Math Assignment Class XI Ch - 2 | Relations & Functions

MATHEMATICS ASSIGNMENT
Chapter 2 Class -11
Relations and Functions

Extra questions of chapter 2 class 11 with answer and hints to the difficult questions. Important and useful math. assignment for the students of class 11

ASSIGNMENT ON RELATIONS & FUNCTIONS

LEVEL-1

Question 1

i) Find x and y if (x + 3, 5) = (6, 2x + y)

Ans (x = 3, y = - 1)

ii) If ordered pair (x, -1) and (5, y) belongs to the set {(a, b): b = 2a - 3}, find the value of x and y

iii) Find the value of a and b If $\left ( \frac{a}{3}+1,\; b-\frac{2}{3} \right )=\left ( \frac{5}{3},\; \frac{1}{3} \right )$

[Ans a = 2, b = 1]

(iv) If a ∈ {-1, 2, 3, 4, 5}, b ∈ {0, 3, 6}, write the set of all ordered pairs (a, b) such that
a + b = 5
Solution
-1 + 6 = 5 ⇒ (-1, 6) ∈ (a, b)
2 + 3 = 5 ⇒ (2, 3) ∈ (a, b)
5 + 0 = 5 ⇒ (5, 0) ∈ (a, b)
⇒ R = {(-1, 6), (2, 3), (5, 0)}

Question 2

i) If A = {1, 3, 5, 6} and B = {2, 4}, find A × B and B × A

ii) If A = {1, 2, 3}, B = {3, 4} and C ={1, 3, 5}, find

(a) A × (B ∩ C)

(b) (A × B) ∩ (A × C)

iii) If A = {1, 3, 5} , B = {x, y} then represent A × B and B × A in arrow diagrams

Question 3

Find the domain and range of the relation R defined by
R = {(x, x³) : x is a prime number less than 10}

Ans [Domain R = {2, 3, 5, 7},
Range R = {8, 27, 125, 343}]

Question 4

(i) n(A) = 3, n(B) = 4, then find n(A × A × B)

Ans: 36

(ii) If A = {1, 2, 4}, B = {2, 4, 5) and C = {2, 5}, write (A - C) × (B - C)

Ans : {(1, 4), (4, 4)}

(iii) If A = {1, 2} and B = {3, 4}. Find A × B and total number of subsets of A × B. Also find the total number of relations from A to B

[Ans: 2^{2 ✕ 2} = 2⁴=16 ]

(iv) If A = {1, 2, 3, 5}, B = {4, 6, 9} and R be a relation from A to B defined by

R = {(x, y) : |x - y| is odd}. Write R in roster form.

Ans: R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)

Question 5

Express the following functions as set of ordered pair and determine their ranges

(i) f : A➝ R, f(x) = x²+ 1, where A = {-1, 0, 2, 4}

Ans(i): Range = {1, 2, 5, 17}

(ii) g : A➝ N, g(x) = 2x, where A = { x : x∈ N, x ≤ 7}

Ans(ii) : Range = {2, 4, 6, 8, 10, 12, 14}

Question 6

Let f : R - {2} ➝ R be defined by

${"type":"$","id":"11","backgroundColor":"#ffffff","code":"$f\\left(x\\right)\\,=\\frac{x^{2}-4}{x-2}$","aid":null,"font":{"family":"Arial","size":14,"color":"#000000"},"backgroundColorModified":false,"ts":1614492207057,"cs":"Mpdh6G8bUfZVmkS9hkijVQ==","size":{"width":112,"height":28}}$

and g : R ➝ R be defined by g(x) = x + 2. Find whether f = g or not.

Solution:

$f(x)=\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$

Also g(x) = x + 2 ⇒ f(x) = g(x)

But Domain of f(x) = R - {2} and the domain of g(x) = R and two functions

are equal only if their domains are equal.

Hence f(x) ≠ g(x)

Question 7

Find the domain for which the functions f(x) = 2x² - 1 and g(x) = 1 - 3x are equal

Solution:

Here f(x) = g(x)

2x² - 1 = 1 - 3x ⇒ 2x² + 3x – 2 = 0

⇒ x = -2, 1/2

⇒ f(x) and g(x) are equal on the set {-2, 1 / 2 }

Question 8

If f, g, h are three functions defined from R to R as follows

(i) f(x) = x² (ii) f(x) = sinx (iii) f(x) = x² + 1

Find the range of each function

Solution

(i) For all values of x, f(x) takes only +ve value. Also if x = 0, then f(x) = 0 . So Range of f(x) = [0, ∞)

(ii) Since -1 ≤ sinx ≤ 1 for all values of x. So Range of f(x) = [-1,1]

(iii) Since x² is ≥ 0 ⇒ x²+ 1 ≥ 1 ⇒ Range of f(x) = [1, ∞)

Question 9 If $f(x)=\frac{x-1}{x+1}, x\neq -1,$ then find f(f(x))

Solution:

$f(f(x))=\frac{f(x)-1}{f(x)+1}=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{-1}{x}$

Question 10

Find the domain of the function f(x) defined by

$f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}$

Solution

f(x) is defined for all x satisfying

4 - x ≥ 0 and x²- 1 > 0

x ≤ 4 and (x - 1)(x + 1) > 0
For (x - 1)(x + 1) = 0, the critical points are x = 1 and x = -1

(x - 1)(x + 1) > 0

x ∈ (- ∞, -1) U (1, ∞), But x ≤ 4

Therefore : D_f = (- ∞, -1) U (1, 4]

Question 11

Find the domain and range of the functions $f(x)=\frac{x}{1+x^{2}}$

Solution

$f(x)\; is\; defined\; \forall \; x\; \epsilon\; R\Rightarrow D_{f}=R$

$f(x)\; =y=\frac{x}{1+x^{2}}\Rightarrow yx^{2}-x+y=0$

By using quadratic formula we get

$x=\frac{1\pm \sqrt{1-4y^{2}}}{2y}$

Now x is defined (real) if 1 - 4y² ≥ 0 and y ≠ 0

⇒ 4y² - 1 ≤ 0 ⇒ (2y + 1)(2y - 1) ≤ 0

Critical points are y = -1/2 and 1/2

(2y + 1)(2y - 1) ≤ 0 and y ≠ 0 ⇒ y ∈ [-1/2, 1/2] - {0}

⇒R_f = [-1/2, 1/2] - {0}

(ii) Find the domain and range of the functions

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

Solution Hint:

$Domain:R-\left\{-\sqrt{2},\sqrt{2}\right\}$

For range find x in terms of y we get

$x=\sqrt{\frac{2y-3}{y}}$

Range = (-∝, 0) ⋃ [3/2, ∝)

(iii) Find the domain and range of the functions

$f(x)=\frac{x^{2}-9}{x-3}$

Solution (iii)

f(x) is defined for all real numbers except at x = 3

Therefore D_f = R - 3

$Let\; \; y=f(x)=\frac{x^{2}-9}{x-3}$

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

But x = 3 ∉ D_f ⇒ y = 3 + 3 = 6 ∉ R_f ⇒ R_f = R - {6}

Question 12

Find the domain and range of the following functions

i) f(x) = |x - 1| Ans : [D = R] R = [0, ∞)

ii) x² + y² = 25 , D_f = [-5, 5], R_f = [0, 5]

iii) $f(x)=\frac{ax-b}{cx-d}\; \; \; \;$

Ans: $D_{f}=R-\left \{ \frac{d}{c} \right \}, R_{f}= R-\left \{ \frac{a}{c} \right \}$

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

Ans: Domain = R - {2}, Range = R - {-1}

v) $f(x)=\sqrt{16-x^{2}}$

Ans: Domain = [-4, 4], Range = [0, 4]

vi) $f(x)=\frac{3}{\sqrt{9-x^{2}}}$

Ans: Domain = (-3, 3) Range = [1, ∞)

Solution (vi)

f(x) is defined if $9-x^{2}>0\Rightarrow x<\pm 3$

$-3<x<3\Rightarrow x\epsilon (-3,3)\Rightarrow D_{f}=(-3,3)$

$y=\frac{3}{\sqrt{9-x^{2}}}\Rightarrow y^{2}=\frac{9}{9-x^{2}}$

⇒ 9y²- x²y²= 9

⇒ x²y² = 9y² - 9

$\Rightarrow x^{2}=\frac{9y^{2}-9}{y^{2}} \Rightarrow x=\sqrt{\frac{9y^{2}-9}{y^{2}}}$
x is defined if 9y²- 9 ≥ 0 and y ≠ 0
x is defined if y²- 1 ≥ 0 and y ≠ 0

x is defined if (y + 1)(y - 1) ≥ 0 and y ≠ 0

x is defined if y ≤ -1, y ≥ 1 and y ≠ 0

⇒ y ∈ (-∞, -1] ⋃ [1, ∞)
⇒ R_f= (-∞, -1] ⋃ [1, ∞)

(vi) Find the domain of the function

$f(x)=\sqrt{\frac{9-x^{2}}{x+1}}\; \; \; \; \; \; \; \; \; Ans: (-\infty ,-3]\cup (-1,3]$

Ans: D_f= (-∞, -3] ⋃ (-1, 3]

Question 13

Draw the graph of the following write its range

$f(x)=\left \{ \begin{matrix} 1-x, & x<0\\ 1,&x=0 \\ 1+x, & x>0 \end{matrix} \right.$

Domain of this function is R

Rang of the given function (1, ∞)

Question 14

If $f(x)=\left\{\begin{matrix} 1-x, &x\leq 1 \\ x-1, & x>1 \end{matrix}\right.$ then evaluate f(-2) + f(2)

Ans: 4

Question 15

Find the domain of the following function

$f(x)=\frac{x^{2}+3x+5}{x^{2}-5x+4}$ Ans : D_f= R - {1, 4}

Question 16: Solve : $\frac{1}{|x|-3}\leq \frac{1}{2}$

Ans: $(-\infty ,-5]\cup (-3,3)\cup [5,\infty )$

Solution: Let |x| = y

$\frac{1}{|x|-3}\leq \frac{1}{2}\; \; =\; \; \frac{1}{y-3}-\frac{1}{2}\leq 0$

$\Rightarrow \frac{5-y}{2y-6}\leq 0\; \; \Rightarrow \; \; \frac{y-5}{2y-6}\geq 0$

Critical points are y = 5, y = 3

y = |x| = 5 and y = |x| = 3 ⇒ x = 土 5, x = 土 3

$x\;\epsilon\; (-\infty ,-5]\cup (-3,3)\cup [5,\infty )\]$

$D_{f}= (-\infty ,-5]\cup (-3,3)\cup [5,\infty )$

Question 17:

Let f = {(1,1), (2,3), (0,-1), (-1, -3)} be a linear function from Z ⟶ Z. Find the function f as a linear function of x.

Solution:

As given f is a linear function, let f(x) ax + b. Also (1,1) and (2,3) ∊ f

f(1) =1 and f(2) = 3

⇒ a + b = 1 and 2a + b = 3

On solving these equations we get a = 2, b = -1

⇒ f(x) = 2x - 1

Question 18 (DAV Final Paper, 2023)

Find the domain and range of $f(x) =\frac{3}{2-x^{2}}$

Ans: Domain of f = R-{√2, -√2}

Range of f = [3/2, ∞) ∪ (-∞,0]

Question 19 (DAV Final Paper, 2023)

Draw the graph of the function f defined by f(x) = x + |x + 1|. Hence find the range from the graph.

Solution: The given function can be written as

$f(x)=\left\{\begin{matrix}x-x-1&x\leq-1\\x+x+1&x>-1\\\end{matrix}\right.$

$f(x)=\left\{\begin{matrix}-1&x\leq-1\\2x+1&x>-1\\\end{matrix}\right.$

Find few points on the graph as follows

X	-3	-2	-1	0	1	2
y	-1	-1	-1	1	3	5

From the above graph we find

Range = [-1, ∞)

Question 20 : (DAV SP 2023)

Find the domain of $f(x)=\frac{1}{\sqrt{x+|x|}}$

Solution

$|x|=\begin{cases}x,&\text{if}\;\;x\geq 0\\-x,&\text{if}\;\;x<0\end{cases}$

$x+|x|=\begin{cases}x+x,&\text{if}\;\;x\geq 0\\x-x,&\text{if}\;\;x<0\end{cases}$

$f(x)=x+|x|=\begin{cases}2x,&\text{if}\;\;x\geq 0\\0,&\text{if}\;\;x<0\end{cases}$

Also f(x) is not defined at x = 0

⇒ f(x) has real values if x > 0

Domain of f(x) = (0, ∝)

PDF form of this assignment

Questions deleted from CBSE syllabus

Question 1

Find the inverse relation R^-1 in each of the following

(i) R = {(1,2), (1,3), (2, 3), (3, 2), (5, 6)

Ans {(2,1), (3,1), (3,2), (2,3), (6,5)}

(ii) R = {(x, y) : x, y ∈ N, x + 2y = 8

Ans {(3, 2), (2, 4), (1,6)

Solution
If x = 2, y = 3 ⇒ 2 + 2 × 3 = 8 ⇒ (2, 3) ∈ R

If x = 4, y = 2 ⇒ 4 + 2 × 2 = 8 ⇒ (4, 2) ∈ R

If x = 6, y = 1 ⇒ 6 + 2 × 1 = 8 ⇒ (6, 1) ∈ R

R = {(2, 3), (4, 2), (6, 1)} ⇒ {(3, 2), (2, 4), (1, 6)}

Question 2

Let f : R➝R be a function given by f(x) = x²+ 1. Find f ^-1{10, 37}

Solution

If f(x) = y then x = f ^-1(y)

f ^-1(10) = x ⇒ f(x) = 10 ⇒ x²+ 1 = 10

⇒ x²= 9 ⇒ x = 土 3

f ^-1(37) = x ⇒ f(x) = 37 ⇒ x²+ 1 = 37

⇒ x²= 36 ⇒ x = 土 6

f ^-1{10, 37} = {-3, 3, - 6, 6}

Question 3

Let A = {-2, -1, 0, 1, 2} and f : A ➝ Z be a function defined by f(x) = x²- 2x - 3. Find

(i) Range of f i.e. f(A) (ii) Pre - image of 6, - 3 and 5

Solution

(i) f(A) = {f(-2), f(-1), f(0), f(1), f(2) } = {5, 0, - 3, - 4, - 3} = {- 4,- 3, 0, 5}

(ii) Let Pre - image of 6 = x ⇒ f ^-1(6) = x ⇒ f(x) = 6

⇒ x² - 2x - 3 = 6 ⇒ x²- 2x - 9= 0

There is no real value of x which satisfies this equation.

So Pre - image of 6 = Φ

Let Pre - image of - 3 = x ⇒ f ^-1(-3) = x ⇒ f(x) = - 3

⇒ x² - 2x - 3 = - 3 ⇒ x² - 2x = 0 ⇒ x = {0, 2}

Let Pre - image of 5 = x ⇒ f ^-1(5) = x ⇒ f(x) = 5

⇒ x² -2x - 3 = 5 ⇒ x² - 2x - 8 = 0 ⇒ x = {-2, 4}

Question 4

(i) f(x) = 3x⁴- 5x²+ 9, find f(x - 1)

Ans [3x⁴- 12x³+ 13x² - 2x + 7]

(ii) Write the domain of f(x) = x²+ 1 and draw its graph. Also find the value of x for which f(x) = f(x + 1)

(iii) If f(x) = x³- 3x + 4, then find the value of x such that f(x) = f(2x + 1)

[Ans x = -1, 2/3]

(iv) F(x) = 4x - x², x ∈ R, then find f(a + 1) – f(a - 1)

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