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.

“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=

Let p(n) be the given statement that

 1² + 2² + 3² + …… + n²=

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 = 

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=

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

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

                                  a = 2, d = 1

                                 a_n = 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

                                 a_n = 3 + (n – 1) d

                                       = 3 + (n – 1) (1)

                                       = 3 + n – 1

                                       = n + 2

n^th 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) =

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) =  ………… (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)  

  =  + (k + 2) (k + 3)

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 =

Visit my Youtube Channel: Click on Below Logo

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) =

given function is f (x) =

 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) =

 Given function is f (x) =

 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) =

Given function is f (x) =

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) =

Given function is f (x) =

f (x) is defined when x ∈ R      

∴ Domain of f(x) is R

Functions Exercise 1c

(v)

f (x) =

Given function is f (x) =

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) =

 Given function is f (x) =

 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) =

 Given function is f (x) =

  f (x) is defined when x + 1≠ 0   

  ⟹ x ≠ – 1  

  ∴ Domain of f(x) is R – {– 1}

(viii)

f(x) =

  Given function is f (x) =

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

  ⟹ (x – 5) (x + 5) ≥ 0  

  ⟹ x ∈ (–∞, –5] ∪ [5, ∞)

  ⟹ x ∈ R – (– 5, 5)

 ∴ Domain of f(x) is R – (– 5, 5)

Functions Exercise 1c

(ix)

f(x) =

Given function is f (x) =

f (x) is defined when x – [x] ≥ 0   

  ⟹ x ≥ [x]

 ⟹ x ∈ R   

∴ Domain of f(x) is R  

(x)

f(x) =

Given function is f (x) =

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) =

Given function is f (x) =

Let y =

  ⟹  |4 – x²| = e^y

∵ e^y > 0 ∀ y ∈ R

∴ Range of f(x) is R

(ii) 

f(x) =

Given function is f (x) =

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) =

Given function is f (x) =

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) =

           Given function is f (x) =

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

          ⟹ x ≠ 2

         Domain of f(x) is R – {2}

         Let y =

            =

             = x + 2

       If x = 2 ⟹ y = 2 + 2 = 4

        ∴ Range of f(x) is R – {4}

Functions Exercise 1c

(v) 

f (x) =

let y =

      y² = 9 + x²

          x² = y² – 9

       x =

     it is defined when y² – 9 ≥ 0

      ⟹ (y – 3) (y + 3) ≥ 0

    y ∈ (– ∞, – 3] ∪ [3, ∞)

but y = ≥ 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) 

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

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) 

II. 

1.Find the domain of the following real valued functions

(i)

f (x) =

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) =

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) =

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) =

f(x) is defined when ≥ 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) =

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 is an even function

Sol:

Given f(x) =

3. Find the domain and range of the following functions

(i)

f (x) =

Given f(x) =

 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) =

Given f (x) =

 It is defined when 2 – 3x ≠ 0

⟹ 2 ≠ 3x 

⟹ x ≠ 2/3

∴ Domain of f (x) = R – {2/3} 

Let y = f(x)

        y =

  ⟹ y (2 – 3x) = x

          2y – 3xy = x

          2y = x + 3xy  

           2y = x (1 + 3y)

   ⟹ x =

    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

            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, ∞)

Visit my Youtube Channel: Click on Below Logo

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) = e^x and g(x) = , then show that f og = gof and f^-1 = g^-1

Sol:

Given     f (x) = e^x and g(x) =

              (fog) (x) = f (g (x))

                                = f ( )

                                =

                                = x ————- (1)

            (gof) (x) = g (f (x))

                                = g (e^x)

                                =

                                = x

                                = x ————- (2)

From (1) and (2)  f og = gof

let y = f(x)

     x = f^-1(y)

     y = e^x

     x =

    f^-1 (x) =

let g(x) = z

         x = g^-1(z)

         z =

          x = e^z

             g^-1 (x)  = e^x

2. If f (y) = ,  g (y) = then show that fog(y) = y

Sol:

Given f (y) = ,  g (y) =         

Now

     fog(y) = f(g(y))

∴ 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

(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) = and g(x) = for all x ∈ (0, ∞) then find (gof) (x)

Sol:

Given f(x) =  and g(x) =  for all x ∈ (0, ∞)

  (gof) (x) = g (f (x))

                    = g ( )

                    =

               ∴ (gof) (x) =

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

Sol:

   Given f(x) = 2x – 1 and g(x) =  for all x ∈ R

∴ 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 =

            f^-1(y) =

     ∴   f^-1(x) =

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

  Let y = f(x)

         x = f^-1(y)

  now y = 5^x

          x =

             f^-1(y) =

     ∴   f^-1(x) =

(iii) f: (0, ∞) ⟶ R defined by f(x) =

    Let y = f(x)

         x = f^-1(y)

   now y =

          x = 2^y

             f^-1(y) = 2^y

     ∴   f^-1(x) = 2^x

9. If f(x) = 1 + x + x² + … for , then show that f^-1 (x) =      

Sol:

Given, f(x) = 1 + x + x² + … for

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

        a = 1, r = x

  S_∞ =

         =

Now

f (x) =

Let y = f(x)

         x = f^-1(y)

   now y =

           1 – x =

   x = 1 –

       =

             f^-1(y) =

     ∴   f^-1(x) =

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

Sol:

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

Let y = f(x)

         x = f^-1(y)

   now y = 2^{x (x – 1)}

            = x (x – 1)

    x² – x = 

    x² – x –   = 0

Functions Exercise 1b Solutions

II. 

1. If f (x) =   , x ≠ ± 1, then verify (fof^-1) (x) = x

Sol:

   Given f (x) =  , x ≠ ± 1

Let y = f(x)

         x = f^-1(y)

   now y =

            y (x + 1) = x – 1

            xy + y = x – 1

            1 + y = x – xy

            1 + y = x (1 – y)

             x =  

             f^-1(y) =

     ∴   f^-1(x) =

Now

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 =

       f^-1(y) =

       f^-1(x) =

∴ (gof^-1) (2)   =

(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 =

          x =

(fog)^-1(y)=        

(fog)^-1(x)=        

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

Sol:

  Given f(x) = x², g(x) = 2^x

  (fog) (x) = f(g(x))

                    = f (2^x)

                     = (2^x)²

                     = 2^2x

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

                  = g (x²)

                  =

    Now

      (fog) (x) = (gof) (x)

     ⟹ 2^2x =

           2x = x²   [∵ if a^m = 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) = , x ≠ ±1

Visit my Youtube Channel: Click on Below Logo

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

   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

      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) = , then show that f (x) + f (1/x) = 0

Sol:

Given f: R – {0} ⟶ R is defined by f(x) =

       f (1/x)  =

Now

f (x) + f (1/x)  =

∴ f (x) + f (1/x)  = 0

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

Sol:

    Given f: R ⟶ R is defined by f(x) =

       f (tan θ) =

                    = cos 2θ  

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

Sol:

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

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) = , then find range of f.

Sol:

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

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

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

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) =

      let x₁, x₂ ∈ R

     f(x₁) = f(x₂)

      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 =

3y = 2x + 1

3y – 1 = 2x

⟹ x =  ∈ R

Now

∴ f is surjection

f is injection and surjection

∴ f is a bijection

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

let x₁, x₂ ∈ R

     f(x₁) = f(x₂)

       x₁ = x₂

x₁, x₂ ∈ R, f(x₁) = f(x₂) ⟹ x₁ = x₂

∴ f is an injection

Let y = f(x)

       y = 2^x

 x =   ∈ (0, ∞)

Now f(x) = 2^x

                  =

                  = y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

(iii) f: (0, ∞) ⟶ R defined by f(x) =

  let x₁, x₂ ∈ (0, ∞)

     f(x₁) = f(x₂)

       x₁ = x₂

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

∴ f is an injection

Let y = f(x)

       y =

 x = e^y ∈ (0, ∞)

 Now f(x) =

                  =

                  = 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)

       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 =  ∈ [0, ∞)

 Now f(x) = x²

                   =

                  = 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)

       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 = ∈ R

 Now f(x) = x²

                   =

                  = 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) = , 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) =

                                                       = 2 f (x) f (y)

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

and hence reduce the value of

Sol:

Given the function f: R ⟶ R defined by f(x) =

∴ 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)

  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)

  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 = 0

Sol

Given f(x) = cos (log x)

 =

                             cos (log x) cos (log y) – [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) – [2cos (log x) cos (log y)]

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

          ∴    = 0

Visit my Youtube Channel: Click on Below Logo

TS Polycet || Solved Previous Question Papers 2020 Math

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 ( ను 7 భాగిస్తుంది)

  (3) a divide 7 (7 ను a భాగిస్తుంది)

  (4) none (ఏదీ కాదు)  

Answer:  (1)     

2. In the formula , which of the following is true?

    అయిన, 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 =____________

Answer:   (3)             

4. then x =

      అయిన,  x =

     (1) n                   (2) 1                          

     (3)  5                  (4) 2    

Answer:    (4)        

5. , ac = ___

   అయిన, ac = ___

   (1)a²         (2)b²                                 

   (3) c²       (4) None(ఏది కాదు)

Answer:    (2)         

TS 10th class maths concept (E/M)

TS 6th Class Maths Concept

TS 10th Class Maths Concept (T/M)