=xx2+1. Then (gof) (x) is
(a) 3x259x430x2+26

(b) 3x259x46x2+26

(c) 3x2x4+2x24

(d) 3x29x4+30x22

Answer:

Given: f(x) = 3x2 – 5 and gx=xx2+1

gofx=gfx            =g3x2-5            =3x2-53x2-52+1            =3x2-53x22+52-23x25+1            =3x2-59x4+25-30x2+1            =3x2-59x4-30x2+26

​Hence, the correct option is (a).

​

Page No 2.77:

Question 58:

Which of the following functions from Z to Z are bijections?
(a) f(x) = x3
(b) f(x) = x + 2
(c) f(x) = 2x + 1
(d) f(x) = x2 + 1

Answer:

Given: f : Z → Z

(a) f(x) = x3

It is one-one but not onto.

Thus, it is not bijective.

(b) f(x) = x + 2

It is one-one and onto.

Thus, it is bijective.

(c) f(x) = 2x + 1

It is one-one but not onto.

Thus, it is not bijective.

(d) f(x) = x2 + 1

It is neither one-one nor onto.

Thus, it is not bijective.

Hence, the correct option is (b).

Page No 2.77:

Question 59:

Let f : AB and g : BC be the bijective functions. Then, (gof)–1 =
(a) f–1o g–1
(b) fog
(c) g–1of–1
(d) gof

Answer:

Given: f : A → B and g : B → C be the bijective functions

Since, f:ABThus, f-1:BA       ...1Since, g:BCThus, g-1:CB       ...2From 1 and 2, we getf-1og-1:CA          ...3Also, gof:ACgof-1:CA       ...4Therefore, gof-1=f-1og-1

​Hence, the correct option is (a).

​

Page No 2.77:

Question 60:

Let f : NR be the function defined by fx=2x12 and g : QR be another function defined by g(x) = x + 2. Then, (gof) (3/2) is
(a) 1

(b) 2

 (c) 72

(d) none of these

Answer:

Given: fx=2x12 and g(x) = x + 2

gofx=gfx            =g2x-12            =2x-12+2            =2x-1+42            =2x+32gof32=232+32              =3+32              =62              =3

​Hence, the correct option is (d).

​

Page No 2.77:

Question 1:

The total number of functions from the set A = (1, 2, 3, 4 to the set B = a, b, c) is _________.

Answer:

Given: f:AB where A=1, 2, 3, 4 and B=a, b, c

Number of elements in A = 4
Number of elements in = 3
Each element of A have 3 options to form an image.

Thus, Number of functions that can be formed = 3 × 3 × 3 × 3 = 81

​Hence, the total number of functions from the set A = {1, 2, 3, 4} to the set B = {abc} is 81.

​

Page No 2.77:

Question 2:

The total number of one-one functions from the set A = {a, b, c} to the set B = {x, y, z, t} is _________.

Answer:

Given: f:AB where A=a, b, c and B=x, y, z, t

Number of elements in = 3
Number of elements in = 4

To form a one-one function,

Element a ∈ A have 4 options to form an image.
Element b ∈ A have 3 options to form an image.
Element c ∈ A have 2 options to form an image.

Thus, Number of one-one functions that can be formed = 4 × 3 × 2 = 24

​Hence, the total number of one-one functions from the set A = {a, bc} to the set B = {xyzt} is 24.

​

Page No 2.77:

Question 3:

The total number of onto functions from the set A = (1, 2, 3, 4, 5) to the set B = {x, y} is _________.

Answer:

Given: f:AB where A=1, 2, 3, 4, 5 and B=x, y

Number of elements in = 5
Number of elements in = 2


Each Element of A have 2 options to form an image.

Thus, Total number of functions that can be formed = 2 × 2 × 2 × 2 × 2 = 32

Number of functions having only one image i.e., {x} = 1
Number of functions having only one image i.e., {y} = 1

Thus, Number of onto functions that can be formed = 32 − 1 − 1 = 30

​Hence, the total number of onto functions from the set A = {1, 2, 3, 4, 5} to the set B = {xy} is 30.

Page No 2.77:

Question 4:

The domain of the real function fx=16x2 is _________.

Answer:

Given: fx=16x2

To find the domain, we find the real values of x for which the function is defined.

16-x2016x2x216x4 and x-4-4x4x-4, 4

Hence, the domain of the real function fx=16x2 is [−4, 4].

Page No 2.77:

Question 5:

The domain of the real function fx=x9x2 is ___________.

Answer:

Given: fx=x9x2 

To find the domain, we find the real values of x for which the function is defined.

xR and 9-x2>0xR and 9>x2xR and x2<9xR and -3<x<3-3<x<3x-3, 3

Hence, the domain of the real function fx=x9x2 is  (−3, 3).

Page No 2.77:

Question 6:

The range of the function f : RR given by fx=x+x2 is  _________.

Answer:

Given: fx=x+x2 

fx=x+x2      =x+x      =x+x         ,x0x-x         ,x<0      =2x         ,x00           ,x<0

To find the range, we find the real values of y obtained.

y=2x when x0x=y20y0y[0, )         ...1y=0 when x<0    ...2Thus, from 1 and 2,y[0, )

Hence, the range of the function f : R → R given by fx=x+x2 is [0, ).

Page No 2.77:

Question 7:

The range of the function f : R –{–2) → R given by fx=x+2x+2 is _________.

Answer:

Given: fx=x+2x+2

fx=x+2x+2      =x+2x+2                , x+20x+2-x+2          , x+2<0      =1                , x+20-1             , x+2<0

To find the range, we find the real values of y obtained.

Hence, the range of the function f : R –{–2) → R given by fx=x+2x+2 is {–1, 1}.

Page No 2.77:

Question 8:

If f : CC is defined by f(x) = 8x3, then f–1(8) = . _________.

Answer:

Given: f(x) = 8x3

fx=8x3y=8x3x3=y8x=y813Thus, f-1x=x813f-18=8813         =113         =1, ω, ω2        f:CCwhere, ω is the cube root of unity.

Hence, if f : C → C is defined by f(x) = 8x3, then f1(8) = 1, ω, ω2.

Page No 2.77:

Question 9:

If f : RR is defined by f(x) = 8x3 then, f–1(8) = _________.

Answer:

Given: f(x) = 8x3

fx=8x3y=8x3x3=y8x=y813Thus, f-1x=x813f-18=8813         =113         =1                  f:RR

Hence, if f : R → R is defined by f(x) = 8x3 then f1(8) = 1.



Page No 2.78:

Question 10:

If f : R – {0} → R – {0} is defined as fx=23x, then f–1(x) = ___________.

Answer:

Given: A function f : R – {0} → R – {0} is defined as fx=23x

fx=23xy=23x3x=2yx=23yThus, f-1x=23x

Hence, if f : R – {0} → R – {0} is defined as fx=23x, then f1(x) = 23x.

Page No 2.78:

Question 11:

If f : RR is defined by f(x) = 6 – (x – 9)3, then f–1(x) = ___________.

Answer:

Given: A function f : R → R is defined by f(x) = 6 – (x – 9)3

fx=6-x-93y=6-x-93x-93=6-yx-9=6-y13x=9+6-y13Thus, f-1x=9+6-x13

Hence, if f : R → R is defined by f(x) = 6 – (x – 9)3, then f1(x) = 9+6-x13.

Page No 2.78:

Question 12:

Let A = {1, 2, 3, 4} and f : AA be given by f  = {(1, 4), (2, 3), (3, 2), (4, 1)}. Then f–1 = ___________.

Answer:

Given: A function f : A → be given by f  = {(1, 4), (2, 3), (3, 2), (4, 1)}

f=1, 4, 2, 3, 3, 2, 4, 1f-1=4, 1, 3, 2, 2, 3, 1, 4Thus, f-1=4, 1, 3, 2, 2, 3, 1, 4

Hence, f-1=4, 1, 3, 2, 2, 3, 1, 4.

Page No 2.78:

Question 13:

If f : RR be defined by f(x) = (2 – x5)1/5, then fof(x) = ___________.

Answer:

Given: f(x) = (2 – x5)1/5

fofx=ffx         =f2-x515         =2-2-x515515         =2-2-x515×515         =2-2+x515         =x515         =x5×15         =x


Hence, if f : R → R be defined by f(x) = (2 – x5)1/5, then fof(x) = x.

Page No 2.78:

Question 14:

Let A = {1, 2, 3, 4, 5, ..., 10} and f : AA be an invertible function. Then, r=110f1of r=___________.

Answer:

Given: A → is an invertible function, where A = {1, 2, 3, 4, 5, ..., 10}

Since, f is invertibleTherefore, f-1ofx=x       ...1Now,r=110f-1ofr=f-1of1+f-1of2+f-1of3+....+f-1of10                   =1+2+3+....+10             From 1                   =1010+12                        1+2+3+...+n=nn+12                   =511                   =55


Hence, r=110f1of r= 55.

Page No 2.78:

Question 15:

Let A = {1, 2, 3, 4, 5, 6) and B = (2, 4, 6, 8, 10, 12). If f : AB is given by f(x) = 2x, then f–1 as set of ordered pairs, is ___________.

Answer:

Given: A function f : A → B defined as f(x) = 2x, where A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10, 12}

Since, fx=2xTherefore,f=1, 2, 2, 4, 3, 6, 4, 8, 5, 10, 6, 12Hence,f-1=2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6


Hence, if f : A → B is given by f(x) = 2x, then f–1 as set of ordered pairs, is 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6.

Page No 2.78:

Question 16:

Let = {(0, –1), (–1, 3), (2, 3), (3, 5)} be a function from Z to Z defined by f(x) = ax + b. Then, (a, b) = ___________.

Answer:

Given:  = {(0, –1), (–1, –3), (2, 3), (3, 5)} is a function from Z to Z defined by f(x) = axb


 = {(0, –1), (–1, –3), (2, 3), (3, 5)} defined by f(x) = ax + b

f0=-1a0+b=-10+b=-1b=-1          ...1f2=3a2+b=32a+b=32a-1=3             From 12a=3+12a=4a=2               ...2Thus,a=2 and b=-1


Hence, (ab) = â€‹(2, â€‹–1).


Disclaimer: The function f must be equal to  =  {(0, –1), (–1, –3), (2, 3), (3, 5)}.

Page No 2.78:

Question 17:

Let f : RR and g : RR be functions defined by f(x) = 5 – x2 and g(x) = 3x – 4. Then the value of fog (–1) is ___________.

Answer:

Given: f(x) = 5 – x2 and g(x) = 3x – 4


fog-1=fg-1            =f3-1-4            =f-3-4            =f-7            =5--72            =5-49            =-44


Hence, the value of fog (–1) is â€‹–44.

 

Page No 2.78:

Question 18:

Let f be the greatest integer function defined as f(x) = [x] and g be the modules function defined as g(x) = |x|, then the value of gof54 is ___________.

Answer:

Given: f(x) = [x] and g(x) = |x|


gof-54=gf-54              =g-54              =g-2              =-2              =2


Hence, the value of gof54 is â€‹2.
 

Page No 2.78:

Question 19:

If f(x) = cos [e] x + cos [–e] x, then f(π) = ___________.

Answer:

Given: f(x) = cos[e]x + cos[–e]x


fx=cosex+cos-ex              =cos2x+cos-3x        e=2.718 approx      =cos2x+cos3xfπ=cos2π+cos3π           =1-1           =0


Hence, f(π) = â€‹0.

Page No 2.78:

Question 20:

Let A = {1, 2, 3} and B = {a, b} be two sets. Then the number of constant functions from A to B is ___________.

Answer:

Given: Sets A = {1, 2, 3} and B = {ab}

Two constant functions can be formed from A to B.
i.e., one is f(x) = a and other is f(x) = b

Hence, the number of constant functions from A to B is 2.

Page No 2.78:

Question 21:

If f(x) = cos [π2] x + cos [–π2] x, then fπ2=______________.

Answer:

Given: f(x) = cos [π2x + cos [–π2x


fx=cosπ2x+cos-π2x              =cos9x+cos-10x        π2=9.85 approx      =cos9x+cos10xfπ2=cos9π2+cos10π2             =cos9π2+cos5π             =0-1             =-1


Hence, fπ2= â€‹–1.

Page No 2.78:

Question 22:

The number of onto functions from A = {a, b, c} to B = {1, 2, 3, 4} is __________.

Answer:

Given: A function from A = {a, bc} to B = {1, 2, 3, 4}

If a function from A to B is onto, then number of elements of A ≥ number of elements of B.

But here, number of elements of A < number of elements of B

Thus, no onto function exist from A = {a, bc} to B = {1, 2, 3, 4}.

Hence, the number of onto functions from A = {a, bc} to B = {1, 2, 3, 4} is 0.

Page No 2.78:

Question 23:

If f(0, ∞) → R is given by f(x) = log10x, then f–1(x) = ___________.

Answer:

Given: f(x) = log10x

fx=log10xy=log10xx=10yThus, f-1y=10y.

Hence,  f1(x) = 10x.

Page No 2.78:

Question 24:

If f : R+R is defined as f(x) = logx, then f–1(x) = ______________.

Answer:

Given: f(x) = log3x

fx=log3xy=log3xx=3yThus, f-1y=3y.

Hence, f1(x) = 3x.

Page No 2.78:

Question 25:

If f : RR, g : RR are defined by f(x) = 5x – 3, g(x) = x2 + 3, then (gof–1) (3) = _______________.

Answer:

Given: f(x) = 5x – 3 and g(x) = x2 + 3

fx=5x-3y=5x-35x=y+3x=y+35Thus, f-1y=y+35.Now,gof-13=gf-13             =g3+35             =g65             =652+3             =3625+3             =36+7525             =11125

Hence, gof1(3) = 11125.

Page No 2.78:

Question 26:

If f : RR is given by f(x) = 2x + |x|, then f(2x) + f(–x) + 4x = _______________.

Answer:

Given: f(x) = 2x + |x|

fx=2x+xf2x=22x+2xf2x=4x+2x         ...1f-x=2-x+-xf-x=-2x+x        ...2Now,f2x+f-x+4x=4x+2x-2x+x+4x                           =6x+3x                           =32x+x                           =3fx

Hence, f(2x) + f(–x) + 4x = 3f(x).

Page No 2.78:

Question 27:

If fx=1x1+x, then fof(cos 2θ) = ______________.

Answer:

Given: fx=1x1+x

fofcos2θ=ffcos2θ                =f1-cos2θ1+cos2θ                =f2sin2θ2cos2θ                =ftan2θ                =1-tan2θ1+tan2θ                =cos2θ

Hence, fof(cos 2θ) = cos2θ.

Page No 2.78:

Question 28:

Let fx=xx1andfαfα+1=fαk, then k = ______________.

Answer:

Given: fx=xx1andfαfα+1=fαk

fafa+1=aa-1a+1a+1-1             =aa-1a+1a             =a2a-1a+1             =a2a2-1             ...1                It is given that,fafa+1=faka2a2-1=akak-1k=2

Hence, k = 2.

Page No 2.78:

Question 29:

If f (f(x)) = x + 1 for all xR and if f0=12, then f(1) = ____________.

Answer:

Given: f (f(x)) = x + 1 for all x ∈ R and f0=12

ffx=x+1ff0=0+1f12=1           f0=12       ....1Now,ff12=12+1f1=1+22      f12=1f1=32

Hence, f(1) = 32.

Page No 2.78:

Question 30:

If f(x) = 3x + 10 and g(x) = x2 –1, then (fog)–1 is equal to ___________.

Answer:

Given: f(x) = 3x + 10 and g(x) = x2 –1

fogx=fgx         =fx2-1         =3x2-1+10         =3x2-3+10         =3x2+7Thus, fogx=3x2+7y=3x2+73x2=y-7x2=y-73x=±y-73fog-1x=±x-73

Hence, (fog)–1 is equal to ±x-73.

Page No 2.78:

Question 31:

Let f = {(1, 2), (3, 5), (4, 1)) and g = {(2, 3), (5, 1), (1, 3)}. Then, gof = __________ and fog = __________.

Answer:

Given: f = {(1, 2), (3, 5), (4, 1)) and g = {(2, 3), (5, 1), (1, 3)}

fogx=fgxfog1=fg1         =f3         =5fog2=fg2         =f3         =5fog5=fg5         =f1         =2Hence, fog=1, 5, 2, 5, 5, 2gofx=gfxgof1=gf1         =g2         =3gof3=gf3         =g5         =1gof4=gf4         =g1         =3Hence, gof=1, 3, 3, 1, 4, 3.

Hence, gof = 1, 3, 3, 1, 4, 3 and fog = 1, 5, 2, 5, 5, 2.



Page No 2.79:

Question 1:

Which one of the following graphs represents a function?

       

Answer:

In graph (b), 0 has more than one image, whereas every value of x in graph (a) has a unique image.
Thus, graph (a) represents a function.
So, the answer is (a).

Page No 2.79:

Question 2:

Which of the following graphs represents a one-one function?

Answer:

In the graph of (b), different elements on the x-axis have different images on the y-axis.
But in (a), the graph cuts the x-axis at 3 points, which means that 3 points on the x-axis have the same image as 0 and hence, it is not one-one.

Page No 2.79:

Question 3:

If A = {1, 2, 3} and B = {a, b}, write the total number of functions from A to B.

Answer:

Formula:
If set A has m elements and set B has n elements, then the number of functions from A to B is nm.
Given:
A = {1, 2, 3} and B = {a, b}
nA = 3 and  nB = 2
Number of functions from A to B = 23 = 8

Page No 2.79:

Question 4:

If A = {a, b, c} and B = {−2, −1, 0, 1, 2}, write the total number of one-one functions from A to B.

Answer:

Let f:AB be a one-one function.
 Then, fa can take 5 values, fb can take 4 values and fc can take 3 values.

Then, the number of one-one functions = 5 × 4 × 3 = 60

Page No 2.79:

Question 5:

Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {a, b, c}.

Answer:

A has 4 elements and B has 3 elements.
Also, one-one function is only possible from A to B if nAnB.
But, here nA>nB.
So, the number of one-one functions from A to B is 0.

Page No 2.79:

Question 6:

If f : RR is defined by f(x) = x2, write f−1 (25).

Answer:

Let f-125=x            ... 1fx=25x2=25x2-25=0x-5x+5=0x=±5f-125=-5, 5       [from 1]

Page No 2.79:

Question 7:

If f : CC is defined by f(x) = x2, write f−1 (−4). Here, C denotes the set of all complex numbers.

Answer:

Let f-1-4=x                  ... 1fx=-4x2=-4x2+4=0x+2ix-2i=0          using the identity: a2+b2=a-iba+ibx=±2i                           as xCf-125=-2i, 2i         from 1



Page No 2.80:

Question 8:

If f : RR is given by f(x) = x3, write f−1 (1).

Answer:

Let f-11= x           ... 1fx= 1x3= 1x3-1= 0x-1x2+x+1= 0        using the identity:a3-b3=a-ba2+ab+b2x=1          ( as xR) f-11= 1               [from 1]

Page No 2.80:

Question 9:

Let C denote the set of all complex numbers. A function f : CC is defined by f(x) = x3. Write f−1 (1).

Answer:

Let f-11=x                   ... 1fx=1x3=1x3-1=0x-1x2+x+1=0        Using identity: a3-b3=a-ba2+ab+b2x-1x-ωx-ω2=0, where ω=1±i32x=1, ω or ω2                 as xCf-11=1, ω, ω2             [from 1]

Page No 2.80:

Question 10:

Let f  be a function from C (set of all complex numbers) to itself given by f(x) = x3. Write f−1 (−1).

Answer:

Let f-1-1=x                          ... 1fx=-1x3=-1x3+1=0x+1x2-x+1=0        using the identity: a3+b3=a+ba2-ab+b2x+1x+ωx+ω2=0, where ω= 1±i32    x=-1, -ω, -ω2        as xCf-1-1=-1, -ω, -ω2     [from 1]

Page No 2.80:

Question 11:

If f : RR be defined by f(x) = x4, write f−1 (1).

Answer:

Let f-11=x             ... 1fx=1x4=1x4-1=0x2-1x2+1=0               using identity: a2-b2=a-ba+bx-1x+1x2+1=0         using identity: a2-b2=a-ba+bx=±1               as xRf-11=-1, 1       [ from 1]

Page No 2.80:

Question 12:

If f : CC is defined by f(x) = x4, write f−1 (1).

Answer:

Let f-11=x   ... 1fx=1x4=1x4-1=0x2-1x2+1=0                                                   using identity: a2-b2=a-ba+bx-1x+1x-ix+i=0, where i=-1           using identity: a2-b2=a-ba+bx=±1, ±i f-11=-1, 1, i,-i      [from 1]

Page No 2.80:

Question 13:

If f : RR is defined by f(x) = x2, find f−1 (−25).

Answer:

Let f-1-25=xfx=-25x2=-25We cannot find x∈R, such that x2=-25       as x2≥0 for all x∈RSo, f-1-25=ϕ

Page No 2.80:

Question 14:

If f : CC is defined by f(x) = (x − 2)3, write f−1 (−1).

Answer:

Let f-1-1=x                              ... 1fx=-1x-23=-1x-2=-1 or -ω or -ω2         as the roots of -113are -1, -ω and -ω2, where ω=1±i32x=-1+2 or 2-ω or 2-ω2=1,  2-ω, 2-ωf-1-1=1,  2-ω, 2-ω2            [from 1]

Page No 2.80:

Question 15:

If f : RR is defined by f(x) = 10 x − 7, then write f−1 (x).

Answer:

Let f-1x=y                     ... 1fy=x10y-7=x10y=x+7y=x+710f-1x=x+710             From 1

Page No 2.80:

Question 16:

Let f : -π2, π2R be a function defined by f(x) = cos [x]. Write range (f).

Answer:

Domain =-π2, π2=-1.57, 1.57   (as π=227)So, cos x=cos -2=cos 2   x-1.57, 0Also, cos 0=1 for x=0And cos x=cos 1  x0, 1.57Range=1, cos 1, cos 2

Page No 2.80:

Question 17:

If f : RR defined by f(x) = 3x − 4 is invertible, then write f−1 (x).

Answer:

Let f-1x=y                 ... 1fy=x3y-4=x3y=x+4y=x+43f-1x=x+43       [from 1]

Page No 2.80:

Question 18:

If f : RR, g : R → are given by f(x) = (x + 1)2 and g(x) = x2 + 1, then write the value of fog (−3).

Answer:

fog-3=f g -3=f-32+1=f10=10+12=121

Page No 2.80:

Question 19:

Let A = {xR : −4 ≤ x ≤ 4 and x ≠ 0} and f : AR be defined by fx=xx. Write the range of f.

Answer:

fx=xx=±xx=±1 xA, range of f=-1, 1.

Page No 2.80:

Question 20:

Let f : -π2, π2A be defined by f(x) = sin x. If f is a bijection, write set A.

Answer:

f is a bijection,
co-domain of f = range of f
As -1sin x1,
-1y1

So, A = [-1, 1]

Page No 2.80:

Question 21:

Let f : R → R+ be defined by f(x) = ax, a > 0 and a ≠ 1. Write f−1 (x).

Answer:

Let f-1x=y                    ... 1fy=xay=xy=loga xf-1x=log a x            [from 1]

Page No 2.80:

Question 22:

Let f : R − {−1} → R − {1} be given by fx=xx+1. Write f-1 x.

Answer:

Let f-1x=y                ... 1fy=xyy+1=xy=xy+xy-xy=xy1-x=xy=x1-xf-1x=x1-x            [from 1]

Page No 2.80:

Question 23:

Let f : R--35R be a function defined as fx=2x5x+3.

Write f-1 : Range of fR--35.

Answer:

Let f-1x=y              ... 1fy=x2y5y+3=x2y=5xy+3x2y-5xy=3xy2-5x=3xy=3x2-5xf-1x=3x2-5x   [from 1]

Page No 2.80:

Question 24:

Let f : RR, g : RR be two functions defined by f(x) = x2 + x + 1 and g(x) = 1 − x2. Write fog (−2).

Answer:

fog-2=f g -2=f1--22=f-3=-32+-3+1=9-3+1=7

Page No 2.80:

Question 25:

Let f : RR be defined as fx=2x-34. Write fof-1 1.

Answer:

Let f-1x=y                ...1fy=x2y-34=x2y-3=4x2y=4x+3y=4x+32f-1x=4x+32      [from 1]f-1x=4x+32fof-11=f41+32=f72=272-34=7-34=44=1

Page No 2.80:

Question 26:

Let f be an invertible real function. Write

f-1 of 1 +f-1 of 2+...+f-1of 100.

Answer:

Given that f  is an invertible real function.
f-1o f=I,where I is an identity function.So,f-1o f1+f-1o f2+...+f-1o f100=I1+I2+... +I100=1+2+...+100 As Ix=x, xR=100100+12[Sum of first n natural numbers=nn+12]=5050

Page No 2.80:

Question 27:

Let A = {1, 2, 3, 4} and B = {a, b} be two sets. Write the total number of onto functions from A to B.

Answer:

Formula:

When two sets A and B have m and n elements respectively, then the number of onto functions from A to B is
r=1n -1r nCr rm, if mno, if m<n

Here, number of elements in A = 4 = m
Number of elements in B = 2 = n
So, m > n
Number of onto functions
=r=12 -1r 2Cr r4=-11 2C1 14+-12 2C2 24=-2+16=14

Page No 2.80:

Question 28:

Write the domain of the real function fx=x-x.

Answer:

[x] is the greatest integral function.
So, 0≤x-x<1x-x exists for every xR.⇒Domain =R

Page No 2.80:

Question 29:

Write the domain of the real function fx=x-x.

Answer:

[x] is the greatest integer function.
xx, xRx-x0,xRx-x does not exist for any xR.Domain =ϕ

Page No 2.80:

Question 30:

Write the domain of the real function fx=1|x|-x.

Answer:

Case-1: When x>0x=x1x-x=1x-x=10=Case-2: When x<0x=-x1x-x=1-x-x=1-2x exists because when x<0, -2x>0fx is defined when x<0So, domain =-,0

Page No 2.80:

Question 31:

Write whether f : RR, given by fx=x+x2, is one-one, many-one, onto or into.

Answer:

fx=x+x2=x±x=0 or 2xSo, each element x in the domain may contain 2 images.For example,f0=0+02=0f-1=-1+-12=-1+1=-1+1=0Here, the image of 0 and -1 is 0.
Hence, f is may-one.

Page No 2.80:

Question 32:

If f(x) = x + 7 and g(x) = x − 7, xR, write fog (7).

Answer:

fog7=f  g7=f7-7=f 0=0+7=7



Page No 2.81:

Question 33:

What is the range of the function fx=x-1x-1?

Answer:

fx=x-1x-1=±x-1x-1=±1Range of f=-1, 1

Page No 2.81:

Question 34:

If f : RR be defined by f(x) = (3 − x3)1/3, then find fof (x).

Answer:

fof x=f f x=f 3-x313=3-3-x313313=3-3-x313=x313=x

Page No 2.81:

Question 35:

If f : RR is defined by f(x) = 3x + 2, find f (f (x)).

Answer:

ff x=f 3x+2=3 3x+2+2=9x+6+2=9x+8

Page No 2.81:

Question 36:

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. State whether f is one-one or not.

Answer:

f = {(1, 4), (2, 5), (3, 6)}

Here, different elements of the domain have different images in the co-domain.
So, f is one-one.

Page No 2.81:

Question 37:

If f : {5, 6} {2, 3} and g : {2, 3} {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, then find fog.    [NCERT EXEMPLAR]

Answer:

We have,
f : {5, 6}  {2, 3} and g : {2, 3}  {5, 6} are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}

As,
fog(2) = f(g(2)) = f(5) = 2,
fog(3) = f(g(3)) = f(6) = 3,

So,
fog : {2, 3}  {2, 3} is defined as
fog = {(2, 2), (3, 3)}

Page No 2.81:

Question 38:

Let f : R R be the function defined by f(x) = 4x - 3 for all x R. Then write -1.                                                 [NCERT EXEMPLAR]

Answer:

We have,
f : R  R is the function defined by f(x) = 4x - 3 for all x  R

Let fx=y. Then,y=4x-34x=y+3x=y+34So, f-1y=y+34or, f-1x=x+34

Page No 2.81:

Question 39:

Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}                                                                                                         [NCERT EXEMPLAR]

Answer:

As, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}

So, f is a function.

Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}

So, g is not a function.

Page No 2.81:

Question 40:

Write the domain of the real function f defined by f(x) = 25-x2.                                                                                 [NCERT EXEMPLAR]

Answer:

We have,fx=25-x2The function is defined only when 25-x20x2-250x+5x-50x-5, 5So, the domain of the given function is -5, 5.

Page No 2.81:

Question 41:

Let A = {a, b, c, d} and f : A A be given by f = {(a, b), (b, d), (c, a), (d, c)}. Write -1.                                           [NCERT EXEMPLAR]

Answer:

We have,

A = {abcd} and f : A  A be given by f = {(ab), (bd), (ca), (dc)}

Since, the elements of a function when interchanged gives inverse function.

So, -1 = {(ba), (db), (ac), (cd)}

Page No 2.81:

Question 42:

Let f, g : R R be defined by f(x) = 2x + l and g(x) = x2- 2 for all x R, respectively. Then, find gof.                 [NCERT EXEMPLAR]

Answer:

We have,

 fg : R  R are defined by f(x) = 2x + l and g(x) = x2 - 2 for all x  R, respectively

Now,gofx=gfx=g2x+1=2x+12-2=4x2+4x+1-2=4x2+4x-1

Page No 2.81:

Question 43:

If the mapping f : {1, 3, 4} {1, 2, 5} and g : {1, 2, 5} {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, then write fog.                                                                                                                                                                           [NCERT EXEMPLAR]

Answer:

We have,

 f : {1, 3, 4}  {1, 2, 5} and g : {1, 2, 5}  {1, 3}, are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, respectively


As,

fog2=fg2=f3=5,fog5=fg5=f1=2,fog1=fg1=f3=5,So,fog : 1,2,51,2,5 is given byfog=2,5,5,2,1,5

Page No 2.81:

Question 44:

If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = αx+β, then find the values of α and β.                [NCERT EXEMPLAR]

Answer:

We have,

A function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = αx+β

As, g1=1 and g2=3So, α1+β=1α+β=1         .....iand α2+β=32α+β=3       .....iiii-i, we get2α-α=2α=2Substituting α=2 in i, we get2+β=1β=-1

Page No 2.81:

Question 45:

If f(x) = 4 - (x - 7)3, then write f -1(x).                                                                                                                        [NCERT EXEMPLAR]

Answer:

We have,fx=4-x-73Let y=4-x-73x-73=4-yx-7=4-y3x=7+4-y3f-1y=7+4-y3 f-1x=7+4-x3



View NCERT Solutions for all chapters of Class 16