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#### 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

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 $\cup$ {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).

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

Thus, f is not onto.

#### 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.

#### 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).

So, f is one-one.

Surjectivity:

#### 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.

#### 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 $f\left(x\right)=\frac{2x+3}{x-3}$
(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) = $\frac{x}{{x}^{2}+1}$                                                                                                                    [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)

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

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)

${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=±y$

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

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)

${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

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)

${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

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)

$\left|x\right|=\left|y\right|\phantom{\rule{0ex}{0ex}}x=±y$

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

$\left|x\right|=y\phantom{\rule{0ex}{0ex}}x=±y\in \mathbf{Z}$

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)

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

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)

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)

${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1}\in \mathbf{R}$

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)

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

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 $f\left(x\right)=\frac{2x+3}{x-3}$

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)

$\frac{2x+3}{x-3}=\frac{2y+3}{y-3}\phantom{\rule{0ex}{0ex}}\left(2x+3\right)\left(y-3\right)=\left(2y+3\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}2xy-6x+3y-9=2xy-6y+3x-9\phantom{\rule{0ex}{0ex}}9x=9y\phantom{\rule{0ex}{0ex}}x=y$

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

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)

${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$

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

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)

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

$5{x}^{3}+4=y\phantom{\rule{0ex}{0ex}}5{x}^{3}=y-4\phantom{\rule{0ex}{0ex}}{x}^{3}=\frac{y-4}{5}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{\frac{y-4}{5}}\in \mathbf{R}$

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)

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

$3-4x=y\phantom{\rule{0ex}{0ex}}4x=3-y\phantom{\rule{0ex}{0ex}}x=\frac{3-y}{4}\in \mathbf{R}$

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)

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

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

(xvii)  f : R → R, defined by f(x) = $\frac{x}{{x}^{2}+1}$

Injection test:

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

f(x= f(y)

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

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

#### 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.

#### Question 7:

Show that the function f : R − {3} → R − {2} given by $f\left(x\right)=\frac{x-2}{x-3}$ is a bijection.

#### Answer:

f : R − {3} → R − {2} given by
$f\left(x\right)=\frac{x-2}{x-3}$
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
f(x) = f(y)
$⇒\frac{x-2}{x-3}=\frac{y-2}{y-3}\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\left(y-3\right)=\left(y-2\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}⇒xy-3x-2y+6=xy-3y-2x+6\phantom{\rule{0ex}{0ex}}⇒x=y$

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

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.

#### 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) = $\frac{x}{2}$                                (ii) g(x) = |x|                                (iii) h(x) = x2                                                               [NCERT EXEMPLAR]

#### Answer:

(i) f : A $\to$ A, given by f(x) = $\frac{x}{2}$

Injection test:

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

f(x) = f(y)

$\frac{x}{2}$$\frac{y}{2}$

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

$\frac{x}{2}$ = 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.

#### 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.

#### 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 )}

#### 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)
$⇒4{x}^{3}+7=4{y}^{3}+7\phantom{\rule{0ex}{0ex}}⇒4{x}^{3}=4{y}^{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}⇒x=y$
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
$⇒4{x}^{3}+7=y\phantom{\rule{0ex}{0ex}}⇒4{x}^{3}=y-7\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=\frac{y-7}{4}\phantom{\rule{0ex}{0ex}}⇒x=\sqrt[3]{\frac{y-7}{4}}\in R$
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.

#### 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 $R\stackrel{+}{0}$ (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)
$⇒{e}^{x}={e}^{y}\phantom{\rule{0ex}{0ex}}⇒x=y$

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.

#### Question 13:

Show that the logarithmic function is a bijection.

#### Answer:

f:R+R given by , 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)

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

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.

#### 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.

#### 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.

#### 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!

#### 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:

So, f1 is one-one.

Proving f2is one-one:

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.

#### 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.

#### Question 19:

Show that if f1 and f2 are one-one maps from R to R, then the product defined by 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 is not one-one:
Given:
f1×f2

#### Question 20:

Suppose f1 and f2 are non-zero one-one functions from R to R. Is $\frac{{f}_{1}}{{f}_{2}}$ necessarily one-one? Justify your answer. Here, $\frac{{f}_{1}}{{f}_{2}}:R\to R$ is given by for all $x\in R$.

#### 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
${f}_{1}\left(x\right)={f}_{2}\left(y\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=y\phantom{\rule{0ex}{0ex}}⇒x=\sqrt[3]{y}\in R$Let 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
Let f2(x)=f2(y)x=yx=y
So, f2 is one-one.

Proving $\frac{{f}_{1}}{{f}_{2}}$is not one-one:
Given that $\frac{{f}_{1}}{{f}_{2}}\left(x\right)=\frac{{f}_{1}\left(x\right)}{{f}_{2}\left(x\right)}=\frac{{x}^{3}}{x}={x}^{2}$
Let x and y be two elements in the domain R, such that

$\frac{{f}_{1}}{{f}_{2}}\left(x\right)=\frac{{f}_{1}}{{f}_{2}}\left(y\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=±y$
So, $\frac{{f}_{1}}{{f}_{2}}$ is not one-one.

#### 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.

#### Question 22:

Show that f : R $\to$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 $\in$Z

So, f is a many-one function.

Surjection test:

Range (f) = [0, 1) $\ne$R.

So, f is an into function.

Therefore, f is neither one-one nor onto.

#### Question 23:

Let f : N $\to$N be defined by

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

#### 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

(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

#### 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}

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

#### 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}

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

#### 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}

Finding gof:
Co-domain of f is same as the domain of g.
So, fog exists and gof : {a, b, c} {a, b, c}

#### Question 5:

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

#### Question 6:

Let R+ be the set of all non-negative real numbers. If f : R+R+ and g : R+R+ are defined as , 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.

Hence, fog = gof

#### 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.

So,  fog ≠ gof

#### 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.

#### 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.

Hence, the associative property has been verified.

#### 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.

#### 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=-1$\notin N$]

Let us consider g : Ngiven by

#### 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.

Let us take two elements x and y in the domain of gof , such that

So, gof is injective.

#### 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

Hence, gof is one-one.

#### 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.

#### Question 1:

Find fog and gof  if
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)

#### Question 2:

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

#### Question 3:

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

#### Answer:

Domains of  f and fof are same as R.

#### 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.

→→  →

#### 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:

Clearly, fog$\ne$gof

#### 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).

#### Question 7:

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

#### Question 8:

If are two real functions, then describe functions fog and gof.

#### Question 9:

If be defined as
respectively, describe fog and gof.

#### Question 10:

If be two real functions, then find fog and gof.

#### Question 11:

Let f be a real function given by $f\left(x\right)=\sqrt{x-2}$.
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f2

Also, show that foff2 .

#### Question 12:

Let $f\left(x\right)=\left\{\begin{array}{ll}1+x,& 0\le x\le 2\overline{)}\\ 3-x,& 2. Find fof.

#### Question 13:

If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| –$x,\forall x\in \mathrm{R}$. Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).

#### Answer:

Given: f(x) = |x| + x
and g(x) = |x| – $x,\forall x\in \mathrm{R}$

Therefore,

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 $\ge$ 0)

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

#### 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)}

#### 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.

#### 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.

#### 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.

#### 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

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

Finding f  -1:

#### 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.

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

Finding f  -1:

#### 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−1$\left(x\right)=\sqrt{x-4}$, 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)

So, f is one-one.

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

$⇒{x}^{2}+4=y\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=y-4\phantom{\rule{0ex}{0ex}}⇒x=\sqrt{y-4}\in R$

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

Finding f  -1:

#### Question 8:

If , show that fof(x) = x for all $x\ne \frac{2}{3}$. What is the inverse of f?

#### Question 9:

Consider f : R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with ${f}^{-1}\left(x\right)=\frac{\sqrt{x+6}-1}{3}$.

#### Answer:

Injectivity of f :
Let x and y be two elements of domain (R+), such that
f(x)=f(y)

So, f is one-one.

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

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

Finding f  -1:

#### 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),

So, f is one-one.

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

$⇒{x}^{3}-3=y\phantom{\rule{0ex}{0ex}}⇒{x}^{3}=y+3\phantom{\rule{0ex}{0ex}}⇒x=\sqrt[3]{y+3}\in R$

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

Finding f  -1:

#### 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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}+4={y}^{3}+4\phantom{\rule{0ex}{0ex}}⇒{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}⇒x=y$
So, f is one-one.

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

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

Finding f  -1:

#### 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.

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

Finding f  -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.

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

Finding g -1:

Verification of (gof)−1 = f−1og −1:

#### Question 13:

Let A = R $-$ {3} and B = R $-$ {1}. Consider the function f : A $\to$B defined by f(x) = $\frac{x-2}{x-3}$. 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 $\to$ B defined by f(x) = $\frac{x-2}{x-3}$

#### Question 14:

Consider the function f : R+$\to \left[-9,\infty \right]$ given by f(x) = 5x2 + 6x $-$ 9. Prove that f is invertible with f$-$1(y) = $\frac{\sqrt{54+5y}-3}{5}$.      [CBSE 2015]

#### Question 15:

Let f : N$\to$be a function defined as $\left(x\right)=$9${x}^{2}+$6$x-$5. Show that : N$\to$S, where  is the range of f, is invertible. find the inverse of and hence find$-$1(43) and $-$1(163).

#### Answer:

We have,

f : N$\to$is a function defined as f (x) = 9x2 + 6$-$ 5.

Let y(x) = 9x2 + 6$-$ 5

$⇒y=9{x}^{2}+6x-5\phantom{\rule{0ex}{0ex}}⇒y=9{x}^{2}+6x+1-1-5\phantom{\rule{0ex}{0ex}}⇒y=\left(9{x}^{2}+6x+1\right)-6\phantom{\rule{0ex}{0ex}}⇒y={\left(3x+1\right)}^{2}-6\phantom{\rule{0ex}{0ex}}⇒y+6={\left(3x+1\right)}^{2}$

Now,

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

Thus, f is invertible.

So, ${f}^{-1}\left(x\right)=g\left(x\right)=\frac{\sqrt{x+6}-1}{3}$.

Now,

$-$1(43) = $\frac{\sqrt{43+6}-1}{3}=\frac{\sqrt{49}-1}{3}=\frac{7-1}{3}=\frac{6}{3}=2$

And $-$1(163) = $\frac{\sqrt{163+6}-1}{3}=\frac{\sqrt{169}-1}{3}=\frac{13-1}{3}=\frac{12}{3}=4$

#### Question 16:

Let f  R be a function defined as f$\left(x\right)$ $=\frac{4x}{3x+4}$ . Show that
f : R  Rang (f) is one-one and onto. Hence, find $-$1.

#### Answer:

The function .
Injectivity: Let  be such that
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{4x}{3x+4}=\frac{4y}{3y+4}\phantom{\rule{0ex}{0ex}}⇒4x\left(3y+4\right)=4y\left(3x+4\right)\phantom{\rule{0ex}{0ex}}⇒12xy+16x=12xy+16y\phantom{\rule{0ex}{0ex}}⇒16x=16y\phantom{\rule{0ex}{0ex}}⇒x=y$
Hence, f is one-one function.
Surjectivity: Let y be an arbitrary element of $\mathbf{R}-\left\{\frac{4}{3}\right\}$. Then,
f(x) = y
$⇒\frac{4x}{3x+4}=y\phantom{\rule{0ex}{0ex}}⇒4x=3xy+4y\phantom{\rule{0ex}{0ex}}⇒4x-3xy=4y\phantom{\rule{0ex}{0ex}}⇒x=\frac{4y}{4-3y}$
As .
Also, $\frac{4y}{4-3y}\ne -\frac{4}{3}$ because $\frac{4y}{4-3y}=-\frac{4}{3}⇒12y=-16+12y⇒0=-16$, which is not possible.
Thus,
$x=\frac{4y}{4-3y}\in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ such that
$f\left(x\right)=f\left(\frac{4x}{3x+4}\right)=\frac{4\left(\frac{4y}{4-3y}\right)}{3\left(\frac{4y}{4-3y}\right)+4}=\frac{16y}{12y+16-12y}=\frac{16y}{16}=y$, so every element in $\mathbf{R}-\left\{\frac{4}{3}\right\}$ has pre-image in $\mathbf{R}-\left\{-\frac{4}{3}\right\}$.
Hence, f is onto.
Now,
$x=\frac{4y}{4-3y}$
Replacing x by ${f}^{-1}\left(x\right)$ and y by x, we have
${f}^{-1}\left(x\right)=\frac{4x}{4-3x}$

#### Question 17:

If f : R → (−1, 1) defined by $f\left(x\right)=\frac{{10}^{x}-{10}^{-x}}{{10}^{x}+{10}^{-x}}$ is invertible, find f−1.

#### Answer:

Injectivity of f:
Let x and y be two elements of domain (R), such that
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{{10}^{x}-{10}^{-x}}{{10}^{x}-{10}^{-x}}=\frac{{10}^{y}-{10}^{-y}}{{10}^{y}-{10}^{-y}}\phantom{\rule{0ex}{0ex}}⇒\frac{{10}^{-x}\left({10}^{2x}-1\right)}{{10}^{-x}\left({10}^{2x}+1\right)}=\frac{{10}^{-y}\left({10}^{2y}-1\right)}{{10}^{-y}\left({10}^{2y}+1\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{\left({10}^{2x}-1\right)}{\left({10}^{2x}+1\right)}=\frac{\left({10}^{2y}-1\right)}{\left({10}^{2y}+1\right)}\phantom{\rule{0ex}{0ex}}⇒\left({10}^{2x}-1\right)\left({10}^{2y}+1\right)=\left({10}^{2x}+1\right)\left({10}^{2y}-1\right)\phantom{\rule{0ex}{0ex}}⇒{10}^{2x+2y}+{10}^{2x}-{10}^{2y}-1={10}^{2x+2y}-{10}^{2x}+{10}^{2y}-1\phantom{\rule{0ex}{0ex}}⇒2×{10}^{2x}=2×{10}^{2y}\phantom{\rule{0ex}{0ex}}⇒{10}^{2x}={10}^{2y}\phantom{\rule{0ex}{0ex}}⇒2x=2y\phantom{\rule{0ex}{0ex}}⇒x=y$
So, f is one-one.

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

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

Finding f  -1:

#### Question 18:

If f : R → (0, 2) defined by $f\left(x\right)=\frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}+1$ is invertible, find f−1.

#### Answer:

Injectivity of f :
Let x and y be two elements of domain (R), such that
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{{e}^{x}-{e}^{-x}}{{e}^{x}-{e}^{-x}}+1=\frac{{e}^{y}-{e}^{-y}}{e-{e}^{-y}}+1\phantom{\rule{0ex}{0ex}}⇒\frac{{e}^{x}-{e}^{-x}}{{e}^{x}-{e}^{-x}}=\frac{{e}^{y}-{e}^{-y}}{e-{e}^{-y}}\phantom{\rule{0ex}{0ex}}⇒\frac{{e}^{-x}\left({e}^{2x}-1\right)}{{e}^{-x}\left({e}^{2x}+1\right)}=\frac{{e}^{-y}\left({e}^{2y}-1\right)}{{e}^{-y}\left({e}^{2y}+1\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{\left({e}^{2x}-1\right)}{\left({e}^{2x}+1\right)}=\frac{\left({e}^{2y}-1\right)}{\left({e}^{2y}+1\right)}\phantom{\rule{0ex}{0ex}}⇒\left({e}^{2x}-1\right)\left({e}^{2y}+1\right)=\left({e}^{2x}+1\right)\left({e}^{2y}-1\right)\phantom{\rule{0ex}{0ex}}⇒{e}^{2x+2y}+{e}^{2x}-{e}^{2y}-1={e}^{2x+2y}-{e}^{2x}+{e}^{2y}-1\phantom{\rule{0ex}{0ex}}⇒2×{e}^{2x}=2×{e}^{2y}\phantom{\rule{0ex}{0ex}}⇒{e}^{2x}={e}^{2y}\phantom{\rule{0ex}{0ex}}⇒2x=2y\phantom{\rule{0ex}{0ex}}⇒x=y$
So, f is one-one.

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

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

Finding f  -1:

#### 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)}.

#### 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

$⇒$ -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

So, g is one-one.

Surjectivity of g:

$⇒$g is onto.
$⇒$g is a bijection.
So, g-1 exists.

Also,

#### 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

$⇒$f is not one-one.
$⇒$ f is not a bijection.
Thus, f  is not invertible.

#### 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:

Clearly, all these are bijections because they are one-one and onto.

#### 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.

#### 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

So,  fog is an injection.

(ii) Surjectivity of fog:
Let z be an element in the co-domain of fog (A).

So,  fog is a surjection.

#### 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).

#### 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.

#### 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 ${n}^{m}$.
Given:
A = {1, 2, 3} and B = {a, b}
$⇒n\left(A\right)$ = 3 and  $n\left(B\right)$ = 2
$\therefore$ Number of functions from A to B = 23 = 8

#### 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 $f:A\to B$ be a one-one function.

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

#### 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 $n\left(A\right)\le n\left(B\right)$.
But, here $n\left(A\right)>n\left(B\right)$.
So, the number of one-one functions from A to B is 0.

#### Question 6:

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

#### Question 7:

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

#### Question 8:

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

#### 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).

#### 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).

#### Question 11:

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

#### Question 12:

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

#### Question 13:

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

#### Question 14:

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

#### Question 15:

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

#### Question 16:

Let be a function defined by f(x) = cos [x]. Write range (f).

#### Question 17:

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

#### 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).

#### Question 19:

Let A = {xR : −4 ≤ x ≤ 4 and x ≠ 0} and f : AR be defined by $f\left(x\right)=\frac{\left|x\right|}{x}$. Write the range of f.

#### Question 20:

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

#### Answer:

$\because$f is a bijection,
co-domain of f = range of f
As ,
$-1\le y\le 1$

So, A = [-1, 1]

#### Question 21:

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

#### Question 22:

Let f : R − {−1} → R − {1} be given by .

#### Question 23:

Let be a function defined as $f\left(x\right)=\frac{2x}{5x+3}$.

Write .

#### 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).

#### Question 25:

Let f : RR be defined as

#### Question 26:

Let f be an invertible real function. Write

#### Answer:

Given that f  is an invertible real function.

#### 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

Here, number of elements in A = 4 = m
Number of elements in B = 2 = n
So, m > n
Number of onto functions

#### Question 28:

Write the domain of the real function $f\left(x\right)=\sqrt{x-\left[x\right]}.$

#### Answer:

[x] is the greatest integral function.

#### Question 29:

Write the domain of the real function $f\left(x\right)=\sqrt{\left[x\right]-x}.$

#### Answer:

[x] is the greatest integer function.

#### Question 30:

Write the domain of the real function $f\left(x\right)=\frac{1}{\sqrt{|x|-x}}.$

#### Question 31:

Write whether f : RR, given by $f\left(x\right)=x+\sqrt{{x}^{2}}$, is one-one, many-one, onto or into.

#### Answer:

Hence, f is may-one.

#### Question 32:

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

#### Question 33:

What is the range of the function $f\left(x\right)=\frac{\left|x-1\right|}{x-1}?$

#### Question 34:

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

#### Question 35:

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

#### 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.

#### Question 37:

If f : {5, 6} $\to$ {2, 3} and g : {2, 3} $\to$ {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} $\to$ {2, 3} and g : {2, 3} $\to$ {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} $\to$ {2, 3} is defined as
fog = {(2, 2), (3, 3)}

#### Question 38:

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

#### Answer:

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

#### 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.

#### Question 40:

Write the domain of the real function f defined by f(x) = $\sqrt{25-{x}^{2}}$.                                                                                 [NCERT EXEMPLAR]

#### Question 41:

Let A = {a, b, c, d} and f : A $\to$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 $\to$ 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)}

#### Question 42:

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

#### Answer:

We have,

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

$\mathrm{Now},\phantom{\rule{0ex}{0ex}}gof\left(x\right)=g\left(f\left(x\right)\right)\phantom{\rule{0ex}{0ex}}=g\left(2x+1\right)\phantom{\rule{0ex}{0ex}}={\left(2x+1\right)}^{2}-2\phantom{\rule{0ex}{0ex}}=4{x}^{2}+4x+1-2\phantom{\rule{0ex}{0ex}}=4{x}^{2}+4x-1$

#### Question 43:

If the mapping f : {1, 3, 4} $\to$ {1, 2, 5} and g : {1, 2, 5} $\to$ {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} $\to$ {1, 2, 5} and g : {1, 2, 5} $\to$ {1, 3}, are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, respectively

As,

#### Question 44:

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

#### Answer:

We have,

A function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = $\alpha x+\beta$

#### Question 45:

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

#### Question 1:

Let
let . 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

#### Question 2:

(a) injective
(b) surjective
(c) bijective
(d) None of these

#### Answer:

$⇒$f is not a function.
So, the answer is (d).

#### Question 3:

If is a bijection, then
(a)
(b)
(c)
(d) None of these

#### Answer:

(d) None of these

#### Question 4:

The function f : RR defined by
(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
and R is the co-domain of the given function.
Hence, Co-domain$\ne$Range
Therefore, the given function is into.

#### Question 5:

Let the function be defined by

(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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x+a}{x+b}=\frac{y+a}{y+b}\phantom{\rule{0ex}{0ex}}⇒\left(x+a\right)\left(y+b\right)=\left(x+b\right)\left(y+a\right)\phantom{\rule{0ex}{0ex}}⇒xy+bx+ay+ab=xy+ax+by+ab\phantom{\rule{0ex}{0ex}}⇒bx+ay=ax+by\phantom{\rule{0ex}{0ex}}⇒\left(a-b\right)x=\left(a-b\right)y\phantom{\rule{0ex}{0ex}}⇒x=y$
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

$f\left(x\right)=y\phantom{\rule{0ex}{0ex}}⇒\frac{x+a}{x+b}=y\phantom{\rule{0ex}{0ex}}⇒x+a=yx+yb\phantom{\rule{0ex}{0ex}}⇒x-yx=yb-a\phantom{\rule{0ex}{0ex}}⇒x\left(1-y\right)=yb-a\phantom{\rule{0ex}{0ex}}⇒x=\frac{yb-a}{1-y}\in R-\left\{-b\right\}\phantom{\rule{0ex}{0ex}}$
So,  f is onto.

#### Question 6:

The function defined by $f\left(x\right)=-{x}^{2}+6x-8$ is a bijection if
(a)
(b)
(c)
(d)

(a)

#### Question 7:

Let . Then, the mapping 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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x\left|x\right|=y\left|y\right|\phantom{\rule{0ex}{0ex}}⇒x\left(x\right)=y\left(y\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=y\phantom{\rule{0ex}{0ex}}$

Case-2: Let x and y be two negative numbers, such that
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x\left|x\right|=y\left|y\right|\phantom{\rule{0ex}{0ex}}⇒x\left(-x\right)=y\left(-y\right)\phantom{\rule{0ex}{0ex}}⇒-{x}^{2}=-{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=y\phantom{\rule{0ex}{0ex}}$

Case-3: Let x be positive and y be negative.

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)

$⇒$f is onto.
$⇒$f is a bijection.
So, the answer is (c).

#### Question 8:

Let be given by $f\left(x\right)=\left[{x}^{2}\right]+\left[x+1\right]-3$ 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

$f\left(x\right)=\left[{x}^{2}\right]+\left[x+1\right]-3$

It is many one function because in this case for two different values of x
we would get the same value of f(x) .

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 , Codomain$\ne$Range
Hence, the given function is into function.

Therefore, f(x) is many one and into

#### 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:

Injectivity:

So, f is not one-one.

Surjectivity:
Let y be an element of the co-domain, such that

$⇒$f is onto.
So, the answer is (d).

#### Question 10:

The function
(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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x}{x+1}=\frac{y}{y+1}\phantom{\rule{0ex}{0ex}}⇒xy+x=xy+y\phantom{\rule{0ex}{0ex}}⇒x=y$
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that

$⇒$f is not onto.
So, the answer is (b).

#### Question 11:

The range of the function $f\left(x\right){=}^{7-x}{P}_{x-3}$ 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

So, the answer is (d).

#### Question 12:

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

(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).

Surjectivity:

$⇒$f is onto.

#### 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.

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

#### Answer:

So, the answer is (b).

#### Question 14:

Which of the following functions form Z to itself are bijections?
(a) $f\left(x\right)={x}^{3}$
(b) $f\left(x\right)=x+2$
(c) $f\left(x\right)=2x+1$
(d) $f\left(x\right)={x}^{2}+x$

#### Answer:

(b) Injectivity:
Let x and y be two elements of the domain (Z), such that

So, f is one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that

$⇒$f is onto.
So, f is a bijection.

So, the answer is (b).

#### Question 15:

Which of the following functions from to itself are bijections?
(a) $f\left(x\right)=\frac{x}{2}$
(b)
(c) $h\left(x\right)=|x|$
(d) $k\left(x\right)={x}^{2}$

#### Answer:

So, the answer is (b).

#### Question 16:

Let , 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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x\left|x\right|=y\left|y\right|\phantom{\rule{0ex}{0ex}}⇒x\left(x\right)=y\left(y\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=y\phantom{\rule{0ex}{0ex}}$

Case-2: Let x and y be two negative numbers, such that
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒x\left|x\right|=y\left|y\right|\phantom{\rule{0ex}{0ex}}⇒x\left(-x\right)=y\left(-y\right)\phantom{\rule{0ex}{0ex}}⇒-{x}^{2}=-{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=y\phantom{\rule{0ex}{0ex}}$

Case-3: Let x be positive and y be negative.

So, f is one-one.

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

$⇒$f is onto
$⇒$f is a bijection.
So, the answer is (a).

#### Question 17:

If the function is a surjection, then A =
(a) R
(b) [0, 1]
(c) [0, 1)
(d) [0, 1)

#### Answer:

So, the answer is (d).

#### Question 18:

If a function 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

So, the answer is (b).

#### Question 19:

The function defined by
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:

$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$

Injectivity:

So, f is not one-one.

Surjectivity:
Let y be an element in the co domain R, such that

So, the answer is (b).

#### Question 20:

The function , defined by , is
(a) bijection
(b) injection but not a surjection
(c) surjection but not an injection
(d) neither an injection nor a surjection

#### Answer:

$f\left(x\right)={\mathrm{sin}}^{-1}\left(3x-4{x}^{3}\right)\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)=3{\mathrm{sin}}^{-1}x$

Injectivity:
Let x and y be two elements in the domain , such that
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒3{\mathrm{sin}}^{-1}x=3{\mathrm{sin}}^{-1}y\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}x={\mathrm{sin}}^{-1}y\phantom{\rule{0ex}{0ex}}⇒x=y$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain, such that

$⇒$f is onto.
$⇒$f is a bijection.
So, the answer is (a).

#### Question 21:

Let be a function defined by
(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

Therefore, this function is not injective.

Therefore, this function is not surjective .

#### Question 22:

Let be a function defined by

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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x-m}{x-n}=\frac{y-m}{y-n}\phantom{\rule{0ex}{0ex}}⇒\left(x-m\right)\left(y-n\right)=\left(x-n\right)\left(y-m\right)\phantom{\rule{0ex}{0ex}}⇒xy-nx-my+mn=xy-mx-ny+mn\phantom{\rule{0ex}{0ex}}⇒\left(m-n\right)x=\left(m-n\right)y\phantom{\rule{0ex}{0ex}}⇒x=y$
So, f is one-one.

Surjectivity:
Let y be an element in the co domain R, such that

$⇒$f is not onto.
Thus, the answer is (b).

#### Question 23:

Let be a function defined by $f\left(x\right)=\frac{{x}^{2}-8}{{x}^{2}+2}$. 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
$f\left(x\right)=f\left(y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}-8}{{x}^{2}+2}=\frac{{y}^{2}-8}{{y}^{2}+2}\phantom{\rule{0ex}{0ex}}⇒\left({x}^{2}-8\right)\left({y}^{2}+2\right)=\left({x}^{2}+2\right)\left({y}^{2}-8\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}{y}^{2}+2{x}^{2}-8{y}^{2}-16={x}^{2}{y}^{2}-8{x}^{2}+2{y}^{2}-16\phantom{\rule{0ex}{0ex}}⇒10{x}^{2}=10{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=±y$
So, f is not one-one.

Surjectivity:

$⇒$f is not onto.
The correct answer is (d).

#### Question 24:

is defined by
(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

#### Question 25:

The function 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,

${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}⇒x=±y$

So, f is not one-one.

Surjectivity:

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

#### Question 26:

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

is

(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).

Surjectivity:

$⇒$f is onto.
So, the answer is (d).

#### Question 27:

Which of the following functions from to itself are bijections?

(a) $f\left(x\right)=|x|$
(b)
(c)
(d) None of these

#### Answer:

(b)

It is clear that  f(x) is one-one.

⇒ f is onto.
So, f is a bijection.

#### Question 28:

Let be given by
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

So, f is not one-one.

Surjectivity:
Let y be an element in the co-domain (Z), such that

$⇒$f is onto.
So, the answer is (a).

#### Question 29:

The function defined by $f\left(x\right)={6}^{x}+{6}^{|x|}$ 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
and R is the codomain of the given function .
Hence, Codomain$\ne$Range
Therefore, the given function is into .

#### Question 30:

Let . Then, the solution set of the equation is
(a) R
(b) {0}
(c) {0, 2}
(d) none of these

#### Answer:

So, the answer is (c).

#### Question 31:

If
(a) is given by $\frac{1}{3x-5}$
(b) is given by $\frac{x+5}{3}$
(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.

So, the answer is (b).

#### Question 32:

If
(a)
(b)
(c)
(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):

So, the answer is (a).

#### Question 33:

The inverse of the function given by $f\left(x\right)=\frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}$ is
(a)

(b)

(c)

(d) none of these

#### Answer:

So, the answer is (a).

#### Question 34:

Let . The inverse of the function, given by

(a)

(b)

(c)

(d) not defined

#### Answer:

So, the answer is (b).

#### Question 35:

Let be defined as . Then, is
(a) $1+\sqrt{1-x}$
(b) $1-\sqrt{1-x}$
(c) $\sqrt{1-x}$
(d) $1±\sqrt{1-x}$

#### Answer:

The correct answer is (b).

#### Question 36:

Let
(a)
(b)
(c)
(d) none of these

#### Answer:

So, the answer is (c).

#### Question 37:

If the function be such that $f\left(x\right)=x-\left[x\right]$, where [x] denotes the greatest integer less than or equal to x, then is

(a) $\frac{1}{x-\left[x\right]}$

(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 $\ne$Co-domain of ff is not onto.
$⇒$f is not a bijective function.
So,  f -1 does not exist.
Thus, the answer is (c).

#### Question 38:

If is given by equals

(a) $\frac{x+\sqrt{{x}^{2}-4}}{2}$

(b) $\frac{x}{1+{x}^{2}}$

(c) $\frac{x-\sqrt{{x}^{2}-4}}{2}$

(d) $1+\sqrt{{x}^{2}-4}$

#### Answer:

So, the answer is (a).

#### Question 39:

Let , where [x] denotes the greatest integer less than or equal to x. Then for all is equal to
(a) x
(b) 1
(c) f(x)
(d) g(x)

#### Answer:

(b) 1

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

#### Question 40:

Let . Then, for what value of α is
(a) $\sqrt{2}$
(b) $-\sqrt{2}$
(c) 1
(d) −1

(d) −1

#### Question 41:

The distinct linear functions that map [−1, 1] onto [0, 2] are
(a)
(b)
(c)
(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):

So, option (a) is correct.

Option (b):

Here, f(-1) gives -2
So, (b) is not correct.

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

#### Question 42:

Let be defined by $f\left(x\right)=4x-{x}^{2}$. Then, f is invertible if X =
(a)
(b)
(c)
(d)

#### 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

So, the answer is (c).

#### Question 43:

If is defined by equals
(a) $\sqrt{\frac{\left|x\right|}{1-\left|x\right|}}$

(b)

(c) $-\sqrt{\frac{x}{1-x}}$

(d) None of these

(b)

#### Question 44:

Let [x] denote the greatest integer less than or equal to x. If , then
(a) $fogoh\left(x\right)=\frac{\pi }{2}$
(b) $fogoh\left(x\right)=\pi$
(c) $hofog=hogof$
(d) $hofog\ne hogof$

#### Answer:

(c) $hofog=hogof$

#### Question 45:

If , then f(x) is equal to
(a)
(b)
(c)
(d) 2 ${x}^{2}-3x-1$

#### Answer:

We will solve this problem by the trial-and-error method.
Let us check option (a) first.

The given condition is satisfied by (a).
So, the answer is (a).

#### Question 46:

If and the composite function , then g(x) is equal to
(a) $\sqrt{x-1}$
(b) $\sqrt{x}$
(c) $\sqrt{x+1}$
(d) $-\sqrt{x}$

#### Answer:

(b)

So, the answer is (b).

#### Question 47:

If is given by is equal to
(a) ${x}^{1/3}-3$
(b) ${x}^{1/3}+3$
(c) ${\left(x-3\right)}^{1/3}$
(d) $x+{3}^{1/3}$

#### Answer:

(c)

So, the answer is (c).

#### Question 48:

Let $f\left(x\right)={x}^{3}$ 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}

#### Question 49:

Let be given by $f\left(x\right)={x}^{2}-3$. Then, ${f}^{-1}$ is given by
(a) $\sqrt{x+3}$
(b) $\sqrt{x}+3$
(c) $x+\sqrt{3}$
(d) None of these

#### Answer:

(d)

So, the answer is (d).

#### Question 50:

Mark the correct alternative in the following question:

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

(a) $\frac{\mathrm{\pi }}{4}$                              (b) $\left\{n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}:n\in \mathbf{Z}\right\}$                              (c) does not exist                              (d) none of these

#### Answer:

Hence, the correct alternative is option (b).

#### Question 51:

Mark the correct alternative in the following question:

Let f : R $\to$R be defined as f(x) =

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

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

#### Answer:

Hence, the correct alternative is option (a).

#### 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).

#### 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:

Hence, the correct alternative is option (c).

#### 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).

#### Question 55:

Mark the correct alternative in the following question:

Let f : R $-$ $\left\{\frac{3}{5}\right\}$ $\to$R be defined by f(x) = $\frac{3x+2}{5x-3}$. Then,

(a) f $-$l(x) = f(x)                         (b) f $-$1(x) = $-$f(x)                         (c) fof(x) = $-$x                         (d) $-$1(x) = $\frac{1}{19}$f(x)

#### Answer:

We have,

f : R $-$ $\left\{\frac{3}{5}\right\}$ $\to$ R is defined by f(x) = $\frac{3x+2}{5x-3}$

Hence, the correct alternative is option (a).

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