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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 : 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 14:

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

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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]



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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 

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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

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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

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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

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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

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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

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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   

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Question 17:

If f : R → (−1, 1) defined by fx=10x-10-x10x+10-x is invertible, find f−1.

Answer:

Injectivity of f:
Let x and y be two elements of domain (R), such that
fx=fy10x-10-x10x-10-x=10y-10-y10y-10-y10-x102x-110-x102x+1=10-y102y-110-y102y+1102x-1102x+1=102y-1102y+1102x-1102y+1=102x+1102y-1102x+2y+102x-102y-1=102x+2y-102x+102y-12×102x=2×102y102x=102y2x=2yx=y
So, f is one-one.

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

10x-10-x10x+10-x=y10-x102x-110-x102x+1=y102x-1=y×102x+y102x1-y=1+y102x=1+y1-y2x=log 1+y1-yx=12log 1+y1-yR       domain


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

Finding f  -1:
Let f-1x=y               ...1fy=x10y-10-y10y+10-y=x10-y102y-110-y102y+1=x102y-1=x×102y+x102y1-x=1+x102y=1+x1-x2y=log 1+x1-xy=12log 1+x1-xSo, f-1x=12log 1+x1-x      [from 1]

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Question 18:

If f : R → (0, 2) defined by fx=ex-e-xex+e-x+1 is invertible, find f−1.

Answer:

Injectivity of f :
Let x and y be two elements of domain (R), such that
fx=fyex-e-xex-e-x+1=ey-e-ye-e-y+1ex-e-xex-e-x=ey-e-ye-e-ye-xe2x-1e-xe2x+1=e-ye2y-1e-ye2y+1e2x-1e2x+1=e2y-1e2y+1e2x-1e2y+1=e2x+1e2y-1e2x+2y+e2x-e2y-1=e2x+2y-e2x+e2y-12×e2x=2×e2ye2x=e2y2x=2yx=y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain 0, 2, such that f(x) = y.

ex-e-xex+e-x+1=ye-xe2x-1e-xe2x+1+1=ye-xe2x-1e-xe2x+1=y-1e2x-1=y-1e2x+1e2x-1=y×e2x+y-e2x-1e2x=y×e2x+y-e2xe2x2-y=ye2x=y2-y2x=loge y2-yx=12loge y2-yR       domain


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

Finding f  -1:
Let f-1x=y           ...1fy=xey-e-yey+e-y+1=xe-ye2y-1e-ye2y+1+1=xe-ye2y-1e-ye2y+1=x-1e2y-1=x-1e2y+1e2y-1=x×e2y+x-e2y-1e2y=x×e2y+x-e2ye2y2-x=xe2y=x2-x2y=loge x2-xy=12loge x2-xR      domainy=12loge x2-x= f-1x     [from 1]So, f-1x=12loge x2-x

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Question 19:

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

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Question 20:

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]

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Question 21:

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.

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Question 22:

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 23:

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 24:

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:

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 2.73:

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.

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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

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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

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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.

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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]

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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

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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]

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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]

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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]

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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]

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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]

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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

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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]

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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

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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

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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]

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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

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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.

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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]

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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]



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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]

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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]

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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

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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

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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

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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

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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

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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

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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

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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.

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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

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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

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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

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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

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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.

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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)}

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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

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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.

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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.

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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)}

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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

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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

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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

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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



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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.

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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).

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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 .

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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.



 

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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.

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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]

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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).

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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



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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).

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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).

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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).

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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.

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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 .



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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).

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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).

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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.

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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

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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).

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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.

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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).

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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 .
 

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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).

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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).

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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).



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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).

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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).

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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).

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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).

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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).

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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).

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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

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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 2.78:

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 2.78:

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).

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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 2.79:

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 2.79:

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 2.79:

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 2.79:

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 2.79:

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 2.79:

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).

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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 2.79:

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 2.79:

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 2.79:

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 2.79:

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 2.79:

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).



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