# NCERT Solutions for Class 12 Maths Chapter 1 - Relations and Functions - Exercise 1.2

### Access Exercises of Class 12 Maths Chapter 1 - Relations and Functions

Exercise 1.1 Solutions : 16 Questions (14 Short Answers, 2 MCQ)

Exercise 1.2 Solutions : 12 Questions (10 Short Answers, 2 MCQ)

Exercise 1.3 Solutions : 14 Questions (12 Short Answers, 2 MCQ)

Exercise 1.4 Solutions : 13 Questions (12 Short Answers, 1 MCQ)

Miscellaneous Exercise Solutions: 19 Questions (7 Long answers, 9 Short Answer Type, 3 MCQs)

** Exercise 1.2 :**

**1. Show that the function f : R, → R, defined by f( x)=1/x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R, is replaced by N with co-domain being same as R, ? **

$$\textbf{Sol.}\space \text{It is given that f : R → R is defined by}\\f(x)=\frac{1}{x}.\\\text{For one-one,}\space f(x) = f(y)\\⇒\frac{1}{x}=\frac{1}{y}\\⇒ x = y\\\text{Therefore, f is one-one.}\\\textbf{For onto,}\\\text{It is clear that for y ∈ R, there exists}\space x=\frac{1}{y} ∈ R.\\\text{(Exists as} y ≠ 0) \text{such that}\\f(x)=\frac{1}{(\frac{1}{y})}=y\\\text{Therefore, f is onto.}\\\text{Thus, the given function (f) is one-one and onto.}\\\text{ Now, consider function g : N → R* defined by}$$

$$f(x)=\frac{1}{\frac{1}{y}}=y\\\text{Therefore, f is onto. Thus, the given function (f) is one-one and onto.}\\\text{Now, consider function g : N → R defined by}\\g(x)=\frac{1}{x}\\\text{We have}, g(x_1) = g(x_2)\\⇒\frac{1}{x_1}=\frac{1}{x_2}\\⇒ x_1 = x_2\\\text{Therefore, g is one-one.}\\\text{Further, it is clear that g is not onto as for}\\\text{1.2 ∈ R, there does not exist any x in N such that}\\g(x) =\frac{1}{1.2}\\\text{Hence, function g is one-one but not onto.}$$

**2. Check the injectivity and surjectivity of the following functions :****(i) f : N → N given by f(x) = x ^{2}**

**(ii) f : Z → Z given by f(x) = x**

^{2}**(iii) f : R → R given by f(x) = x**

^{2}**(iv) f : N → N given by f(x) = x**

^{3}**(v) f : Z → Z given by f(x) = x**

^{3}**Sol. (i)** f : N → N is given by f(x) = x^{2}

It is seen that for x, y ∈ N, f(x) = f(y)

⇒ x^{2} = y^{2}

⇒ x = y

(∴ x and y are positive numbers) Therefore, f is injective.

Now, 2 ∈ N but there does not exist any x in N such that f(x) = x^{2} = 2.

It means there is some element in co-domain in which do not have any images. Therefore, f is not surjective.

Hence, funciton f is injective but not surjective.

**(ii)** f : Z → Z is given by f(x) = x^{2}

It is seen that f(– 1) = f(1) = 1 but – 1 ≠ 1 Therefore f is not injective.

Now, – 2 ∈ Z. But there does not exist any elements x ∈ Z such that f(x) = x^{2} = – 2.

Therefore, f is not surjective.

Hence, function f is neither injective nor surjective.

**(iii)** f : R → R is given by f(x) = x^{2}.

It is seen that f(– 1) = f(1) = 1 but – 1 ≠ 1.

Therefore, f is not injective.

Now, – 2 ∈ R. But there does not exist any element x ∈ R such that f(x) = x^{2} = – 2.

Therefore, f is not surjective.

Hence, function f is neither injective nor surjective.

**(iv)** f : N → N given by, f(x) = x^{3}

It is seen that for x, y ∈ N, f(x) = f(y)

⇒ x^{3} = y^{3} ⇒ x = y

Therefore, f is injective. Now, 2 ∈ N. But, there does not existany element x in domain N such that f(x) = x^{3} = 2.

Therefore, f is not surjective.

Hence, function f is injective but not surjective.

**(v)** f : Z → Z is given by, f(x) = x^{3}

It is seen that for x, y ∈ Z, f(x) = f(y)

⇒ x^{3} = y^{3} ⇒ x = y

Therefore, f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x^{3} = 2.

Therefore, f is not surjective.

Hence, function f is injective but not surjective.

**3. Prove that the Greatest Integer Function ****f : R → R given by f(x) = [x], is neither one-one ****nor onto, where [x] denotes the greatest integer ****less than or equal to x.**

**Sol.** Here, f : R → R is given by, f(x) = [x]

It is seen that f(1.2) = [1.2] = 1, f(1.9} = [1.9] = 1

Therefore, f(1.2) = f(1.9), but 1.2 ≠ 1.9 Therefore, f is not one-one.

Now, consider 0.7 ∈ R. It is known that f(x) = [x] is always an integer.

Thus, there does not exist any element x ∈ R such that f(x) = 0.7.

Therefore, f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

**4. Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x, is negative.**

**Sol.** Here, f : R → R is given by

$$f(x)=|x|=\begin{cases}x &\text{if\space x≥0 }\\-x &\text{\space if\space x<0}\end{cases}$$

It is seen that f(– 1) = |– 1| = 1, f(1) = |1| = 1

Therefore, f(– 1) = f(1) but – 1 ≠ 1

Therefore, f is not one-one.

Now, consider – 1 ∈ R

It is known that f(x) = |x| is always non-negative.

Thus, there does not exist any element x in domain R such that f(x) = |x| = – 1.

Therefore, f is not onto.

Hence, the modulus functions is neither one-one nor onto.

**5. Show that the Signum function f : R → R, given ****by$$f(x)=\begin{cases}1 &\text{if\space x>0 }\\0, &\text{\space if\space x=0}\\-1 &\text{\space if\space x<0}\end{cases}$$is neither one-one nor onto.**

$$\textbf{Sol.}\space f: R→R,f (x)=\begin{cases}1 &\text{if\space x>0 }\\0, &\text{\space if\space x=0}\\-1 &\text{\space if\space x<0}\end{cases}$$

It is seen that f(1) = f(2) = 1 but 1 ≠ 2. Therefore, f is not one-one.

Now, as f(x) takes only three values (1, 0 or –1) for the element – 2 in co-domain R, there does not exist any x in domain R such that f(x) = – 2.

Therefore, f is not onto.

Hence, the Signum function is neither one-one nor onto.

**6. 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. Show ****that f is one-one.**

**Sol.** Given that A = {1, 2, 3} and B = {4, 5, 6, 7}

Now, f : A → B is defined as f = {(1, 4), (2, 5), (3, 6)}

Therefore, f(1) = 4, f(2) = 5, f(3) = 6

It is seen that the images of distinct elements of A under f are distinct.

Hence, function f is one-one.

**7. In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.****(i) f : R → R defined by f(x) = 3 – 4x****(ii) f : R → R defined by f(x) = 1 + x ^{2 }**

**Sol.** (i) Here, f : R → R is defined by f(x) = 3 – 4x

Let x_{1}, x_{2} ∈ R such that f(x_{1}) = f(x_{2})

⇒ 3 – 4x_{1} = 3 – 4x_{2}

⇒ – 4x_{1} = – 4x_{2}

⇒ x_{1} = x_{2}

Therefore, f is one-one.

For any real number y in R, there exists

$$\frac{3-y}{4}\space\text{in R such that}\\f(\frac{3-y}{4})=3-4(\frac{3-y}{4})=y$$

Therefore, f is onto. Hence, f is bijective.

(ii) Here f : R → R is defined as

f(x) = 1 + x^{2}

Let x_{1}, x_{2} ∈ R such that f(x_{1}) = f(x_{2})

⇒ 1 + x_{1}^{2}

= 1 + x_{2} ^{2}

⇒ x_{1}^{2}

= x_{2}^{2 }

⇒ x_{1} = + x_{2}

For instance, f(1) = f(– 1) = 2

Therefore, f(x_{1}) = f(x_{2}) does not imply that

x_{1} = x_{2}

Therefore, f is not one-one.

Consider an element – 2 in co-domain R.

It is seen that f(x) = 1 + x^{2} is positive for all

x ∈ R.

Thus, there does not exist any x in domain

R such that f(x) = – 2.

Therefore, f is not onto. Hence, f is neither

one-one nor onto.

**8. Let A and B be sets. Show that f : A × B → B × A ****such that f(a, b) = (b, a) is bijective function.**

**Sol.** Here, f : A × B → B × A is defined as f(a, b) = (b, a)

Let (a_{1}, b_{1}), (a_{2}, b_{2}) ∈ A × B

Such that f(a_{1}, b_{1}) = f(a_{2}, b_{2})

⇒ (b_{1}, a_{1}) = (b_{2}, a_{2})

⇒ b_{1} = b_{2} and a_{1} = a_{2}

⇒ (a_{1}, b_{1}) = (a_{2}, b_{2})

Therefore, f is one-one. Now, let (b, a) ∈ B × A of any element.

Then, there exist (a, b) ∈ A × B such that f(a, b)

= (b, a) [definition of f]

Therefore, f is onto. Hence, f is bijective.

**9. Let f : N → N be defined by****$$f (n)=\begin{cases}\frac{n+1}{2}, &\text{if is odd }\\\frac{n}{2}, &\text{if is even}\end{cases}$$**

**Sol.** Here, f : N → N is defined as,

$$f (n)=\begin{cases}\frac{n+1}{2}, &\text{if is odd }\\\frac{n}{2}, &\text{if is even}\end{cases}$$

for all n ∈ N. It can be observed that

$$f (1)=\frac{1+1}{2}=1 \space and f (2)=\frac{2}{2}=1\\\space\text{[By definition of f]}$$

∴ f(1) = f(2), where 1 ≠ 2.

Therefore, f is not one-one. Consider a natural number n in co-domain N.**Case I :** When n is odd.

Therefore, n = 2r + 1 for some r ∈ N.

Then, there exists 4r + 1 ∈ N such that

$$f(4r + 1)=\frac{4r+1+1}{2}=2r+1.$$

Therefore, f is onto.**Case II** : When n is even.

Therefore, n = 2r for some r ∈ N

Then, there exists 4r ∈ N such that

$$f(4r)=\frac{4r}{2}=2r$$

Therefore, f is onto. Hence, f is not a bijective function.**Note :** If f is one-one and onto, then we say that f is bijective function.

**10. Let A = R – {3} and B = R – {1}. Consider the ****function f : A → B defined by$$f(x)=(\frac{x-2}{x-3})$$Is f one-one and onto? Justify your answer.**

**Sol.** Here, A = R – {3}, B = R – {1}

and f : A → B is defined as

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

Let x, y ∈ A such that f(x) = f(y)

$$⇒ \frac{x-2}{x-3}=\frac{y-2}{y-3}$$

⇒ (x – 2)(y – 3) = (y – 2)(x – 3)

⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6

⇒ – 3x – 2y = – 3y – 2x

⇒ 3x – 2x = 3y – 2y

⇒ x = y

Therefore, f is one-one. Let y ∈ B = R – {1}. Then,

y ≠ 1

The function f is onto if there exists x ∈ A such

that f(x) = y.

Now, f(x) = y

$$⇒ \frac{x-2}{x-3}=y$$

⇒ x – 2 = xy – 3y ⇒ x(1 – y) = – 3y + 2

$$⇒ x= \frac{2-3y}{1-y}\epsilon\space A[y=1]\\\text{Thus, for any y}\epsilon B,\text{ there exists}\frac{2-3y}{1-y}\epsilon A\\\text{such that}\\f(\frac{2-3y}{1-y})=\cfrac{(\frac{2-3y}{1-y})-2}{(\frac{2-3y}{1-y})-3}\\=\frac{2-3y-2+2y}{2-3y-3+3y}\\=\frac{-y}{-1}=y$$

Therefore, f is onto. Hence, function f is one-one and onto.

**11. Let f : R → R be defined as f(x) = x4. Choose the ****correct answer.****(a) f is one-one onto****(b) f is many-one onto****(c) f is one-one but not onto****(d) f is neither one-one nor onto**

**Sol.** (d) f is neither one-one nor onto

Function f : R → R is defined as f(x) = x^{4}

Let x, y ∈ R such that f(x) = f(y)

⇒ x^{4} = y^{4}

⇒ x = ± y

Therefore, f(x_{1}) = f(x_{2}) does not imply that x_{1} = x_{2 }For instance, f(1) = f(– 1) = 1

Therefore, f is not one-one.

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.

Therefore, f is not onto. Hence, function f is neither one-one nor onto.

**12. Let f : R → R be defined as f(x) = 3x. Choose the correct answer :****(a) f is one-one onto****(b) f is many-one onto****(c) f is one-one but not onto****(d) f is neither one-one nor onto**

**Sol.** (a) f is one-one onto

f : R → R is defined as f(x) = 3x

Let x, y ∈ R such that f(x) = f(y).

⇒ 3x = 3y ⇒ x = y

Therefore, f is one-one.

Also, for any real number (y) in co-domain R,

$$\text{there exists}\frac{y}{3}\text{in R such that}\\f(\frac{y}{3})=3(\frac{y}{3})=y$$

Therefore, f is onto. Hence, function f is one-one and onto.