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Mathematical Indunction

Mathematical Indunction (M.I) Exerxise wise Solutions

Mathematical Indunction (M.I) Exerxise wise Solutions

Mathematical Indunction: It is a technique for proving results or establishing statements for natural numbers. This part illustrates the method through a variety of examples.

Definition:
Mathematical Indunction  is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.

The technique involves steps to prove a statement, as stated below −
Let P(n) or S(n) be the given statement

Step 1: For n =1
                 we get LHS = RHS
                 then P(n) is true for n= 1
Step 2: Let us assume thet P(n) is true for n =k
Step 3: We have to Prove P(n) is true for n= k + 1

Laplace:

Laplace was a mathematecian and astronomer whose work was pivotal to the development of mathematical astronomy. His most outstanding work was done in the fields of celestial mechonics, probability, differential equations, and geodesy. His five volume work on celestial mechonics earned him the title of the Newton of France.

Laplace

“Analysis and natural philosophy owe their most important discoveries to this fruitful means, which is called indunction” – Pierr Simon de Laplace


Exercise 2(a)

Using Mathematical Induction, Prove each of the following statement for all n ∈ N.

1. 12 + 22 + 32 + …… + n2= Mathematical Induction 1

Let p(n) be the given statement that

 12 + 22 + 32 + …… + n2=Mathematical Induction 1

For n= 1

LHS = 12 = 1

RHS =  =  = 1

LHS = RHS

P(n) is true for n = 1

Let us assume that P(n) is true for n = k

i.e., 12 + 22 + 32 + …… + k2=  ………… (1)

for n = k + 1

add (k +1)2 on both sides of (1)

12 + 22 + 32 + …… + k2 + (k +1)2 =Mathematical Induction 4 

                                                              Mathematical Induction 5   

                                                             Mathematical Induction 6

P(n) is true for n = k+ 1

∴ By the principle of Mathematical induction

   P(n) is true for all n ∈ N

∴ 12 + 22 + 32 + …… + n2=Mathematical Induction 1

2.  2.3 + 3.4 + 4.5 + ……… up to n terms = Mathematical Induction 7

First factors of given series are: 2, 3, 4, 5, …

                                  a = 2, d = 1

                                 an = a + (n – 1) d

                                       = 2 + (n – 1) (1)

                                       = 2 + n – 1

                                       = n + 1

Second factors of given series are: 3, 4, 5,…

                                  a = 3, d = 1

                                 an = 3 + (n – 1) d

                                       = 3 + (n – 1) (1)

                                       = 3 + n – 1

                                       = n + 2

nth term of given series is (n + 1) (n + 2)

let P(n) be the given statement that

2.3 + 3.4 + 4.5 + ……… + (n + 1) (n + 2) = Mathematical Induction 7

For n = 1

LHS = 2.3 = 6

RHS =  =  =  = 6

LHS = RHS

P(n) is true for n = 1

Let us assume that P(n) is true for n = k

i.e., 2.3 + 3.4 + 4.5 + ……… + (k + 1) (k + 2) = Mathematical Induction 8 ………… (1)

for n = k + 1

  add (k + 2) (k + 3) on both sides of (1)

 2.3 + 3.4 + 4.5 + ……… + (k + 1) (k + 2) + (k + 2) (k + 3)  

  =  Mathematical Induction 8 + (k + 2) (k + 3)

Mathematical Induction 9

Mathematical Induction 10

Mathematical Induction 11

P(n) is true for n = k+ 1

∴ By the principle of Mathematical induction

   P(n) is true for all n ∈ N

∴ 2.3 + 3.4 + 4.5 + ……… up to n terms =Mathematical Induction 7

 

 

 

 

 

 


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Functions Exercise 1c

Functions Exercise 1c Solutions ||TS||

Functions Exercise 1c Solutions 

The famous mathematician ” Lejeune Dirichlet”  defined a function.

Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.

Chapter 1 Functions Exercise 1c Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

Exercise 1c

Exercise 1(c) Solutions

I.

1.Find the domains of the following real valued functions

 

(i)

f (x) =Functions 1(c) images 1

given function is f (x) =Functions 1(c) images 1

 f (x) is defined when (x2 – 1) (x + 3) ≠ 0

  ⟹ (x2 – 1) ≠ 0 or (x + 3) ≠ 0

  ⟹ (x + 1) (x – 1) ≠ 0 or (x + 3) ≠ 0

  ⟹ x ≠ 1, x ≠ – 1 or x ≠ – 3

∴ Domain of f(x) is R – {– 1, – 3, 1}

(ii)

f (x) = Functions 1(c) images 2

 Given function is f (x) = Functions 1(c) images 2

 f (x) is defined when (x – 1) (x – 2) (x – 3) ≠ 0    

  ⟹ x ≠ 1, x ≠ 2 or x ≠ 3

∴ Domain of f(x) is R – {1, 2, 3}

(iii)

f (x) = Functions 1(c) images 3

Given function is f (x) = Functions 1(c) images 3

f (x) is defined when 2 – x > 0 and 2 – x ≠ 1                        

  ⟹ 2 > x  and  2 – 1 ≠ x                        

⟹ 2 > x  and  x ≠ 1                        

∴ Domain of f(x) is (– ∞, 2) – {1}

(iv)

f (x) = Functions 1(c) images 4

Given function is f (x) = Functions 1(c) images 4

f (x) is defined when x ∈ R      

∴ Domain of f(x) is R

Functions Exercise 1c

(v)

f (x) =Functions 1(c) images 5

Given function is f (x) = Functions 1(c) images 5

f (x) is defined when 4x – x2 ≥ 0      

⟹ x (4 – x) ≥ 0

⟹ x (x – 4) ≤ 0

⟹ (x – 0) (x – 4) ≤ 0

⟹ x ∈ [0, 4]

∴ Domain of f(x) is [0, 4]

(vi) 

f (x) =Functions 1(c) images 6

 Given function is f (x) =Functions 1(c) images 6

 f (x) is defined when 1 – x2 > 0   

 ⟹ x2 – 1 < 0

⟹ (x – 1) (x + 1) < 0

⟹ x ∈ (– 1, 1)

∴ Domain of f(x) is (– 1, 1)

(vii)

f (x) =Functions 1(c) images 7

 Given function is f (x) = Functions 1(c) images 7

  f (x) is defined when x + 1≠ 0   

  ⟹ x ≠ – 1  

  ∴ Domain of f(x) is R – {– 1}

(viii)

f(x) =Functions 1(c) images 8

  Given function is f (x) =Functions 1(c) images 8

  f (x) is defined when x2 – 25 ≥ 0   

  ⟹ (x – 5) (x + 5) ≥ 0  

  ⟹ x ∈ (–∞, –5] ∪ [5, ∞)

 Functions 1(c) images 11

  ⟹ x ∈ R – (– 5, 5)

 ∴ Domain of f(x) is R – (– 5, 5)

Functions Exercise 1c

(ix)

f(x) = Functions 1(c) images 12

Given function is f (x) = Functions 1(c) images 12

f (x) is defined when x – [x] ≥ 0   

  ⟹ x ≥ [x]

 ⟹ x ∈ R   

∴ Domain of f(x) is R  

(x)

f(x) =Functions 1(c) images 13

Given function is f (x) =Functions 1(c) images 13

f (x) is defined when [x] – x ≥ 0   

  ⟹ [x] ≥ x

 ⟹ x ∈ Z   

∴ Domain of f(x) is Z

Functions Exercise 1c

2. find the ranges of the following real valued functions

 

(i)

f(x) =Functions 1(c) images 14

Given function is f (x) = Functions 1(c) images 14

Let y = Functions 1(c) images 14

  ⟹  |4 – x2| = ey

∵ ey > 0 ∀ y ∈ R

∴ Range of f(x) is R

(ii) 

f(x) =Functions 1(c) images 13

Given function is f (x) =Functions 1(c) images 13

f (x) is defined when [x] – x ≥ 0   

  ⟹ [x] ≥ x

 ⟹ x ∈ Z   

Domain of f(x) is Z

Range of f = {0}

(iii) 

f(x) =Functions 1(c) images 15

Given function is f (x) =Functions 1(c) images 15

f (x) is defined when x ∈ R   

             Domain of f(x) is R

           For x ∈ R   [x] is an integer

           Since sin nπ = 0, ∀ n ∈ z

            ⟹ sin π[x] = 0

          ∴ Range of f = {0}

(iv) 

f (x) =Functions 1(c) images 16

           Given function is f (x) = Functions 1(c) images 16

          f (x) is defined when x – 2 ≠ 0

          ⟹ x ≠ 2

         Domain of f(x) is R – {2}

         Let y = Functions 1(c) images 16

            =Functions 1(c) images 17

             = x + 2

       If x = 2 ⟹ y = 2 + 2 = 4

        ∴ Range of f(x) is R – {4}

Functions Exercise 1c

(v) 

f (x) =Functions 1(c) images 18

let y =Functions 1(c) images 18

      y2 = 9 + x2

          x2 = y2 – 9

       x = Functions 1(c) images 19

     it is defined when y2 – 9 ≥ 0

      ⟹ (y – 3) (y + 3) ≥ 0

    y ∈ (– ∞, – 3] ∪ [3, ∞)

but y = Functions 1(c) images 18 ≥ 0

∴ Range of f(x) is [3, ∞)

3. If f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2 then   
    find

Sol:

    Given f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2

(i)

(3f – 2g) (x) = 3 f(x) – 2g (x)

                         = 3 (2x – 1) – 2(x2)

                         = 6x – 3 – 2x2

                          = – 2x2 + 6x – 3                   

∴ (3f – 2g) (x) =– 2x2 + 6x – 3

(ii)

(fg) (x) = f (x) g (x)

         = (2x – 1) (x2)

                = 2x3 + x2   

 ∴ (fg) (x) = 2x3 + x2  

(iii) 

Functions 1(c) images 20   

4. If f = {(1, 2), (2, – 3) (3, – 1)} then find (i) 2f (ii) (fog) 2 + f iii) f2 (iv) Functions 1(c) images 21

Sol:

Given f = {(1, 2), (2, – 3) (3, – 1)}

(i)

(2f) (1) = 2 f (1) = 2 × 2 = 4

(2f) (2) = 2 f (2) = 2 × – 3 = – 6

(2f) (3) = 2 f (3) = 2 × – 1 = – 2

∴ 2f = {(1, 4), (2, – 6) (3, – 2)}    

(ii)

(2 + f) (1) = 2 + f (1) = 2 + 2 = 4

(2 + f) (2) = 2 + f (2) = 2 + (– 3) = – 1

(2 + f) (3) = 2 + f (3) = 2 + (– 1) = 1

∴ 2 + f = {(1, 4), (2, – 1) (3, 1)}

(iii) 

(f2) (1) = [f (1))]2 = 22 = 4

(f2) (2) = [f (2))]2 = (– 3)2 = 9

(f2) (3) = [f (1))]2 = (– 1)2 = 1

∴ f2= {(1, 4), (2, 9) (3, 1)}

(iv) 

Functions 1(c) images 22

II. 

1.Find the domain of the following real valued functions

(i)

f (x) = Functions 1(c) images 23

f(x) is defined when x2 – 3x + 2 ≥ 0       

   x2 – 2x – x + 2 ≥ 0

   x (x – 2) – 1(x – 2) ≥ 0

  (x – 1) (x – 2) ≥ 0

   x ∈ (– ∞, 1] ∪ [2, ∞)

∴ Domain of f(x) is R – (1, 2)

(ii)

f(x) = log (x2 – 4x + 3)

f(x) is defined when x2 – 4x + 3 > 0       

   x2 – 3x – x + 3 > 0

   x (x – 3) – 1(x – 3) > 0

  (x – 1) (x – 3) > 0

   x ∈ (– ∞, 1) ∪ (3, ∞)

∴ Domain of f(x) is R – [1, 3]

(iii)

f(x) =Functions 1(c) images 24

f(x) is defined when 2 + x ≥ 0, 2 – x ≥ 0  and x ≠ 0 

 x ≥ – 2, x ≤ 2 and x ≠ 0   

  – 2 ≤ x ≤ 2 and x ≠ 0   

   x ∈ [–2, 2] – {0}

∴ Domain of f(x) is [–2, 2] – {0}

(iv)

f(x) =Functions 1(c) images 26

f(x) is defined in two cases as follows:

case (i) 4 – x2 ≥ 0 and [x] + 2 > 0    

               x2 – 4 ≤ 0 and [x] > –2 

               (x – 2) (x + 2) ≤ 0 and [x] > –2 

                x ∈ [– 2, 2] and x ∈ [– 1, ∞)

                x ∈ [– 1, 2]

case (ii) 4 – x2 ≤ 0 and [x] + 2 < 0       

                x2 – 4 ≥ 0 and [x] < –2 

               (x – 2) (x + 2) ≥ 0 and [x] < –2 

               x ∈ (– ∞, –2] ∪ [2, ∞) and x ∈ (–∞, –2)

                x ∈ (–∞, –2)

from case (i) and case (ii)

   x ∈ (–∞, –2) ∪ [– 1, 2]

∴ Domain of f(x) is (–∞, –2) ∪ [– 1, 2]

(v)

f(x) = Functions 1(c) images 27

f(x) is defined when Functions 1(c) images 28≥ 0 and x – x2 > o

 x – x2 ≥ (0.3)0   and x2 – x < 0

x – x2 ≥ 1 and x (x – 1) < 0

   x2 –x + 1 ≤ 0 and (x – 0) (x – 1) < 0

 it is true for all x ∈ R and x ∈ (0, 1)

∴ domain of f(x) =  R∩ (0, 1) = (0, 1)

(vi)

f(x) =Functions 1(c) images 29

f(x) is defined when x +|x| ≠ 0       

  |x| ≠ – x      

  |x| ≠ – x      

 ⟹|x| = x      

 ⟹ x > 0   

   x ∈ (0, ∞)

∴ Domain of f(x) is (0, ∞)

2. Prove that the real valued function Functions 1(c) images 30 is an even function

Sol:

Given f(x) = Functions 1(c) images 30

Functions 1(c) images 31 

                         Functions 1(c) images 32

3. Find the domain and range of the following functions

 

(i)

f (x) = Functions 1(c) images 33

Given f(x) =Functions 1(c) images 33

 Since [x] is an integer

  sin π[x] = tan π[x] = 0 ∀ x ∈ R

∴ domain of f(x) is R

and

since tan π[x] = 0

 Range of f(x) = {0}

(ii)

f(x) = Functions 1(c) images 34

Given f (x) = Functions 1(c) images 34

 It is defined when 2 – 3x ≠ 0

⟹ 2 ≠ 3x 

⟹ x ≠ 2/3

∴ Domain of f (x) = R – {2/3} 

Let y = f(x)

        y =Functions 1(c) images 34

  ⟹ y (2 – 3x) = x

          2y – 3xy = x

          2y = x + 3xy  

           2y = x (1 + 3y)

   ⟹ x = Functions 1(c) images 35

    It is defined when 1 + 3y ≠ 0

                                        1 ≠ –3y

                                      y ≠ – 1/3

∴ Range of f (x) = R – {– 1/3} 

(iii) 

f(x) = |x| + | 1 + x|

Given function is f (x) = |x| + |1 + x|

             f (x) is defined for all x ∈ R

           ∴ domain of f(x) = R

        Functions 1(c) images 36

            f (– 3) = |– 3| + |1 – 3|

                        =|– 3| + |– 2|

                       = 3 + 2 = 5 

            f (– 2) = |– 2| + |1 – 2|

                        =|– 2| + |– 1|

                        = 2 + 1 = 3

            f (– 1) = |– 1| + |1 – 1|

                        =|– 1| + |0|

                        = 1 + 0 = 1

            f (0) = |0| + |1 + 0| = 1

            f (1) = |1| + |1 + 1|

                      = 1 + |2|

                      = 1 + 2 = 3

            f (2) = |2| + |1 + 2|

                      = |2| + |3|

                      = 2 + 3 = 5

            f (3) = |3| + |1 + 3|

                      = |3| + |4|

                      = 3 + 4 = 7

∴ Range of f(x) = [1, ∞)


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Functions Exercise 1b

Functions Exercise 1b Solutions

Functions Exercise 1b Solutions 

The famous mathematician ” Lejeune Dirichlet”  defined a function.

Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.

Chapter 1 Functions Exercise 1b Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

Exercise 1b

Exercise 1(b) Solutions

I.

1.If f (x) = ex and g(x) = Functions 1(b) images 1, then show that f og = gof and f-1 = g-1

Sol:

Given     f (x) = ex and g(x) =Functions 1(b) images 1

              (fog) (x) = f (g (x))

                                = f ( Functions 1(b) images 1)

                                =Functions 1(b) images 2

                                = x ————- (1)

            (gof) (x) = g (f (x))

                                = g (ex)

                                =Functions 1(b) images 3

                                = xFunctions 1(b) images 4

                                = x ————- (2)

From (1) and (2)  f og = gof

let y = f(x)

     x = f-1(y)

     y = ex

     x = Functions 1(b) images 5

    f-1 (x) = Functions 1(b) images 1

let g(x) = z

         x = g-1(z)

         z =Functions 1(b) images 1

          x = ez

             g-1 (x)  = ex

2. If f (y) =Functions 1(b) images 6 ,  g (y) =Functions 1(b) images 7 then show that fog(y) = y

Sol:

Given f (y) =Functions 1(b) images 6 ,  g (y) = Functions 1(b) images 7        

Now

     fog(y) = f(g(y))

Functions 1(b) images 8

∴ fog(y) = y

3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2x2 + 3 ang g (x) = 3x – 2 then find
       (i)  (fog) (x)    (ii) (gof) (x)       (iii) (fof) (0)      (iv) go(fof)(3)

Sol:

    Given f: R ⟶ R is, g: R ⟶ R is defined by f(x) = 2x2 + 3 ang g (x) = 3x – 2

(i)  (fog) (x) = f(g(x))

                 = f(3x – 2)

                 = 2 (3x – 2)2 + 3

                 = 2 (9x2 – 12x + 4) + 3

                 = 18 x2 – 24x + 8 + 3

                = 18 x2 – 24x + 11

∴ (fog) (x) = 18 x2 – 24x + 11

 

(ii)  (gof) (x) = g (f (x))

                 = g (2x2 + 3)

                 = 3(2x2 + 3) – 2  

                 = 6x2 + 9 – 2

                 = 6 x2 + 7

 ∴ (gof) (x) = 6 x2 + 7

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(iii)  (fof) (0) = f (f (0))

                 = f (2(0)2 + 3)

                = f (2(0) + 3)

                = f (3)

                = 2 (3)2 + 3

                = 2 (9) + 3

                = 18 + 3 = 21

(iv)  go(fof) (3) = go (f (f (3)))

                     = go (f (2 (3)2 + 3))

                     = go (f (21))

                     = g (f (21))

                     = g (2 (21)2 + 3))

                     = g (2 (441) + 3))

                     = g (882 + 3)

                     = g (885)

                     = 3 (885) – 2

                     = 2655 – 2

                     = 2653

∴ go(fof) (3) = 2653

 

Functions Exercise 1b

Ts Inter Maths IA Concept

4. If f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x2 + 1, then find     
     (i) (fof) (x2 + 1)    (ii) (fog) (2)        (iii) (gof) (2a – 3)

Sol:

Given f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x2 + 1

(i) (fof) (x2 + 1) = f (f (x2 + 1))

                          = f (3 (x2 + 1)– 1)

                          = f (3x2 + 3– 1)

                          = f (3x2 + 2)

                          = 3(3x2 + 2) – 1

                          = 9x2 + 6 – 1

                          = 9x2 + 5                

(ii) (fog) (2) = f (g (2))

                  = f (22 + 1)

                  =f (4 + 1)

                  = f (5)

                  = 3(5) – 1

                  = 15 – 1

                  = 14

(iii) (gof) (2a – 3) = g (f (2a – 3))

                            = g (3(2a – 3) – 1)

                            = g (6a – 9 – 1)

                            = g (6a – 10)

                            = (6a – 10)2 + 1

                            = 362 – 120a + 100 + 1

                            = 362 – 120a + 101

 

5. If f(x) = Functions 1(b) images 9 and g(x) =Functions 1(b) images 10 for all x ∈ (0, ∞) then find (gof) (x)

Sol:

Given f(x) = Functions 1(b) images 9 and g(x) = Functions 1(b) images 10 for all x ∈ (0, ∞)

  (gof) (x) = g (f (x))

                    = g (Functions 1(b) images 10 )

                    =Functions 1(b) images 11

               ∴ (gof) (x) =Functions 1(b) images 12

6. If f(x) = 2x – 1 and g(x) =Functions 1(b) images 13 for all x ∈ R then find (gof) (x)

Sol:

   Given f(x) = 2x – 1 and g(x) = Functions 1(b) images 13 for all x ∈ R

  Functions 1(b) images 14           

∴ gof(x) = x

7. If f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)

Sol:

       Given f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R

         (fo(goh)) (x) = fo (g (h (x))

                                   = fo g(2x)

                                   = f (g(2x))

                                   = f((2x)2)

                                   = f(4x2)

                                   = 2                            

       ∴ (fo(goh)) (x) = 2

 

Functions Exercise 1b Solutions

8. Find the inverse of the following functions

(i) a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b, (a ≠ 0)

Given a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b

 Let y = f(x)

         x = f-1(y)

now y = ax + b

          ax = y – b

             x =Functions 1(b) images 15

            f-1(y) =Functions 1(b) images 15

     ∴   f-1(x) =Functions 1(b) images 16

(ii) f: R ⟶ (0, ∞) defined by f(x) = 5x

  Let y = f(x)

         x = f-1(y)

  now y = 5x

          x =Functions 1(b) images 17

             f-1(y) =Functions 1(b) images 17

     ∴   f-1(x) =Functions 1(b) images 18

(iii) f: (0, ∞) ⟶ R defined by f(x) =Functions 1(b) images 19

    Let y = f(x)

         x = f-1(y)

   now y =Functions 1(b) images 19

          x = 2y

             f-1(y) = 2y

     ∴   f-1(x) = 2x

9. If f(x) = 1 + x + x2 + … for Functions 1(b) images 20 , then show that f-1 (x) = Functions 1(b) images 21     

Sol:

Given, f(x) = 1 + x + x2 + … for

        1 + x + x2 + … is an infinite G.P

        a = 1, r = x

  S =Functions 1(b) images 22

         =Functions 1(b) images 23

Now

f (x) =Functions 1(b) images 23

Let y = f(x)

         x = f-1(y)

   now y =Functions 1(b) images 23

           1 – x = Functions 1(b) images 24

   x = 1 –Functions 1(b) images 24

       =Functions 1(b) images 25

             f-1(y) =Functions 1(b) images 25

     ∴   f-1(x) =Functions 1(b) images 26

10 . If f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1) then find f-1 (x)

Sol:

Given f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1)

Let y = f(x)

         x = f-1(y)

   now y = 2x (x – 1)

            Functions 1(b) images 27= x (x – 1)

    x2 – x =  Functions 1(b) images 27

    x2 – x – Functions 1(b) images 27  = 0

Functions 1(b) images 28

Functions 1(b) images 29

 

Functions Exercise 1b Solutions

II. 

1. If f (x) = Functions 1(b) images 30  , x ≠ ± 1, then verify (fof-1) (x) = x

Sol:

   Given f (x) =Functions 1(b) images 30  , x ≠ ± 1

Let y = f(x)

         x = f-1(y)

   now y =Functions 1(b) images 30

            y (x + 1) = x – 1

            xy + y = x – 1

            1 + y = x – xy

            1 + y = x (1 – y)

             x = Functions 1(b) images 31 

             f-1(y) =Functions 1(b) images 31

     ∴   f-1(x) =Functions 1(b) images 32

Now

        Functions 1(b) images 33

2.  If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)-1 = f-1og-1

Sol:

Given A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, p), (β, r), (γ, p)}

A = {1, 2, 3}, B = {α, β, γ}

f: A ⟶ B; and f = {(1, α), (2, γ), (3, β)}

 ⟹ f (1) = α; f (2) = γ; f (3) = β

       ⟹    Every element of set A has a unique image in set B

          f is injection (one – one)

range of f = codomain of f

⟹ f is surjection (on to)

∴ f is bijective

B = {α, β, γ}, C = {p, q, r}

g: B ⟶ C is defined by g = {(α, q), (β, r), (γ, p)}

⟹ g (α) =q; g (β) = r; g (γ) = p

       ⟹    Every element of set B has a unique image in set C

          g is injection (one – one)

range of g = codomain of g

⟹ g is surjection (on to)

∴ g is bijective

Now

f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}

gof = {(1, q), (2, p), (3, r)

(gof)-1 = {(q, 1), (p, 2), (r, 3)} ———– (1)

f-1 = {(α, 1), (γ, 2), (β, 3)

g-1 = {(q, α), (r, β), (p, γ)}

f-1og-1 = {(q, 1), (p, 2), (r, 3)} ———– (2)

From (1) and (2)

(gof)-1 = f-1og-1

3. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x2 + 1          then find
 (i) (gof-1) (2)           (ii) (gof) (x – 1)

 

Sol:

    Given f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x2 + 1

     Let y = f(x)

         x = f-1(y)

   now y = 3x – 2

            3x = y + 2

               x = Functions 1(b) images 34

       f-1(y) =Functions 1(b) images 34

       f-1(x) =Functions 1(b) images 35

Functions 1(b) images 36

∴ (gof-1) (2)   =Functions 1(b) images 37

(ii)  (gof) (x – 1) = g (f (x – 1))

                             = g (3(x – 1) – 2)

                             = g (3x – 3 – 2)

                             = g (3x – 5)

                             = (3x – 5)2 + 1

                             = 9x2 – 30x + 25 + 1

                             = 9x2 – 30x + 26

∴(gof) (x – 1) = 9x2 – 30x + 26

 

4. Let f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)}, then show that (gof)-1 = f-1og-1

 

Sol:

Given f = {(1, a), (2, c), (4, d), (3, b)} and g-1 = {(2, a), (4, b), (1, c), (3, d)}

⟹ f -1 = {(a, 1), (c, 2), (d, 4), (b, 3)} and g = {(a, 2), (b, 4), (c, 1), (d, 3)}

Now gof = {(1, 2), (2, 1), (4, 3), (3, 4)}

          (gof)-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(1)

g-1 = {(2, a), (4, b), (1, c), (3, d)} ; f -1 = {(a, 1), (c, 2), (d, 4), (b, 3)}

f-1og-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(2)

from (1) and (2)

(gof)-1 = f-1og-1      

 

5. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5 then find (fog)-1 (x)

 

Sol:

      Given R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5

       (fog) (x) = f (g(x))

                         = f (x3 + 5)

                          = 2 (x3 + 5) – 3

                         = 2 x3 + 10 – 3 = 2x3 + 7

Let y = (fog) (x)

     ⟹ x = (fog)-1(y)

       y = 2x3 + 7

       2x3 = y – 7

          x3 =Functions 1(b) images 38

          x =Functions 1(b) images 39

(fog)-1(y)=  Functions 1(b) images 39      

(fog)-1(x)=Functions 1(b) images 40        

6. Let f(x) = x2, g(x) = 2x then solve the equation (fog) (x) = (gof) (x)

 

Sol:

  Given f(x) = x2, g(x) = 2x

  (fog) (x) = f(g(x))

                    = f (2x)

                     = (2x)2

                     = 22x

(gof) (x) = g(f(x))

                  = g (x2)

                  =Functions 1(b) images 41

    Now

      (fog) (x) = (gof) (x)

     ⟹ 22x =Functions 1(b) images 41

           2x = x2   [∵ if am = an , then m = n]

           x2 – 2x = 0

         ⟹ x (x – 2) = 0

         ⟹x = 0 or x – 2 = 0

∴ x = 0 or x = 2

7.  If f (x) = , x ≠ ±1 then find (fofof) (x) and (fofofof) (x)

 

Sol:

Given f (x) = Functions 1(b) images 43, x ≠ ±1

Functions 1(b) images 42

Functions 1(b) images 44

 

PDF Files || Inter Maths 1A &1B || (New)

6th maths notes|| TS 6 th class Maths Concept

TS 10th class maths concept (E/M)

Ts Inter Maths IA Concept

TS 10th Class Maths Concept (T/M)

 


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Functions Exercise 1a Solutions

Functions Exercise 1a Solutions

Functions Exercise 1a Solutions

The famous mathematician ” Lejeune Dirichlet”  defined a function.

Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.

Chapter 1 Functions Exercise 1a Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

Exercise 1a

 if ( function_exists( ‘pgntn_display_pagination’ ) ) pgntn_display_pagination( ‘multipage’ );

I.

1. If the function f is defined by

        Functions 1(a) images 1

   then find the values of (i) f (3)    (ii) f (0)      (iii) f (– 1.5)       (iv) f (2) + f (– 2)       (v) f (– 5 )

Sol:

Given Functions 1(a) images 1

      Domain of f(x) is (– 3, ∞)

(i) f (3)

3 lies in the interval x > 1

⟹ f(x) = x + 2

     f(3) = 3 + 2 = 5

     ∴ f (3) = 5

 

(ii) f (0)

0 lies in interval – 1 ≤ x ≤ 1

 ⟹f(x) = 2 

      ∴ f (0) = 2

 

(iii) f (– 1.5)

         – 1.5 lies in interval – 3 < x < – 1 

⟹f(x) = x – 1 

     f (– 1.5) = – 1.5 – 1 = – 2.5

      ∴ f (– 1.5) = – 2.5

 

(iv) f (2) + f (– 2)

        2 lies in the interval x > 1 

⟹ f (x) = x + 2

     f (3) = 2 + 2 = 4

      f (2) = 4

         – 2 lies in interval – 3 < x < – 1 

⟹f(x) = x – 1 

     f (– 2) = – 2 – 1 = – 3

     f (– 2) = – 2 – 1 = – 3

now f (2) + f (– 2) = 4 – 3 = 1

          ∴ f (2) + f (– 2) = 1

(v) f (– 5)

since domain of f(x) is (– 3, ∞)

f (– 5) is not defined

2. If f: R – {0} ⟶ R is defined by f(x) = Functions 1(a) images 2, then show that f (x) + f (1/x) = 0

Sol:

Given f: R – {0} ⟶ R is defined by f(x) = Functions 1(a) images 2

       f (1/x)  = Functions 1(a) images 3

Now

f (x) + f (1/x)  = Functions 1(a) images 4

∴ f (x) + f (1/x)  = 0

3. If f: R ⟶ R is defined by f(x) = Functions 1(a) images 5, then show that f (tan θ) = cos 2θ

Sol:

    Given f: R ⟶ R is defined by f(x) =Functions 1(a) images 5

       f (tan θ) =Functions 1(a) images 6

                    = cos 2θ   Functions 1(a) images 7

 

4. If f: R – {±1} ⟶ R is defined by f(x) = Functions 1(a) images 8 , then show that f Functions 1(a) images 9 = 2 f (x)

Sol:

Given f: R – {±1} ⟶ R is defined by f(x) =

            Functions 1(a) images 10

5. If A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection (onto function) defined by f(x) = x2 + x + 1, then find B

Sol:

Given A = {– 2, – 1, 0, 1, 2} and   f: A ⟶ B is a surjection defined by f(x) = x2 + x + 1

f(– 2) = (–2)2 + (–2) + 1

            = 4 – 2 + 1 = 3

f(– 1) = (–1)2 + (–1) + 1

            = 1 – 1 + 1 = 1

f(0) = (0)2 + (0) + 1

            = 0 + 0 + 1 = 1

f(1) = (1)2 + (1) + 1

            =1 +1 + 1 = 3

f( 2) = (2)2 + (2) + 1

            = 4 + 2 + 1 = 4

∴ B = {1, 3, 7}

TS 10th class maths concept (E/M)

Functions Exercise 1a

 

 

 

6. If A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) = Functions 1(a) images 11, then find range of f.

Sol:

Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) =Functions 1(a) images 11

    Functions 1(a) images 12

7. If f (x + y) = f (xy) ∀ x, y ∈ R, then prove that f is a constant function

Sol:

       Given f (x + y) = f (xy) ∀ x, y ∈ R

        Let x = 0 and y = 0

         f (0 + 0) = f (0 × 0) = f (0)

         f (1) = f (0 + 1)

                  = f (0 × 1)

                   = f (0)

         f (2) = f (1 + 1)

                  = f (1 × 1)

                 = f (1)

                 = f (0)

       f (3) = f (1 + 2)

                = f (1 × 2)

                = f (2)

                = f(0)

Similarly, f(4) = 0

                    f(5) = 0

and so on.

∴ f is a constant function

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6th maths notes|| TS 6 th class Maths Concept

II.

1. If A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2, g(x) = x3, which of the following are surjections?

      (i) f: A ⟶ A                 (ii) g: A ⟶ A

Sol:

(i) Given A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2

A = {– 1, 0, 1}; f: A ⟶ A                

f (x) = x2

f (– 1) = (– 1)2 = 1

f (0) = (0)2 = 0

f (1) = (1)2 = 1

∵ range is not equal to co domain of f

f is nor a surjection

(ii) A = {x/ – 1 ≤ x ≤ 1}, g (x) = x3

A = {– 1, 0, 1}; g: A ⟶ A

g (x) = x3

g (– 1) = (– 1)3 = – 1

g (0) = (0)3 = 0

g (1) = (1)3 = 1

∵ range is equal to co domain of f

f is a surjection      

2. Which if the following are injection, surjection or bijection? Justify your answer

(i) f: R ⟶ R defined by f(x) =Functions 1(a) images 13

      let x1, x2 ∈ R

     f(x1) = f(x2)

Functions 1(a) images 14

      2x1 + 1 = 2x2 + 1

      2x1 = 2x2

       x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y =Functions 1(a) images 15

3y = 2x + 1

3y – 1 = 2x

⟹ x = Functions 1(a) images 16 ∈ R

Now

         Functions 1(a) images 17           

∴ f is surjection

f is injection and surjection

∴ f is a bijection

 

(ii) f: R ⟶ (0, ∞) defined by f(x) = 2x

let x1, x2 ∈ R

     f(x1) = f(x2)

       Functions 1(a) images 27            

       x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y = 2x

 x = Functions 1(a) images 18  ∈ (0, ∞)

 

Now f(x) = 2x

                  = Functions 1(a) images 19

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(iii) f: (0, ∞) ⟶ R defined by f(x) =Functions 1(a) images 21

  let x1, x2 ∈ (0, ∞)

     f(x1) = f(x2)

          Functions 1(a) images 20

       x1 = x2

x1, x2 ∈ (0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y =Functions 1(b) images 1

 x = ey ∈ (0, ∞)

 Now f(x) =Functions 1(a) images 21

                  = Functions 1(a) images 22

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(iv) f: [0, ∞) ⟶ [0, ∞) defined by f(x) = x2

  let x1, x2 ∈ [0, ∞)

     f(x1) = f(x2)

     Functions 1(a) images 28           

       x1 = x2 [∵ x1, x2 ∈ [0, ∞)]

 x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

       y = x2

 x = Functions 1(a) images 23 ∈ [0, ∞)

 Now f(x) = x2

                   =Functions 1(a) images 24

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(v) f: R ⟶ [0, ∞) defined by f(x) = x2

  let x1, x2 ∈ R

     f(x1) = f(x2)

        Functions 1(a) images 28 - 1        

       x1 = ± x2 [∵ x1, x2 ∈ R]

 x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

Let y = f(x)

       y = x2

 x = Functions 1(a) images 23∈ R

 Now f(x) = x2

                   =Functions 1(a) images 24

                  = y

∴ f is surjection

f is not an injection but surjection

∴ f is not a bijection

(vi) f: R ⟶ R defined by f(x) = x2

  let x1, x2 ∈ R

     f(x1) = f(x2)                

       x1 = ± x2 [∵ x1, x2 ∈ R]

 x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

 f (1) = 12 = 1

 f (– 1) = (–1)2 = 1

here ‘– 1’ has no pre image 

∴ f is not a surjection

f is not an injection and not surjection

∴ f is not a bijection

 

Ts Inter Maths IA Concept

hai

 

3. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}? If this is given by the formula g(x) = ax + b, then find a and b

Sol:

Given A = {1, 2, 3, 4}, B = {1, 3, 5, 7} and g = {(1, 1), (2, 3), (3, 5), (4, 7)}

g (1) = 1; g(2) = 3; g(3) = 5 ; g(4) = 7

here, every element of set A has a unique image in set B

  ∴ g: A ⟶ B is a function

And also given g(x) = ax + b

g (1) = 1

⟹ a (1) + b = 1

       a + b = 1

       b = 1 – a _______________   (1)

g (2) = 3

      a (2) + b = 3

   2a + b = 3

    2a + 1 – a = 3 (from (1))

     a+ 1 = 3

      a = 2

     b = 1 – 2

     b = – 1

∴ a = 2, b = – 1

4. If the function f: R ⟶ R defined by f(x) = Functions 1(a) images 29 , then show that f (x + y) + f (x – y) = 2 f (x) f (y).

Sol:

 Given the function f: R ⟶ R defined by f(x) =Functions 1(a) images 29

Functions 1(a) images 30

                                                       = 2 f (x) f (y)

5. If the function f: R ⟶ R defined by f(x) = Functions 1(a) images 31 , then show that f (1 – x) = 1 – f (x)

and hence reduce the value of Functions 1(a) images 32

Sol:

Given the function f: R ⟶ R defined by f(x) =Functions 1(a) images 31

Functions 1(a) images 33 

∴ f (1 – x) = 1 – f(x)

 

TS 10th Class Maths Concept (T/M)
TS 10th class maths concept (E/M)
 

6. If the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection, then find a and b

Sol:

Given the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection

Case(i)

         Functions 1(a) images 34 

  If f(– 1) = 0 and f(1) = 2

            a (– 1) + b = 0

           – a + b = 0

              b = a ————(1)

       and

        a (1) + b = 2

         a + b = 2

        a + a = 2         [ from (1)]

         2a = 2

           a = 1

            b = 1

Case (ii)

        Functions 1(a) images 35

  If f(– 1) = 2 and f(1) = 0

            a (– 1) + b = 2

           – a + b =  

              b = 2 + a ————(2)

       and

        a (1) + b = 0

         a + b = 0

        a + 2 + a = 0         [ from (2)]

         2a + 2 = 0

           2a = – 2

            a = – 1

            b = 1

From Case(i) and  Case (ii) a = ±1 and b = 1

7. If f(x) = cos (log x), then show that Functions 1(a) images 36= 0

Sol

Given f(x) = cos (log x)

Functions 1(a) images 37

Functions 1(a) images 38

Functions 1(a) images 36 =

                             cos (log x) cos (log y) – Functions 1(a) images 39 [cos (log x) cos (log y) – sin (log x) sin (log x)

                         + cos (log x) cos (log y) + sin (log x) sin (log x)]

                          = cos (log x) cos (log y) – Functions 1(a) images 39 [2cos (log x) cos (log y)]

                          = cos (log x) cos (log y) – cos (log x) cos (log y)

          ∴ Functions 1(a) images 36   = 0


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Ts Polyset 2020 feature image

Maths TS Polycet || Solved Previous Question Papers 2020

TS Polycet || Solved Previous Question Papers 2020 Math

BIM feature image Maths
The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology.

The Main Subjects in this Exam are Maths, Physics, Chemistry and Biology. Here we are providing Previous Maths  Papers Questions and Solutions.

The syllabus of  Maths, Physics, Chemistry and Biology for POLYCET-2022 is the same as that of SSC Examination conducted by the Board of Secondary Education, Telangana.

TS Polycet || Solved Previous Question Papers 2021 Mathematics gives an idea becuase,  it is very helpful to solve the problems in POLYCET entence  examination

 

 

TS VI – IX Maths Concepts


Chapter 1: Real Numbers

1.If 7 divides a2 then

a2  ను 7 భాగించినచో

   (1)  7 divides a (a ను 7 భాగిస్తుంది)

   (2) 7 divides TS POLYCET 2020 - Real Numbers 1 ( TS POLYCET 2020 - Real Numbers 1ను 7 భాగిస్తుంది)

  (3) a divide 7 (7 ను a భాగిస్తుంది)

  (4) none (ఏదీ కాదు)  

Answer:  (1)     

2. In the formula TS POLYCET 2020 - Real Numbers 2, which of the following is true?

    TS POLYCET 2020 - Real Numbers 2అయిన, ఈ క్రింది వాటిలో ఏది సత్యము.

    (1)  x > 0, y > 0, a = 1       

    (2) x < 0, y < 0, a = 1  

    (3)  a > 0, y > 0, x = 1       

    (4) x > 0, y > 0, a ≠ 1                                    

Answer:   (4)      

                 

3. 5 =____________
  TS POLYCET 2020 - Real Numbers 3

Answer:   (3)             

                                      

4. TS POLYCET 2020 - Real Numbers 5 then x =
      TS POLYCET 2020 - Real Numbers 5అయిన,  x =
     (1) n                   (2) 1                          
     (3)  5                  (4) 2    

Answer:    (4)        

             

5. TS POLYCET 2020 - Real Numbers 10 , ac = ___
   TS POLYCET 2020 - Real Numbers 10అయిన, ac = ___
   (1)a2         (2)b2                                 
   (3) c2       (4) None(ఏది కాదు)

Answer:    (2)         

         

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