# 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.

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.

### 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.

##### 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.

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.

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.

## 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) =

Sol:   given f(x) =

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) =

Sol: Given f(x) =

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) =

Sol: Given f(x) =

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) =

Given f(x) =

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) =

Given f(x) = 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) =

Given f(x) =

It is defined when  – x > 0

⇒ > x

⇒ x < 0

∴ domain = (–∞, 0)

(viii) f(x) =

Given f(x) =

It is defined when  +x ≠0

⇒ ≠ – 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) ()  (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)  ()= (g (f ())

=  g(4() – 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) (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)

(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) (viii)   (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)}

(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) =

f(π/4) = cos (π/4) =

f(π/3) = cos (π/3) = 1/2

f(π/2) = cos (π/2) = 0

∴ B = {1, , , 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) = then find B.

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

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 =

f-1(y) =

∴ f-1(x) =

(iii) f: (0, ∞) → R defined by f(x) =

Given function is f(x) =

Let y = f(x) ⟹ x = f-1(y)

y =

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 =

4x =   – 7

x = (  – 7)/4

f-1(y) = (  – 7)/4

∴ f-1(x) = (  – 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) = , then show that f (tan θ) = cos 2θ

Sol: Given function is f(x) =

f (tan θ) =

∴  f (tan θ) = cos 2θ

QUESTION 13

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

Sol: Given f(x) =

=

=

=

=

=

= 2

∴ = 2f (x)

QUESTION 14

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 function is f(x) =

f (x + y) + f (x – y) =

=

=

=

=

=2

∴ f (x + y) + f (x – y) = 2 f(x) f(y)

QUESTION 15

If f(x) = cos (log x), then show that = 0

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

= 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) =

Given function is f(x) =

Let y =

⟹ ey =

ey > 0

∴ The range of f(x) is R

(ii) f(x) =

It is defined when x – 2 ≠ 0

⟹ x ≠ 2

Domain = R – {2}

let y =

=

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) =

It is defined when 1 + x2 ≠ 0

⟹ x2 ≠ – 1

x ∈ R

∴ domain of f(x) is R

Let y =

y (1 + x2) = x

y + x2y = x

x2 y – x + y = 0

x =

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) =

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 =

y2 = 9 – x2

x2 = 9 – y2

x =

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) =

clearly, x ∈ R

∴ domain of f(x) is R

Let y =

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