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Mathematics Lab Manual Class XII | 14 Activities

    Mathematics Lab Manual Class XII 14 lab activities for class 12 with complete observation Tables strictly according to the CBSE syllabus also very useful & helpful for the students and teachers. General instructions All these activities are strictly according to the CBSE syllabus. Students need to complete atleast 12 activity from the list of 14 activities. Students can make their own selection.

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  

[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,  x) : 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:    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 5

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}   

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 

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

Solution:    

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  x+ 1 ≥ 1   Range of f(x) = [1, ∞)

Question 9  If  then find f(f(x))

Solution: 

Question 10

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

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

∈ (- ∞, -1) U (1, ∞),   But   x ≤ 4

Therefore : Df  = (- ∞, -1) U (1, 4]

Question 11

Find the domain and range of the functions  

Solution

By using quadratic formula we get 

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 

Solution Hint: 

equation 

For range find x in terms of y we get

equation 
Range =  (-∝, 0) ⋃ [3/2, ∝) 

(iii)  Find the domain and range of the functions

Solution (iii)

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

Therefore Df = R - 3

 y = f(x) =  

 But x = 3  Df    y = 3 + 3 = 6  Rf    Rf = 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 ,  Df = [-5, 5],  Rf = [0, 5]

iii)  

Ans:  

iv) 

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

v) 

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

vi)  

Ans: Domain = (-3, 3) Range = [1, ∞)

Solution (vi)

f(x) is defined if  

⇒ 9y- x2y= 9

⇒ x2y2 = 9y2 - 9


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 




Ans: Df  = (-∞, -3] ⋃ (-1, 3]

Question 13

Draw the graph of the following write its range

  


Domain of this function is R
Rang of the given function (1, ∞)

Question 14

If     then evaluate f(-2) + f(2)

Ans: 4

Question 15

Find the domain of the following function

    Ans : Df  = R - {1, 4}

Question 16:  Solve :    

Ans:  

Solution: Let |x| = y 

 Critical points are  y = 5, y = 3

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









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    
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

equation 

equation 
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 equation

Solution


equation

equation


equation 
Also f(x) is not defined at x = 0
⇒ f(x) has real values if  x > 0
Domain of f(x) = (0, ∝)


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 : 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 3

 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}

Question 4

(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)


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