Test: Graphs Of Trigonometric Functions - JEE MCQ

Period of sin x is 2π.
Period of sin 5x is 2π/5

15cosec(x/3)= 2π/1/3=6 π

y = sec x is periodic function. The period of y = sec x is 2π.

Period = 2π/|b|
As it is given a =5, b = 1/2
period = 2π/|1/2|
⇒ 4π

Range of cotx is (−∞,+∞)

Explanation:

The cotangent function can take up any values depending on the value of x , the independent variable.
And thus, the range (−∞,+∞) is justified.

The domain is all real numbers other than integral multiples of π where, the function is not defined.

This occurs in either the I or the III quadrant. The only way for tan θ positive and sin θ negative is the III quadrant.

The Magnitude of y = 3sinx is 3.
graph{y=-3*sinx [-10, 10, -5, 5]}
Amplitude is the height of a periodic function, aka the distance from the center of the wave to it's highest point (or lowest point). You can also take the distance from the highest point to the lowest point of the graph and divide it by two.

The period of y=tanx is π.
Since the function y = tan x is periodic of period π hence each branch is simply a repetition of the branch from - 
(-π/2 , π/2).

(i) The graph of the function y = cos x is continuous and extends on either side in symmetrical wave form.
(ii) Since the graph of y = cos x intersects the x-axis at the origin and at points where x is an odd multiple of 90°, hence cos x is zero at x = (2n + 1)/π² where n = 0, ±1, ±2, ±3, ±4, ……………... .

The sine function has no domain restrictions. That means that the domain is (−∞,+∞)
However, the range of a since function is restricted as such : [−1,+1].