important questions

ts inter trigonometric equations 4 marks important questions 2024

TS Inter Trigonometric Equations – 4-M important Questions

TS Inter Trigonometric Equations – 4-Marks important Questions

 

“Trigonometric Equations 4-Mark Important Questions” would typically refer to a collection of questions worth four marks each that focus on solving equations involving trigonometric functions. These questions are likely intended for students studying trigonometry at the intermediate level or equivalent.

multiplication of vectors 4m imp qs link

Trigonometric equations involve expressions containing trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. The goal in solving these equations is typically to find the values of the variable(s) that satisfy the given equation within a specified interval.

inter maths 1a trigonometric equations - 1-1

 

These types of questions may cover various topics within trigonometric equations, including:

Solving basic trigonometric equations: These equations involve single trigonometric functions and can often be solved using algebraic techniques such as factoring, substitution, or trigonometric identities.

Solving trigonometric equations involving multiple angles: Equations may involve multiple angles, such as sums, differences, or multiples of trigonometric functions. Students may need to apply trigonometric identities or properties to simplify the equations before solving them.

addition of vectors 4m imp qs link

 

inter maths 1a trigonometric equations - 2

Solving trigonometric equations involving trigonometric identities:

Students may encounter equations where trigonometric identities need to be applied to rewrite the equation in a more simplified form before solving it.

ts inter trigonometric equations 4 marks important questions 2024

Solving trigonometric equations with restrictions: Some equations may have restrictions on the domain, such as finding solutions within a specific interval or range of values.

Solving trigonometric equations involving transformations: Equations may involve transformations of trigonometric functions, such as amplitude changes, phase shifts, or vertical and horizontal translations.

inter maths 1a trigonometric equations - 3

matrices 4m imp qs link

 

These important questions serve as a means for students to practice and demonstrate their understanding of trigonometric equations, their ability to apply various problem-solving techniques, and their proficiency in manipulating trigonometric functions to find solutions. 

 

Additionally, these questions may also help students prepare for assessments or examinations where solving trigonometric equations is a key component of the curriculum.


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Functions vsaqs questions and solutions - 1

Functions (2M Questions &Solutions)|| V.S.A.Q.’S||

Functions (2M Questions &Solutions)|| V.S.A.Q.’S||

Functions (2M Questions &Solutions): This note is designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  


Functions

QUESTION 1

Find the Domain of the following real-valued functions.

(i) f(x) =Functions VSAQS 12 

Sol:   given f(x) = Functions VSAQS 12

         It is defined when 6x – x2 – 5 ≠ 0

              ⇒ x2 – 6x + 5 ≠ 0

                  x2 – 5x – x + 5 ≠ 0 

                  x (x – 5) –1(x – 5) ≠ 0

                     (x – 5) (x – 1) ≠ 0

                      x ≠ 5 or x ≠ 1

                      ∴ domain = R – {1, 5}

(ii) f(x) = Functions VSAQS 13

Sol: Given f(x) =Functions VSAQS 13

  It is defined when 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0

 ⇒ x ≥ –3, x ≤ 3 and x ≠ 0

       ⇒   –3≤ x, x ≤ 3 and x ≠ 0

       ⇒   –3≤ x ≤ 3 and x ≠ 0

              x ∈ [–3, 3] – {0}

        ∴ domain = [–3, 3] – {0}

(iii) f(x) = Functions VSAQS 14 

Sol: Given f(x) =Functions VSAQS 14

 It is defined when x + 2 ≥ 0, 1 – x > 0 and 1 – x ≠ 0

 ⇒ x ≥ –2, x < 1 and x ≠ 0 

  x ∈ [–2, ∞) ∩ (– ∞, 1) – {0}

 ⇒ x ∈ [–2, 1) – {0}

    ∴ domain = [–2, 1) – {0}

(iv) f(x) =Functions VSAQS 15

         Given f(x) =Functions VSAQS 15

         It is defined when 4x – x2 ≥ 0

               ⇒ x2 – 4x ≤ 0

                   x (x – 4) ≤ 0

                  (x – 0) (x – 4) ≤ 0

                  x ∈ [0, 4]

               ∴ domain = [0, 4]

(v) f(x) = log (x2 – 4x + 3)

Given f(x) = log (x2 – 4x + 3)

 It is defined when x2 – 4x + 3 > 0

      ⇒   x2 – 3x – x + 3 > 0

             x (x – 3) –1(x – 3) > 0

            (x – 3) (x – 1) > 0

          x ∈ (–∞, 1) ∪ (3, ∞)

          x ∈ R – [1, 3]

         ∴ domain = R – [1, 3]

(vi) f(x) = Functions VSAQS 16

    Given f(x) =Functions VSAQS 16 It is defined when x2 – 1 ≥ 0 and x2 – 3x + 2 > 0  

(x + 1)(x – 1) ≥ 0 and x2 – 2x – x + 2 > 0  

(x + 1) (x – 1) ≥ 0 and x (x – 2) (x – 1) > 0 

x∈ (–∞, –1) ∪ (1, ∞) and x ∈ (–∞, –1) ∪ (2, ∞)

            ∴ domain = R – (–1, 2]

(vii)  f(x) = Functions VSAQS 17

Given f(x) = Functions VSAQS 17

It is defined when Functions VSAQS 18 – x > 0

                                 ⇒Functions VSAQS 18 > x

                                 ⇒ x < 0

 ∴ domain = (–∞, 0)  

(viii) f(x) =Functions VSAQS 19

           Given f(x) =Functions VSAQS 19

        It is defined when Functions VSAQS 18 +x ≠0

                                 ⇒ Functions VSAQS 18 ≠ – x

                                   ⇒ x > 0

         ∴ domain = (0, ∞)          

QUESTION 2

If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x2 + 2 then find (i) (gof) (x) (ii) (gof) (Functions VSAQS 1)  (iii) (fof) (x)  (iv) go(fof) (0).

Sol: Given f(x) = 4x – 1, g(x) = x2 + 2

(i)  (gof) (x) = g (f (x))

= g (4x – 1)

= (4x – 1)2 + 2

= 16x2 – 8x + 1 + 2

= 16x2 – 8x + 3

(ii) (gof)  (Functions VSAQS 1)= (g (f (Functions VSAQS 1))

=  g(4(Functions VSAQS 1) – 1)

= g (a + 1 – 1)

= g(a)

= a2 + 2

(iii) (fof) (x) = f (f (x))

= f (4x – 1)

= 4 (4x – 1) – 1

= 16x – 4 – 1

= 16x – 5

(iv) go(fof) (0) = g(fof) (0)

= g (f (f (0)))

= g (f (– 1))

= g (– 4 – 1)

= g (– 5)

= (– 5)2 + 2

= 25 + 2 = 27

QUESTION 3

If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x2, then find (i) (3f – 2g) (x) (ii) (fg)(x) (iii) Functions VSAQS 3 (x) (iv) (f + g+ 2) (x)

Sol: Given f(x) = 2x – 1 and g(x) = x2

(i) (3f – 2g) (x) = 3f(x) – 2 g(x)

= 3(2x – 1) – 2(x2)

= 6x – 3 – 2x2

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

= (2x – 1) (x2)

= 2x3 – 3x2

(iii)  Functions VSAQS 2

(iv) (f + g+ 2) (x) = f(x) + g(x) + 2

= 2x – 1 + x2 + 2

= x2 + 2x  + 1

QUESTION 4

If f = {(4, 5), (5, 6), (6, – 4)} g = {(4, – 4), (6, 5), (8,5)}, then find (i) f + g (ii) f – g (iii) 2f +4g (iv) f + 4 (v) fg (vi) f/g (vii)Functions VSAQS 4 (viii)Functions VSAQS 5   (ix) f2 (x) f3

Sol: Given f = {(4, 5), (5, 6), (6, – 4)}

  g = {(4, – 4), (6, 5), (8,5)} 

The domain of f ∩ The Domain of g = {4, 6}

(i) (f + g) (4) = f (4) + g (4)

                     = 5 – 4 = 1

     (f + g) (6) = f (6) + g (6)

                     =– 4 + 5 = 1

           ∴ f + g = {(4, 1), (6, 1)}                    

(ii) (f – g) (4) = f (4) – g (4)

                   = 5 – (– 4) = 5 + 4 = 9

(f – g) (6) = f (6) – g (6)

                   = – 4– 5   = – 9

         ∴ f – g   = {(4, 9), (6, – 9}

(iii) (2f +4g) (4) = 2 f (4) + 2 g (4)

                         = 2(5) + 4 (– 4)

                         = 10 – 16

                          =– 6

    (2f +4g) (6) = 2 f (6) + 2 g (6)

                         = 2(– 4) + 4 (5)

                         = – 8 + 20

                          =12

     ∴ (2f +4g) = {(4, – 6), (6, 12)}

(iv) (f + 4) (4) = f (4) + 4 = 5 + 4 = 9

       (f + 4) (5) = f (5) + 4 = 6 + 4 = 10

      (f + 4) (6) = f (6) + 4 = – 4 + 4 = 0

      ∴ (f + 4) = {(4, 9), (5, 10), (6, 0)}

(v) fg (4) = f (4) g (4) = (5) (– 4) =– 20

      fg (6) = f (6) g (6) = (– 4) (5) =– 20

      ∴ fg = {(4, – 20), (6, – 20)}

(vi) f/g (4) = f(4)/g(4) = 5/ – 4 = – 5/4

       f/g (6) = f(6)/g(6) = – 4/ 5

     ∴ f/g = {(4, – 5/4), (6, – 4/ 5)}

Functions VSAQS 6

Functions VSAQS 7

(ix) f2(4) = (f (4))2 = (5)2 = 25

       f2(5) = (f (5))2 = (6)2 = 36  

       f2(6) = (f (6))2 = (– 4)2 = 16

     ∴ f2 = {(4, 25), (5, 36), (6, 16)}

(x) f3(4) = (f (4))3 = (5)3 = 125

     f3(5) = (f (5))3 = (6)3 = 216  

     f3(6) = (f (6))3 = (– 4)3 = –64

   ∴ f3 = {(4, 125), (5, 216), (6, –64)}

QUESTION 5

If A = {0, π/6, π/4, π/3, π/2} and f: A→ B is a surjection defined by f(x) = cos x, then find B.

 Sol: Given A = {0, π/6, π/4, π/3, π/2}

            f(x) = cos x

          f (0) = cos (0) = 1

          f(π/6) = cos (π/6) =Functions VSAQS 8

          f(π/4) = cos (π/4) =Functions VSAQS 9

          f(π/3) = cos (π/3) = 1/2

          f(π/2) = cos (π/2) = 0

∴ B = {1,Functions VSAQS 8 ,Functions VSAQS 9 , ½, 0}

QUESTION 6

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

Sol: Given A = {–2, –1, 0, 1, 2} and 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 = 7

        ∴ B = {1, 3, 7}

QUESTION 7

If A = {1, 2, 3, 4} and f: A→ B is a surjection defined by f(x) =Functions VSAQS 10 then find B.

 Sol: Given A = {1, 2, 3, 4} and f(x) =Functions VSAQS 10

    Functions VSAQS 11

QUESTION 8

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

Sol: Given f(x) = 2, g(x) = x2, h(x) = 2x

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

                          = f (g (h (x))

                           = f(g(2x)

                           = f((2x)2)

                           = f(4x2) = 2

QUESTION 9

If f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x2 +1 then find (i) (gof-1) (2) (ii) (gof) (x – 1) 

Sol: Given f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x2 +1

(i) Let y = f(x) ⟹ x = f-1(y)

      y = 3x – 2

     y + 2 = 3x

    x = (y + 2)/3

f-1(y) = (y + 2)/3

∴ f-1(x) = (x + 2)/3

  Now 

(gof-1) (2) = g(f-1(2))

                = g ((2 + 2)/3)

               = g (4/3)

              = (4/3)2 + 1 = 16/9 + 1 = 25/9

(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

QUESTION 10

If f: N→ N defined by f (x) = 2x + 5, Is onto? Explain with reason.

Sol: Given f (x) = 2x + 5

 Let y = f(x) ⟹ x = f-1(y)

y = 2x + 5

2x = y – 5

 x = (y – 5)/2 ∉ N

∴ f(x) is not onto 

QUESTION 11

Find the inverse of the following functions

(i) If a, b ∈ R, f: R→ R defined by f(x) = ax + b

    Given function is f(x) = ax + b

   Let y = f(x) ⟹ x = f-1(y)

      y = ax + b

     y – b = ax

    x = (y – b)/a

f-1(y) = (y – b)/a

∴ f-1(x) = (x – b)/a

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

    Given function is f(x) = 5x

   Let y = f(x) ⟹ x = f-1(y)

        y = 5x

        x =Functions VSAQS 20

   f-1(y) =Functions VSAQS 20

    ∴ f-1(x) =Functions VSAQS 21

(iii) f: (0, ∞) → R defined by f(x) = Functions VSAQS 22

    Given function is f(x) =Functions VSAQS 22

     Let y = f(x) ⟹ x = f-1(y)

    y =Functions VSAQS 22

   x = 2y

  f-1(y) = 2y

  ∴ f-1(x) = 2x

(iv) f: R→ R defined by f(x) = e4x + 7

     Given function is f(x) = e4x + 7

   Let y = f(x) ⟹ x = f-1(y)

  y = e4x + 7

  4x + 7 =Functions VSAQS 23

  4x = Functions VSAQS 23  – 7

  x = ( Functions VSAQS 23 – 7)/4

  f-1(y) = (Functions VSAQS 23  – 7)/4

 ∴ f-1(x) = ( Functions VSAQS 24 – 7)/4

(v) f: R→ R defined by f(x) = (2x + 1)/3

   Given function is f(x) = (2x + 1)/3

  Let y = f(x) ⟹ x = f-1(y)

 y= (2x + 1)/3

 3y = 2x + 1

 2x = 3y – 1

  x = (3y – 1)/2

 f-1(y) = (3y – 1)/2

 ∴ f-1(x) = (3x – 1)/2

QUESTION 12

If f: R→ R defined by f(x) =Functions VSAQS 25 , then show that f (tan θ) = cos 2θ

Sol: Given function is f(x) =Functions VSAQS 25

        f (tan θ) =Functions VSAQS 26

   ∴  f (tan θ) = cos 2θ                           

QUESTION 13  

If f: R – {±1} → R defined by f(x) =Functions VSAQS 27 , then show that Functions VSAQS 28= 2f (x)

Sol: Given f(x) =Functions VSAQS 27

         Functions VSAQS 28 =Functions VSAQS 29

                          = Functions VSAQS 30

                          = Functions VSAQS 31

= Functions VSAQS 32

                          =Functions VSAQS 33

                           = 2Functions VSAQS 27

    ∴ Functions VSAQS 28= 2f (x)

QUESTION 14

If the function f: R→ R defined by f(x) =Functions VSAQS 35 , then show that f (x + y) + f (x – y) = 2 f(x) f(y).

Sol: Given function is f(x) =Functions VSAQS 35

f (x + y) + f (x – y) =  Functions VSAQS 36

 = Functions VSAQS 37 

=Functions VSAQS 38

=Functions VSAQS 39 

=Functions VSAQS 40

=2 Functions VSAQS 41

∴ f (x + y) + f (x – y) = 2 f(x) f(y)

QUESTION 15

If f(x) = cos (log x), then show thatFunctions VSAQS 42 = 0

Sol: Given function is f(x) = cos (log x)

Functions VSAQS 43

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

= 0

Hence proved

QUESTION 16

Find the range of the following real-valued functions

Sol:

(i) f(x) =Functions VSAQS 44

     Given function is f(x) =Functions VSAQS 44

    Let y = Functions VSAQS 44

    ⟹ ey = Functions VSAQS 45

       ey > 0

∴ The range of f(x) is R

(ii) f(x) =Functions VSAQS 46

     It is defined when x – 2 ≠ 0

     ⟹ x ≠ 2

    Domain = R – {2}                 

    let y = Functions VSAQS 46

          = Functions VSAQS 52

         y = x + 2

  if x = 2 ⟹ y = 4

 ∴ Range of f(x) is R – {4}

QUESTION 17

Find the domain and range of the following real-valued functions

Sol:

(i) f(x) =Functions VSAQS 47

It is defined when 1 + x2 ≠ 0

  ⟹ x2 ≠ – 1

    x ∈ R

   ∴ domain of f(x) is R

  Let y = Functions VSAQS 47

      y (1 + x2) = x

     y + x2y = x

     x2 y – x + y = 0

   x = Functions VSAQS 48

It is defined when 1 – 4y2 ≥ 0 and 2y ≠ 0

    ⟹ 4y2 – 1 ≤ 0 and y ≠ 0     

        (2y – 1) (2y + 1) ≤ 0 and y ≠ 0 

        (y – 1/2) (y + 1/2) ≤ 0 and y ≠ 0 

  –  ½≤ y ≤ ½ and y ≠ 0 

    ∴ Range of f(x) is [– ½, ½] – {0}    

(ii) f(x) =  Functions VSAQS 49

      It is defined when 9 – x2 ≥ 0

     ⟹ x2 – 9 ≤ 0

       (x + 3) (x – 3) ≤ 0

   –  3 ≤ y ≤ 3  

∴ the domain of f (x) is [– 3, 3]

  Let y =Functions VSAQS 49

       y2 = 9 – x2

       x2 = 9 – y2

       x =Functions VSAQS 50

      It is defined when 9 – y2 ≥ 0

     ⟹ y2 – 9 ≤ 0

       (y + 3) (y – 3) ≤ 0

     –  3 ≤ y ≤ 3  

         y ∈ [– 3, 3]

  ∴ Range of f (x) is [0, 3] (∵ y ≥ 0)

(iii) f(x) =Functions VSAQS 51

     clearly, x ∈ R

    ∴ domain of f(x) is R

        Let y = Functions VSAQS 51

       If x = 0, then y = 1

       If x = – 1, then y = 1

       If x = 1, then y = 3

      If x = – 2, then y = 3

      If x = 2, then y = 5

      ∴ Range of f (x) is [1, ∞)

(iv) f(x) = [x]

clearly Domain = R  and Range = Z     

QUESTION 18.

Define (i) One – One function (ii) Onto function (iii) Bijection (iv) Even and Odd functions

     (i) One – One Function: one – one, if every element of A has a unique image in B.

    (ii) Onto function: A function f: A→ B is said to be onto if ∀ y ∈ B there exists x ∈ A such that f(x) = y.

    (iii) Bijection: A function f: A→ B is said to be a Bijection if it is both one-one and onto.

    (iv) Even and Odd functions:

     If f(–x) = f(x), then f(x) is even function

          If f(– x)

= – f(x), then f(x) is odd function

 


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