Students often turn to ML Aggarwal Maths for Class 11 Solutions Chapter 3 Trigonometry Ex 3.4 to clarify doubts and improve problem-solving skills.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4

Question 1.

Draw the graphs of the following functions :

(i) sin 3x

(ii) 3 sin x

(iii) 2 sin 2x.

Also write their range and period.

Solution:

(i) Given y = sin 3x

= sin (3x + 2π)

= sin 3 (x + \(\frac{2 \pi}{3}\))

∴ The period of sin 3x is \(\frac{2 \pi}{3}\).

So it is suffices to draw the graph in the interval 0 to \(\frac{2 \pi}{3}\) and repeat it over other intervals.

We construct the table of values are given as under:

A portion of graph is given as under

since – 1 ≤ sin 3x ≤ 1

∴ Range = {y ; y ∈ R : – 1 ≤ y ≤ 1}

(ii) Let y = 3 sin x 3 sin (x + 2π), it is defined for all x ∈ R.

Thus period of 3 sin x be 2π.

So it is sufficient to draw the graph in the interval 0 to 2π and repeat it over other intervals.

Now construct the table of values is as under :

A portion of graph is given as under :

since, – 1 ≤ sin x ≤ 1 ∀ x ∈ R

⇒ – 3 ≤ 3 sin x ≤ 3

Thus, R_{f} = {y ; y ∈ R, – 3 ≤ y ≤ 3}

= [- 3, 3]

(iii) Let y = 2 sin 2x. it is defined for all x ∈ R.

⇒ y = 2 sin (2x + 2π)

= 2 sin 2 (x + π)

Thus period of 2 sin 2x be π.

So it is sufficient to draw the graph in the interval 0 to π and repeat it over other intervals.

Now construct the table of values is as under:

The portion of graph is represented as under :

since – 1 ≤ sin 2x ≤ 1 ∀ x ∈ R

⇒ – 2 ≤ y ≤ 2 ∀ x ∈ R

∴ range of f (x) = R_{f}

= {y : y ∈ R – 2 ≤ y ≤ 2} = [- 2, 2]

Question 2.

Draw the graph or the following functions:

(i) cos \(\frac{x}{2}\)

(ii) 3 cos 2x

(iii) 2 cos 3x

Also write their range and period.

Solution:

(i) Let y = cos \(\frac{x}{2}\)

which is defined for all x ∈ R.

Now y = cos \(\frac{x}{2}\)

= cos (\(\frac{x}{2}\) + 2π)

= cos \(\frac{1}{2}\) + (x + 4π)

∴ period of y be 4π.

So it is sufflcient to draw the graph in the interval 0 to 4π and repeat it over other intervals.

We construct the table of values is as under:

A portion of graph is given as under:

since, – 1 ≤ cos \(\frac{x}{2}\) ≤ 1 ∀ x ∈ R

Thus R_{f} = [- 1, 1]

(ii) Let y = 3 cos 2x

= 3 cos (2x + 2π)

= 3 cos 2 (x + π), which is defined ∀ x ∈ R.

The period of 3 cos 2x be π.

So it is sufficient to draw the graph in the interval 0 to π and repeat it over other intervals.

We construct the table of values is as under:

The portion of graph is given as under:

since, – 1 ≤ cos 2x ≤ 1 ∀ x ∈ R

⇒ – 3 ≤ 3 cos 2x ≤ 3

⇒ – 3 ≤ y ≤ 3

∴ Range = {y : y ∈ R ; – 3 ≤ y ≤ 3} = [- 3, 3]

(iii) Let y = 2 cos 3x

= 2 cos (3x + 2π)

= 2 cos 3 (x + \(\frac{2 \pi}{3}\))

which is defined for all x ∈ R.

Thus, the period of 2 cos 3x be \(\frac{2 \pi}{3}\).

So it is sufficient to draw the graph in the interval 0 to \(\frac{2 \pi}{3}\) and then repeat it over other intervals.

We construct the table of values is given as under:

The portion of graph is given as under :

since – 1 ≤ cos 3x ≤ 1 ∀ x ∈ R – 2 ≤ 2 cos 3x ≤ 2

∴ Range = [- 2, 2]