### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

## MATHEMATICS ASSIGNMENTChapter 2 Class -11Relations 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 x B and B x A

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

(a) A x (B ∩ C)

(b)  (A x B) ∩ (A x C)

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

Question 3

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), (5,6)}

(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 x 3 = 8  (2, 3)  R
If  x = 4, y = 2
4 + 2 x 2 = 8  (4, 2)  R
If  x = 6, y = 1
6 + 2 x 1 = 8  (6, 1)  R
R = {(2, 3), (4, 2), (6, 1)}
{(3, 2), (2, 4), (1, 6)}

Question 4

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

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

Question 5

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

Ans: 36

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

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

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

[Ans:    22 ✕ 2  = 2=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 6

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

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

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

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

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

Question 7

Let f : R - {2}  R be defined by

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 8

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 9

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

Solution

If f(x) = y then  x = f -1(y)

-1(10) = x  f(x) = 10  x+ 1 = 10

x= 9   x =  3

f -1(37) = x  f(x) = 37  x+ 1 = 37

x= 36   x =  6

f -1{10, 37} = {-3, 3, - 6, 6}

Question 10

Let A = {-2, -1, 0, 1, 2} and f : A  Z be a function defined by  f(x) = x- 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   -1(6) = x  f(x) = 6

x2 - 2x - 3 = 6     x- 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   -1(-3) = x  f(x) = - 3

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

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

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

## ASSIGNMENT ON RELATIONS & FUNCTIONS

Level - 2

Question 11

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  x+ 1 ≥ 1   Range of f(x) = [1, ∞)

Question 12  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 13

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 x- 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 : Df  = (- ∞, -1) U (1, 4]

Question 14

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}

Rf = [-1/2, 1/2] - {0}

(ii)  Find the domain and range of the functions

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

(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 Df = 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  Df    y = 3 + 3 = 6  Rf    Rf = R - {6}

Question 15

Find the domain and range of the following functions

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

ii) x2 + y2 = 25

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 = {-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

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

⇒ 9y- x2y= 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  9y- 9 ≥ 0 and  y ≠ 0
x is defined if  y- 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, ∞)
⇒ Rf  = (-∞, -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: Df  = (-∞, -3] ⋃ (-1, 3]

Question 16

(i) f(x) = 3x- 5x+ 9, find  f(x - 1)

Ans[3x- 12x+ 13x2  - 2x + 7]

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

(iii) If f(x) = x- 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)

Question 17

Draw the graph of the following write its range

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

Question 18

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

Question 19

Find the domain of the following function

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

Question 20:  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 )\]$

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