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NCERT exemplar Class 12 Maths solutions chapter 2 contains the topic that requires some time and effort on the student’s part. It is an important chapter in Calculus Mathematics, which helps in understating the integrals and their existence. We will help you to find the solutions of all the NCERT exemplar questions of this chapter. Through NCERT Exemplar Class 12 Math chapter 2 solutions, the students will learn about inverse trigonometric functions. Students looking for NCERT exemplar Class 12 Maths solutions chapter 2 PDF download can use the online webpage to PDF tool.
More about NCERT exemplar Class 12 Maths solutions chapter 2 Inverse Trigonometric Functions
In NCERT exemplar solutions for Class 12 Maths chapter 2, one will learn about how to find integrals and what are the ranges and domains of these inverses functions . Some other concepts including what makes these functions inverse, how these functions behave and Various properties are explained in NCERT exemplar Class 12 Maths chapter 2 solutions. Thus, this chapter of NCERT Class 12 Maths Solutions will help the students to solve the questions at a later stage of the chapter.
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Question:1
Answer:
we know that
and
[since , ]
[since , ]
Question:7
Find the real solution of the equation:
Answer:
We have ,
Let
On Putting the value of in Eq. (i), We get
we know that,
So,(ii) becomes,
For real solution , we have x=0,-1.
Question:11
Answer:
We have
Let
And
From (c),(d);(i) becomes
[From (a),(b)]
On squarinting both Sides, we get
Question:15
Answer:
Solving LHS,
Again , let
We know that, [from (i),(ii)]
Since , LHS=RHS
Hence Proved.
Question:16
Answer:
Solving LHS,
Let
Squaring both sides,
Again,
Let
Squaring both sides,
Since,
We know that,
Since , LHS=RHS
Hence Proved
Question:18
Show that and, justify why the other value is ignored.
Answer:
Solving LHS,
{ but as we can see , , since }
Note: Scince
Question:19
Answer:
We have
And,
Given that,
Question:20
Which of the following in the principal value branch of
A.
B.
C.
D.
Answer:
Answer : (c)
We know that the principal value branch of is
Question:21
Which of the following in the principal value branch of .
Answer:
Answer :(D)
We know that the principal value branch of is
Question:22
If , then x equals to
A. 0
B. 1
C. -1
D. 1/2
Answer:
Answer : B
Given That,
Cross multiplying
Here only x=1 satifies the given equation.
Note:- Here ,Putting x=-1 in the given equation we get,
Hence , x=-1 does not satisfy the given equation.
Question:24
The domain of the function is
A.[0,1]
B.[-1,1]
C.(-1,1)
D.[0,]
Answer:
Answer:(A)
We Have
Scince
Question:25
The domain of the function defined by is
A.[1,2]
B.[-1,1]
C.[0,1]
D. None of these
Answer:
Answer: (A)
Question:26
If then x is equal to
C.0
D.1
Answer:
Answer: (B)
Given,
Let
So
Principal value is .......(2)
Also , we know that
From (1) ,,(2), and (3) we have
But
So,
We know that
As
so
Question:27
The value of sin (2tan^{–1} (.75)) is equal to
A. 0.75
B. 1.5
C. 0.96
D. sin 1.5
Answer:
Answer :(c)
sin (2tan^{–1} (.75))
Let, tan^{–1} (.75) = θ
As, so
Now,
sin (2tan^{–1} (.75)) = sin 2θ
= 2 sin θ cos θ
So, sin (2tan^{–1 }(.75)) = 0.96.
Question:28
Answer:
We have
We know that,
So,
Let
⇒ cos θ = 0
Principal value of cos^{-1} x is [0, π]
For, cos θ = 0
so,
Question:29
The value of the expression is
D.1
Answer:
Answer :(B)
We have,
Principal value of sin^{-1} x is
Principal value of sec^{-1} x is [0, π]
Let
So, … (1)
Let sec^{-1} 2 = B
⇒ sec B = 2
So, 2 sec^{-1} 2 = 2B
...(2)
So, the value of from (1) and (2) is
So,
Question:30
If tan^{–1} x + tan^{–1 }y = 4π/5, then cot^{–1}x + cot^{–1} y equals
Answer:
Answer :(A)
We know that,
We have,
tan^{–1} x + tan^{–1} y = 4π/5 … (1)
Let, cot^{–1}x + cot^{–1} y = k … (2)
Adding (1) and (2) –
...(3)
Now, tan^{–1} A + cot^{–1} A = π/2 for all real numbers.
So, (tan^{–1} x + cot^{–1} x) + (tan^{–1}y + cot^{–1} y) = π … (4)
From (3) and (4), we get,
Question:31
If where a, then the value of x is
A. 0
B. a/2
C. a
D.
Answer:
Answer:(D)
We have
we know that
From (1) and (2) we have,
L.H.S-
From (3) R.H.S
So, we have 4 tan^{-1} a = 2 tan^{-1} x
⇒ 2 tan^{-1} a = tan^{-1} x
But from (3)
So
Question:32
The value of
A. 25/24
B.25/7
C.24/25
D.7/24
Answer:
Answer :(d)
We have to find
Let
Also,
As,
So
We need to find cot A
So
Question:33
The value of the expression is
Answer:
Answer:(B)
We need to find ,
Let,
Also we need to find
We know that
so,
on rationalizing,
Again rationalizing
Question:34
If |x| ≤ 1, then is equal to
A. 4 tan–1 x
B. 0
C.
D. π
Answer:
Answer(A)
We need to find,
We know that
So,
=
Question:35
If cos^{–1}α + cos^{–1}β + cos^{–1}γ = 3π, then α(β + γ) + β (γ + α) + γ (α + β) equals
A. 0
B. 1
C. 6
D. 12
Answer:
Answer :(C)
Given, cos^{–1}α + co^{s–1}β + cos^{–1}γ = 3π … (1)
Principal value of cos^{-1} x is [0, π]
So, maximum value which cos^{-1} x can have is π.
So, if (1) is correct then all the three terms i.e,
cos^{–1}α, cos^{–1}β, cos^{–1}γ should be equal to π
So, cos^{–1}α = π
cos^{–1}β = π
cos^{–1}γ = π
So, α = β = γ = -1
So, α(β + γ) + β (γ + α) + γ (α + β)
= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)
= 3(-1)(-2)
= 6
Question:36
The number of real solutions of equarion is
A. 0
B. 1
C. 2
D. Infinite
Answer:
Answer:(A)
We have , , x is in
R.H.S
So,
Squaring both side , we get,
Now plotting cos 2x and 2x^{2}-1, we get,
As , there is no point of intersection in , so therre is no
solution of the given equation in
Question:37
If , then
D.x>0
Answer:
Answer :(C)
Plotting cos^{-1} x and sin^{-1} x, we get,
As, graph of cos^{-1 }x is above graph of sin^{-1} x in .
So, cos^{–1}x > sin^{–1} x for all x in .
Question:38
Fill in the blanks The principle value of is ___________.
Answer:
The principal value of is .
Principal value cos^{-1} x is [0,]
Let,
As,
So,
Question:39
Fill in the blanks The value of is_______.
Answer:
The value of is
Principal value of is
now, should be in the given range
is outside the range
As, sin (π – x) = sin x
So,
Question:40
Fill in the blanks
If cos (tan^{–1}x + cot^{–1} √3) = 0, then value of x is _________.
Answer:
If cos (tan^{–1}x + cot^{–1} √3) = 0, then value of x is
Given, cos (tan^{–1}x + cot^{–1}) = 0
we know that,
so, x=
Question:41
fill in blanks the set of value of is______________.
Answer:
Fill in the blanks the set of value of is
Domain of sec^{-1} x is R – (-1,1).
As, is outside domain of sec^{-1} x.
Which means there is no set of value of
So, the solution set of is null set or
Question:42
Fill in the blanks
The principal value of tan^{–1} √3 is _________.
Answer:
The Principal value of is
Principal value of tan^{-1} x is
Let,
As
so,
Question:43
Answer:
The value of is
We needd,
Principal value of cos^{-1} x is [0,π]
Also, cos (2nπ + θ) = cos θ for all n ? N
Question:44
Fill in the blanks
The value of cos (sin^{–1 }x + cos^{–1} x), |x| ≤ 1 is ________.
Answer:
The value of cos (sin^{–1} x + cos^{–1} x) for |x| ≤ 1 is 0.
cos (sin^{–1} x + cos^{–1} x), |x| ≤ 1
We know that, (sin^{–1} x + cos^{–1} x), |x| ≤ 1 is
So,
= 0
Question:45
The value of expression when is___________.
Answer:
The value of expression When is 1
When
We know that, (sin^{–1} x + cos^{–1} x) for all |x| ≤ 1 is
As, lies in domain
So =
=1
Question:46
Fill in the blanks if for all x, then ______<y<_____.
Answer:
Fill in the blanks if for all x, then
We know that,
so
=4 tan^{-1} x
So, y = 4 tan^{-1} x
As, principal value of tan-1 x is
So,
Hence, -2π < y < 2π
Question:47
The result is true when value of xy is _________.
Answer:
The result is true when value of xy is > -1.
We have,
Principal range of tan^{-1}a is
Let tan^{-1}x = A and tan^{-1}y = B … (1)
So, A,B
We know that, … (2)
From (1) and (2), we get,
Applying, tan^{-1} both sides, we get,
As, principal range of tan^{-1}a is
So, for tan^{-1}tan(A-B) to be equal to A-B,
A-B must lie in – (3)
Now, if both A,B < 0, then A, B
∴ A and -B
So, A – B
So, from (3),
tan^{-1}tan(A-B) = A-B
Now, if both A,B > 0, then A, B
∴ A and -B
So, A – B
So, from (3),
tan^{-1}tan(A-B) = A-B
Now, if A > 0 and B < 0,
Then, A and B
∴ A and -B
So, A – B (0,π)
But, required condition is A – B
As, here A – B (0,π), so we must have A – B
Applying tan on both sides,
As,
So, tan A < - cot B
Again,
So,
⇒ tan A tan B < -1
As, tan B < 0
xy > -1
Now, if A < 0 and B > 0,
Then, A and B
∴ A and -B
So, A – B (-π,0)
But, required condition is A – B
As, here A – B (0,π), so we must have A – B
Applying tan on both sides,
As,
So, tan B > - cot A
Again,
So,
⇒ tan A tan B > -1
⇒xy > -1
Question:48
Fill in the blanks
The value of cot^{–1}(–x) for all x ? R in terms of cot^{–1} x is _______.
Answer:
The value of cot^{–1}(–x) for all x ? R in terms of cot^{–1} x is
π – cot^{-1} x.
Let cot^{–1}(–x) = A
⇒ cot A = -x
⇒ -cot A = x
⇒ cot (π – A) = x
⇒ (π – A) = cot^{-1} x
⇒ A = π – cot-1 x
So, cot^{–1}(–x) = π – cot-1 x
Question:49
Answer:
True.
It is well known that all trigonometric functions have inverse over their respective domains.
Question:50
Answer:
As, cos^{-1} x is not equal to sec x. So, (cos^{–1}x)^{2} is not equal to sec^{2} x.
Question:51
Answer:
As, all trigonometric and their corresponding inverse functions are periodic so, we can obtain the inverse of a trigonometric ratio in any branch in which it is one-one and onto.
Question:52
Answer:
True
We know that the smallest value, either positive or negative of angle θ is called principal value of the inverse trigonometric function
Question:53
Answer:
True.
Graph of any inverse function can be obtained by interchanging x and y-axis in the graph of the corresponding function. If (p, q) are two points on f(x) then (q, p) will be on f^{-1}(x).
Question:54
State True or False for the statement
The minimum value of n for which is valid is 5.
Answer:
false
As , tan is an increasing function so applying tan on both side
we get,
As, and
so
⇒ n > π
⇒ n > 3.14
As, n is a natural number, so least value of n is 4.
Question:55
State True or False for the statement
The principal value of is
Answer:
True
Principal value of sin^{-1} x is
Principal value of cos^{-1} x is [0, π]
We have,
As, so
As, so,
The sub-topics that are covered in this chapter of inverse trigonometric functions are:
Chapter 1 | |
Chapter 2 | Inverse Trigonometric Functions |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 |
Introduction to Inverse Trigonometric Functions, The Basic Concepts of Inverse Trigonometric Functions and Properties of Inverse Trigonometric Functions are important topics of this chapter.
Yes, the NCERT exemplar Class 12 Maths chapter 2 solutions are helpful for you to prepare for board exams.
There is only 1 exercise in this chapter with 55 problem solving questions.
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