RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 6 - Graphs Of Trigonometric Functions

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RD Sharma XI 2020 2021 Volume 1 Solutions for Class 12 Commerce Maths Chapter 6 Graphs Of Trigonometric Functions are provided here with simple step-by-step explanations. These solutions for Graphs Of Trigonometric Functions are extremely popular among class 12 Commerce students for Maths Graphs Of Trigonometric Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020 2021 Volume 1 Book of class 12 Commerce Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020 2021 Volume 1 Solutions. All RD Sharma XI 2020 2021 Volume 1 Solutions for class 12 Commerce Maths are prepared by experts and are 100% accurate.

Page No 6.13:

Question 1:

Sketch the graphs of the following functions:
f(x) = 2 cosec πx

Answer:

f(x) = 2 cosec πx

Page No 6.13:

Question 2:

Sketch the graphs of the following functions:
f(x) = 3 sec x

Answer:

f(x) = 3 sec x

Page No 6.13:

Question 3:

Sketch the graphs of the following functions:
f(x) = cot 2x

Answer:

f(x) = cot 2x

Page No 6.13:

Question 4:

Sketch the graphs of the following functions:
f(x) = 2 sec πx

Answer:

f(x) = 2 sec πx

Page No 6.13:

Question 5:

Sketch the graphs of the following functions:
f(x) = tan²x

Answer:

f(x) = tan²x

Page No 6.13:

Question 6:

Sketch the graphs of the following functions:
f(x) = cot²x

Answer:

f(x) = cot²x

Page No 6.13:

Question 7:

Sketch the graphs of the following functions:
$f (x) = \cot \frac{π x}{2}$

Answer:

$f (x) = \cot \frac{π x}{2}$

Page No 6.13:

Question 8:

Sketch the graphs of the following functions:
f(x) = sec²x

Answer:

f(x) = sec²x

Page No 6.13:

Question 9:

Sketch the graphs of the following functions:
f(x) = cosec²x

Answer:

f(x) = cosec²x

Page No 6.13:

Question 10:

Sketch the graphs of the following functions:
f(x) = tan 2x

Answer:

Step I- We find the value of c and a by comparing y = 2 tan 2x with y = c tan ax, i.e. c = 1 and a = 2.
Step II- Then, we draw the graph of y = tan x and mark the point where it crosses the x-axis.
Step III- Divide the x-coordinates of the points where y = tan x crosses x-axis by 2(i.e. a = 2) and mark the maximum value (i.e. c = 1) and minimum value (i.e. $-$ c = $-$ 1).
Then , we obtain the following graph:

Page No 6.5:

Question 1:

Sketch the graphs of the following functions:

(i) f(x) = 2 sin x, 0 ≤ x ≤ π
(ii) $g (x) = 3 \sin (x - \frac{π}{4}), 0 \leq x \leq \frac{5 π}{4}$
(iii) $h (x) = 2 \sin 3 x, 0 \leq x \leq \frac{2 π}{3}$
(iv) $ϕ (x) = 2 \sin (2 x - \frac{π}{3}), 0 \leq x \leq \frac{7 π}{5}$
(v) $ψ (x) = 4 \sin 3 (x - \frac{π}{4}), 0 \leq x \leq 2 π$
(vi) $θ (x) = \sin (\frac{x}{2} - \frac{π}{4}), 0 \leq x \leq 4 π$
(vii) $u (x) = \sin^{2} x, 0 \leq x \leq 2 π v (x) = |\sin x|, 0 \leq x \leq 2 π$
(viii) f(x) = 2 sin πx, 0 ≤ x ≤ 2

Answer:

The graphs of the following functions are:

(i) f(x) = 2 sin x, 0 ≤ x ≤ π

x	0	π
f(x) = 2 sin x	0	0

(ii)

g (x) = 3 \sin (x - \frac{π}{4}), 0 \leq x \leq \frac{5 π}{4}

x	$\frac{π}{4}$	$\frac{5 π}{4}$
$g (x) = 3 \sin (x - \frac{π}{4})$	0	0

(iii)

h (x) = 2 \sin 3 x, 0 \leq x \leq \frac{2 π}{3}

x	0	$\frac{π}{3}$	$\frac{2 π}{3}$
$h (x) = 2 \sin 3 x$	0	0	0

(iv)

ϕ (x) = 2 \sin (2 x - \frac{π}{3}), 0 \leq x \leq \frac{7 π}{5}

x	$\frac{π}{6}$	$\frac{4 π}{6}$
$ϕ (x) = 2 \sin (2 x - \frac{π}{3})$	0	0

(v)

ψ (x) = 4 \sin 3 (x - \frac{π}{4}), 0 \leq x \leq 2 π

x	$\frac{π}{4}$	$\frac{7 π}{12}$
$ψ (x) = 4 \sin 3 (x - \frac{π}{4})$	0	0

(vi)

θ (x) = \sin (\frac{x}{2} - \frac{π}{4}), 0 \leq x \leq 4 π

x	$\frac{π}{2}$	$\frac{5 π}{2}$
$θ (x) = \sin (\frac{x}{2} - \frac{π}{4})$	0	0

(vii)

u (x) = \sin^{2} x, 0 \leq x \leq 2 π v (x) = |\sin x|, 0 \leq x \leq 2 π

x	0	$π$
$u (x) = \sin^{2} x$	0	0

x	0	$π$
$v (x) = \|\sin x\|$	0	0

(viii) f(x) = 2 sin πx, 0 ≤ x ≤ 2

x	0	1
f(x) = 2 sin πx	0	0

Page No 6.5:

Question 2:

Sketch the graphs of the following pairs of functions on the same axes:
(i) $f (x) = \sin x, g (x) = \sin (x + \frac{π}{4})$
(ii) f(x) = sin x, g(x) = sin 2x
(iii) f(x) = sin 2x, g(x) = 2 sin x
(iv) $f (x) = \sin \frac{x}{2}, g (x) = \sin x$

Answer:

(i) $f (x) = \sin x, g (x) = \sin (x + \frac{π}{4})$
Clearly, sin x and $\sin (x + \frac{π}{4})$ is a periodic function with period 2π.

The graphs of $f (x) = \sin x and g (x) = \sin (x + \frac{π}{4})$ on different axes are shown below:

If these two graphs are drawn on the same axes, then the graph is shown below.

(ii) f(x) = sin x, g(x) = sin 2x

Clearly, sin x and sin 2x is a periodic function with period 2π and π, respectively.

The graphs of f(x) = sin x and g(x) = sin 2x on different axes are shown below:

If these two graphs are drawn on the same axes, then the graph is shown below.

(iii) f(x) = sin 2x, g(x) = 2 sin x

Clearly, sin 2x and 2 sin x is a periodic function with period π and 2π, respectively.

The graphs of f(x) = sin 2x and g(x) = 2 sin x on different axes are shown below:

If these two graphs are drawn on the same axes, then the graph is shown below.

(iv) $f (x) = \sin \frac{x}{2}, g (x) = \sin x$

Clearly, sin $\frac{x}{2}$ and sin x is a periodic function with period 4π and 2π, respectively.

The graphs of f(x) = sin $\frac{x}{2}$ and g(x) = sin x on different axes are shown below:

If these two graphs are drawn on the same axes, then the graph is shown below.

Page No 6.8:

Question 1:

Sketch the graphs of the following trigonometric functions:
(i) $f (x) = \cos (x - \frac{π}{4})$
(ii) $g (x) = \cos (x + \frac{π}{4})$
(iii) h(x) = cos² 2x
(iv) $ϕ (x) = 2 \cos (x - \frac{π}{6})$
(v) ψ(x) = cos 3x
(vi) $u (x) = \cos^{2} \frac{x}{2}$
(vii) f(x) = cos π x
(viii) g(x) = cos 2π x

Answer:

(i)
$y = \cos (x - \frac{π}{4}) \Rightarrow y - 0 = \cos (x - \frac{π}{4}) . . . (i) O n shifting the origin at (\frac{π}{4}, 0), we g e t : x = X + \frac{π}{4} and y = Y + 0 O n subsititut i n g the values in (i) we get : Y = \cos X Then, we draw the graph of Y = \cos X and shift it by \frac{π}{4} to the right .$
Then, we obtain the following graph:

(ii)
$y = \cos (x + \frac{π}{4}) \Rightarrow y - 0 = \cos (x + \frac{π}{4}) . . . (i) O n shifting the origin at (- \frac{π}{4}, 0), we g e t : x = X - \frac{π}{4} and y = Y + 0 On subsitituting the values in (i), we get : Y = \cos X Then, we draw the graph of Y = \cos X and shift it by \frac{π}{4} to the left .$
Then, we obtain the following graph:

(iii)
$y = \cos^{2} 2 x$

The following graph is:

(iv)
$y = 2 \cos (x - \frac{π}{6}) \Rightarrow y - 0 = 2 \cos (x - \frac{π}{6}) . . . (i) On shifting the origin at (\frac{π}{6}, 0), we get : x = X + \frac{π}{6} and y = Y + 0 On subsitituting the values in (i), we get : Y = 2 \cos X Then, we draw the graph of Y = \cos X and shift it by \frac{π}{6} to the right .$
Then, we obtain the following graph:

(v)
$y = \cos 3 x$

The following graph is:

(vi)
$y = \cos^{2} \frac{x}{2}$

The following graph is:

(vii)
$y = \cos π x$

The following graph is:

(viii)
$y = \cos 2 π x$

The following graph is:

Page No 6.8:

Question 2:

Sketch the graphs of the following curves on the same scale and the same axes:
(i) $y = \cos x and y = \cos (x - \frac{π}{4})$
(ii) $y = \cos 2 x and y = \cos 2 (x - \frac{π}{4})$
(iii) $y = \cos x and y = \cos \frac{x}{2}$
(iv) $y = \cos^{2} x and y = \cos x$

Answer:

(i)
First, we draw the graph of y = cos x.
Let us now draw the graph of $y = \cos (x - \frac{π}{4})$ .
$y = \cos (x - \frac{π}{4}) \Rightarrow y - 0 = \cos (x - \frac{π}{4}) . . . (i) O n shifting the origin at (\frac{π}{4}, 0), we g e t : x = X + \frac{π}{4} and y = Y + 0 O n subsititut i n g the values in (i), we get : Y = \cos X Then, we draw the graph of Y = \cos X and shift it by \frac{π}{4} to the right .$
Then, we will obtain the following graph:

(ii)
First, we draw the graph of y = cos 2x.
Let us now draw the graph of $y = \cos 2 (x - \frac{π}{4})$ .
$y = \cos 2 (x - \frac{π}{4}) \Rightarrow y - 0 = \cos 2 (x - \frac{π}{4}) . . . (i) O n shifting the origin at (\frac{π}{4}, 0), we g e t : x = X + \frac{π}{4} and y = Y + 0 O n subsititut i n g the values in (i), we get : Y = \cos 2 X Then, we draw the graph of Y = \cos 2 X and shift it by \frac{π}{4} to the right .$
Then, we will obtain the following graph:

(iii)

First, we draw the graph of y = cos x.
Let us now draw the graph of $y = \cos (\frac{x}{2})$ .
$\Rightarrow y = \cos \frac{1}{2} (x)$
Then, we will obtain the following graph:

(iv)
First, we draw the graph of y = cos²x.
Let us now draw the graph of y = cos x.

Then, we will obtain the following graph:

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