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Trigonometric Graph Worksheet

Trigonometric Graph Worksheet
  • Page 1
 1.  
Which of the following is a point on the curve y = cot x?
a.
(π6, 13)
b.
(0, 0)
c.
(π3, 13)
d.
(π2, 1)


Solution:

f(x) = cot x

Since f(0) = cot 0 is not defined ,

f(π6) = cot(π6) = 3 ,

f(π3) = cot(π3) = 13

f(π2) = cot(π2) = 0 ,

The graph of cot x passes through the points (π6, 3), (π3, 13) and (π2, 0)

So, the point (π3, 13) is on the curve y = cot x.


Correct answer : (3)
 2.  
The graph of cos x
a.
Passes through the origin
b.
Does not pass through (π4, 0)
c.
Passes through the point (π2, 1)
d.
Does not pass through the origin


Solution:

f(x) = cos x

Since f(0) = cos 0 = 1,

f( π2) = cos(π2) = 0

cos x is passing through (0, 1), (π2, 0).

So, graph of cos x does not pass through origin (0, 0).


Correct answer : (4)
 3.  
Which of the following is a point on the curve y = tan x?
a.
(0, 1)
b.
(- π2, 1)
c.
(π2, 1)
d.
(π4, 1)


Solution:

f(x) = tan x

Since f(0) = tan 0 = 0,

f( π4) = tan(π4) = 1

f( π2) = tan(π2) is not defined.

f(- π2) = tan(- π2) is not defined.

The curve passes through (0, 0), (π4, 1)

So, (π4, 1) is a point on y = tan x.


Correct answer : (4)
 4.  
Through which of the following points does the graph of sin x pass?
a.
(π6, 32)
b.
(π6, 12)
c.
(π2, 0)
d.
(0, -1)


Solution:

f(x) = sin x

Since f(0) = sin 0 = 0 ,

f(π6) = sin(π6) = 12 and

f(π2) = sin π2 = 1.

The graph of sin x passes through the points (0, 0), (π6, 12) and (π2, 1).


Correct answer : (2)
 5.  
Which of the following is correct?
a.
The graph of sin x passes through (0, 0)
b.
The graph of sin x does not pass through the origin.
c.
The graph of sin x does not pass through (π2, 1)
d.
The graph of sin x does not pass through (-π2, -1)


Solution:

The graph of sinx passes through the points (0, 0), (π2, 1), (- π2, -1).


Correct answer : (1)
 6.  
The y - intercept of the graph y = sin x is
a.
-1
b.
2
c.
1


Solution:

Substitute x = 0 in y = sin x to get its y- intercept.

y = sin 0 = 0
[Substitute x = 0.]

So, y-intercept of the graph y = sin x is 0.


Correct answer : (2)
 7.  
The zeros of the graph y = cos x are existing at x =
a.
nπ2 for all integer values of n
b.
nπ for all integer values of n
c.
(2n + 1) π2 for all real values of n
d.
(2n + 1) π for all integer values of n


Solution:

The zeros of y = sin x are the solutions of sin x = 0

The solutions of cos x = 0 are .....- 3π , - 2π , - π , 0, π , 2π , 3π , ..... that is the multiples of π which are in short n π for all integer values of n.


Correct answer : (2)
 8.  
The y - intercept of y = tan x is:
a.
2
b.
-1
c.
1


Solution:

Substitute x = 0 in y = tan x to get its y-intecept.

y = tan 0 = 0
[Substitute x = 0.]

So, y-intercept of the graph y = tan x is 0.


Correct answer : (1)
 9.  
Write the amplitude of y = 5sin(x4).
a.
1 3
b.
5
c.
4


Solution:

The amplitude of y = bsin(xa) is | b |
[Definition.]

On comparing y = 5sin (x4) with y = bsin (xa) we have b = 5, a = 4

So, the amplitude of y = 5in(x3) is | b | = | 5 | = 5


Correct answer : (3)
 10.  
Evaluate: limx0 46x21-cos 6x
a.
23
b.
1 9
c.
23 9
d.
23 3


Solution:

limx0 46x21-cos 6x
[Undefined at x = 0.]

= limx0 46x22sin2 3x
[Use 1 - cos 2x = 2sin2 x.]

= 23 limx0 1sin2 3xx2
[Divide both the numerator and the denominator by x2.]

= 23 1(limx0sin 3xx)2
[Use quotient law of limits.]

= 23 1(3)2 = 23 / 9
[Use limx0 sin kxx = k.]


Correct answer : (3)

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