Period Of Trigonometric Functions - Solved Exercise 12.1 - Unit 12: Graphs Of Trigonometry And Inverse Trigonometric Functions And Solutions Of Trigonometric Equations - Mathematics for XI (Science Group)

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Mathematics for XI (Science Group)
Unit 12: Graphs Of Trigonometry And Inverse Trigonometric Functions And Solutions Of Trigonometric Equations
Solved Exercise 12.1

Special Thanks To Sir Adnan Ullah Khan
1.Find the domain and range of each of the following functions:
(i) 2sin 3x
SOLUTION:
Here, y = 2sin 3x
∵ Given function is defined for all real numbers
∴ function of 2sin 3x is defined for all real numbers.
So,
Domain of 2sin 3x is same as the Domain of sin x = R OR (-∞⁰, +∞⁰)
As Range of sin 3x is same as the range of sin x = {y | y ∈ R ⋀ -1 ≤ y ≤ 1} OR [-1, +1]
So, the Range of 2sin 3x = {y | y ∈ R ⋀ -2 ≤ y ≤ 2} OR [-2, +2]
Ans: Domain = R, Range = -2 ≤ y ≤ 2 ∀ y ∈ R or [-2, +2]

(ii) 5cos 4x
SOLUTION:
Here, y = 5cos 4x
∵ Given function is defined for all real numbers
∴ function of 5cos 4x is defined for all real numbers.
So,
Domain of cos 4x is same as the Domain of cos x = R OR (-∞⁰, +∞⁰)
As Range of cos 4x is same as the range of cos x = {y | y ∈ R ⋀ -1 ≤ y ≤ 1} OR [-1, +1]
So, the Range of 5cos 4x = {y | y ∈ R ⋀ -5 ≤ y ≤ 5} OR [-5, +5]
Ans: Domain = R, Range = -5 ≤ y ≤ 5 ∀ y ∈ R OR [-5, +5]

(ii) 8tan 2x
SOLUTION:
Here y = 8tan 2x
∵ Given function is defined for all real numbers
∴ function of 5cos 4x is defined for all real numbers.
So,
Domain of cos 4x is same as the Domain of cos x = R OR (-∞⁰, +∞⁰)
As Range of cos 4x is same as the range of cos x = {y | y ∈ R ⋀ -1 ≤ y ≤ 1} OR [-1, +1]
So, the Range of 5cos 4x = {y | y ∈ R ⋀ -5 ≤ y ≤ 5} OR [-5, +5]
Ans: Domain = R, Range = -5 ≤ y ≤ 5 ∀ y ∈ R OR [-5, +5]

Q.2: Determine whether the following trigonometric functions are even, odd or neither.
i. f(x) = sinx cosx
SOLUTION:
Replacing x by -x
we get,
f(-x) = sin(-x) cos(-x)
f(-x) = [-sin(x)][cos(x)] {∵ sin(-x) = -sinx and cos(-x) = cox(x)
f(-x) = -sinx cosx
f(-x) = -f(x) {∴ f(x) = sinx cos x}
Hence f(x) is an odd function
Ans: sinx cosx is an odd function.

ii. k(x) = x³(sinx + cosx)
SOLUTION:
Replacing x by -x
we get,
k(-x) = (-x)³ sin(-x) + cos(-x)
k(-x) = -x³[(-sinx) + cosx] {∵ sin(-x) = -sinx and cos(-x) = +cox(x)
k(-x) = -x³(-sinx + cosx)  
≠ - k(x) or k(x)
Hence k(x) is neither
Ans: k(x) = x³(sinx + cosx) is neither Even nor Odd.

Q.3: Find the period of the following functions:

Q.4: Find the maximum and minimum values of each of the following:
i) y = 4 + 3 sin𝛉
SOLUTION:
y = 4 + 3 sin𝛉
⇒Here a = 4, b = 3
Maximum value of y = a + |b|
= 4 + |3|
= 4 + 3
= 7
Minimum value of y = a -|b|
= 4 - |3|
= 4 - 3
=1
Ans: Maximum value of y = 7 and Minimum value of y = 1.

iv) y = 8 + 5cos(𝛉 - 25)
SOLUTION:
y = 8 + 5cos(𝛉 - 25)
⇒Here a = 8, b = 5
Maximum value of y = a + |b|
= 8 + |5|
= 4 + 5
= 13
Minimum value of y = a -|b|
= 8 - |5|
= 8 - 5
= 3
Ans: Maximum value of y = 13 and Minimum value of y = 3.

SOLUTION:
y = ¹ / _{25 - 12 sin(3𝛉 - 2)}
⇒Here a = 25, b = -12
Maximum value of y =  a + |b|
M = 25 + |-12|
M = 25 + 12
M =  37
Minimum value of y =  a -|b|
m = 25 - |-12|
m = 25 - 12
m = 13
Let M' and m' represents the maximum and minimum value of the reciprocal of the functions respectively
∵ M > 0 and  m > 0 then
∴ M' = ¹ /_m and
m' = ¹/_M
Ans: Maximum value of y = ¹ / ₁₃ and
Minimum value of y = ¹ /₃₇.

SOLUTION:
y = ¹ / _{1 + 16cos(5𝛉 - 4}
⇒Here a = 1, b = 6
Maximum value of y = a + |b|
M = 1 + |6|
M = 1 + 6
M = 7
Minimum value of y = a -|b|
m = 1 - |6|
m = 1 - 6
m = -5
Let M' and m' represents the maximum and minimum value of the reciprocal of the functions respectively
∵ M > 0 and m > 0  then
∴ M' = ¹/_M and
m' = ¹/_m

Ans: Maximum value of y = ¹/₇ and
Minimum value of y = ¹/_-5

EXAMPLES

DOMAIN AND RANGE

The Domain and Range for the Trigonometric Functions sinθ, cosθ, tanθ, cosecθ, secθ and cotθ

EVEN, ODD & NEITHER FUNCTIONS

PERIODICITY OF TRIGONOMETRIC FUNCTIONS

MAXIMUM & MINIMUM VALUES OF FUNCTIONS