Class 11 Maths NCERT Solutions Chapter 2: Relations and Functions Explained with PDF

Class 11 Math Chapter 2 Relation and Function Exercise 2.3 Solutions

Q1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

(iii) {(1,3), (1,5), (2,5)}.

A.1. (i)Domain of the given relation ={2,5,8,11,14,17}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain ={2,5,8,11,14,17}

range ={1}

(ii)Domain of the given relation ={2,4,6,8,10,12,14}

Since every element of the domain has one and only one image, the given relation is a fxⁿ.

So, domain = {2, 4, 6, 8, 10, 12, 14}

range ={1,2,3,4,5,6,7}

(iii)Domain of the given relation ={1,2}

As element 1 has more than one image i.e., 3 and 5, the given relation is not a fxⁿ.

Q2. Find the domain and range of the following real functions:

(i) f(x) = – |x|                            (ii)

Q3. A function f is defined by f(x) = 2x –5. Write down the values of

(i) f (0), (ii) f (7), (iii) f (–3)

A.3. Give, f(x) = 2x – 5.

(i)f(0)=(2 × 0) –5=0 – 5= –5

(ii)f(7)=(2 × 7) –5=14 – 5=9

(iii)f(–3)=2 ×(–3) –5= –6 – 5= –11.

Q4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = $\frac{9}{}$  C / 5   +   3   2  .

Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212.

Q5. Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x  $\in$ R, x > 0.

(ii) f (x) = x²+ 2, x is a real number.

(iii) f (x) = x, x is a real number.

A.5. (i)f(x)=2 – 3x, x $\in$ R, x>0.

Given, x>0

3x>3 × 0

3x>0

(–1) × 3x<(–1) × 0.

–3x<0

2 – 3x<0+2

2 – 3x<2

i.e., f(x) < 2

Hene, range of f(x) = (– ꝏ, 2)

(ii)Given, f(x) = x²+2, x is a real number.

Since, x is a real number,

x² ≥ 0 (x²=0 for x=0)

x²+2 ≥ 0+2

x²+2 ≥ 2

f(x) ≥ 2

⸫Range of f(x) = [2, ꝏ) 

(iii)Given, f(x) = x, x is a real number.

As, f(x) = x, the range of f(x) is also real.

i.e., Range of f(x) = R.

Q1. The relation f is defined by $f\left(x\right)=\{\begin{array}{l}{x}^2,0\le x\le 3\\ 3x,3\le x\le 10\end{array}$

The relation g is defined by $g\left(x\right)=\{\begin{array}{l}{x}^2,0\le x\le 2\\ 3,2\le x\le 10\end{array}$

Show that f is a function and g is not a function.

A.1. Given, f(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 3\hfill \\ 3x,\hfill & 3\le x\le 10\hfill \end{array}$

f(x)={(0,0),(1,1),(2,4),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the elements in domain of f  has one and only one image.

 ⸫          f(x) is a function.

Given, g(x)= $\{\begin{array}{ll}{x}^2,\hfill & 0\le x\le 2\hfill \\ 3x,\hfill & 2\le x\le 10\hfill \end{array}$ .

g(x)={(0,0),(1,1),(2,4),(2,6),(3,9),(4,12),(5,15),(6,18),(7,21),(8,24),(9,27),(10,30)}

So, the element 2 of the domain has more than one image i.e., 4 and 6.  

⸫ g(x) is not a function.

Q2. If f (x) = x², find $\frac{f\left(1.1\right)-f\left(1\right)}{\left(1.1-1\right)}$

A.2.      Given, f(x)=x².

$\frac{f(1.1)-f(1)}{1.1-1}=\frac{{(1.1)}^2-1^2}{1.1-1}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$

Q3. Find the domain of the function f (x) $\frac{x^2+2x+1}{x^2-8x+12}$ .

A.3. Given, f(x)= $\frac{x^2+2x+1}{x^2-8x+12}$

            The given function is valid if denominator is not zero.

            So, if x² – 8x+12=0.

            ⇒        x² – 2x – 6x+12=0

            ⇒        x(x – 2) –6(x – 2)=0

            ⇒        (x – 2)(x – 6)=0

            ⇒        x=2 and x=6.

            So, f(x) will be valid for all real number x except x=2,6.

            ⸫          Domain of f(x)=R – {2,6}

Q4. Find the domain and the range of the real function f defined by

Given, f(x)=

The given fxⁿ is valid for all x such that x – 1 ≥ 0 ⇒x≥ 1

⸫          Domain of f(x)=[1, ꝏ)

As x ≥ 1

⇒        x – 1 ≥ 1 – 1

⇒        x – 1 ≥ 0

⇒         ≥ 0

⇒        f(x) ≥ 0

 So, range of f(x)=[0, ꝏ)

A.5. Given, f(x)=|x – 1|.

The given function is defined for all real number x.

Hence, domain of f(x)=R.

As f(x)=|x – 1|, x  $\in$ R is a non-negative no.

Range of f(x)=[0, ꝏ), if positive real numbers.

Q6. Let $\left\{\left(x,\frac{x^2}{1+x^2}\right):x\in R\right\}$ be a function from R into R. Determine the range of f.

A.6.      Given, f(x)= $\left\{\left(x,\frac{x^2}{1+x^2}\right):x\in R\right\}$

            We know that, for x $\in$ R.

            So, x²≥ 0 ⇒ $\frac{x^2}{x^2+1}\ge \frac{0}{x^2+1}\Rightarrow \frac{x^2}{x^2+1}\ge 0\Rightarrow f(x)\ge 0$

            and x²+1>x²

            ⇒ $1>\frac{x^2}{x^2+1}$

            ⇒1 > f(x).

            So, 0 ≤ f(x) < 1

            ⸫          Range of f(x) = [0,1).

A.7. Given, f(x) = x+1. and g(x) = 2x – 3.

So, (f +g)(x) = f(x)+g(x) = (x+1)+(2x – 3) = x+1+2x – 3 = 3x – 2

(f – g)(x) = f(x) –g(x) = (x+1)–(2x–3) = x+1 – 2x+3 = 4 – x

$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2x-3} such that x\ne \frac{3}{2}$

A.8.  Given, f(x) = (ax + b)

= {(1,1),(2,3),(0, – 1),(–1, –3)} .

As (1,1)  $\in$ f.

Then, f(1)=1   [⸪ f(x) = y for (x, y)]

a × 1+b=1

a+b=1…… (1)

and (0, – 1)  $\in$ f .

Then, f(0)= –1

a× 0+b= –1

b= –1……..(2)

Putting value of (2) in (1) we gets

a – 1=1

a=1+1

a=2

So, (a, b)=(2, –1)

Q9. Let R be a relation from N to N defined by R = {(a, b) : a, b $\in$ N and a = b²}. Are the following true?

(i) (a,a)  $\in$ R, for all a _ N (ii) (a,b)  $\in$ R, implies (b,a)  $\in$ R

(iii) (a,b)  $\in$ R, (b,c)  $\in$ R implies (a,c)  $\in$ R.

Justify your answer in each case.

A.9. Given, R={(a, b): a, b  $\in$ N and a = b²}

(i)Let a = 2  $\in$ N

Then b = 2² = 4 $\in$ N

but a ≠ b.

Hence the given statement is not true.

(ii)For a=b² the inverse b=a² may not hold true

Example (4,2)  $\in$ R, a=4, b=2 and a=b²

but (2,4)  $\notin$ R.

Hence, the given statement is not true.

(iii)If (a,b)  $\in$ R

a=b²……(1)

and (b, c)  $\in$ R

b=c²…….(2)

so for (1) and (2),

a=(c²)²=c⁴.

is, a ≠c²,

Hence, (a, c)  $\notin$ R.

⸫The given statement is false.

Q10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?

(i) f is a relation from A to B (ii) f is a function from A to B.

Justify your answer in each case.

A.10. Given,

A={1,2,3,4}

B={1,5,9,11,15,16}

f={(1,5),(2,9),(3,1),(4,5),(2,11)}.

(i)As every element of f is an element of A × B

We can clearly say that f  $\subset$ A × B.

⸫f is a relation from A to B.

(ii)As the element 2 of the domain has two image i.e., 9 and 11. f is not a function.

A.11. Given, f={(ab, a+b): a, b $\in$ z}

Let a=1 and b=1;                   a, b $\in$ z.

So, ab=1 × 1=1

a+b=1+1=2.

So, we have the order pair (1,2).

Now, let a= –1 and b= –1; a, b  $\in$ z

So, ab=(–1) × (–1)=1

a+b=(–1)+(–1)= –2

So, the ordered pair is (1, –2).

⸫The element 1 has two image i.e., 2 and –2.

Hence, f is not a function.

A.12. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [⸪ prime factor of 9=3]

f (10)=5 [⸪ prime factor of 10=2,5]

f(11)=11 [⸪ prime factor of 11 = 11]

f(12)=3 [⸪ prime factor of 12 = 2, 3]

f(13)=13 [⸪ prime factor of 13 = 13]

⸫Range of f=set of all image of f(x) = {3,5,11,13}.