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RBSE Solutions for Class 12 Maths Chapter 2 प्रतिलोम वृत्तीय फलन Ex 2.1

May 3, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 12 Maths Chapter 2 प्रतिलोम वृत्तीय फलन Ex 2.1 is part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Class 12 Maths Chapter 2 प्रतिलोम वृत्तीय फलन Exercise 2.1.

Rajasthan Board RBSE Class 12 Maths Chapter 2 प्रतिलोम वृत्तीय फलन Ex 2.1

प्रश्न 1.
निम्नलिखित कोणों के मुख्य मान ज्ञात कीजिए:
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 1
हल :
(i) sin-1 (1)
sin-1 की मुख्य मान शाखा \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) है।
माना sin-1 (1) = x
⇒ sin x = 1
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 2
(ii) cos-1 \(\left( -\frac { 1 }{ 2 } \right) \)
cos-1 की मुख्य मान शाखा [0, π] है।
माना cos-1 \(\left( -\frac { 1 }{ 2 } \right) \) = x
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 3
(iii) sec-1 (-√2)
sec-1 का मुख्य मान शाखा [0, π] – \(\left\{ \frac { \pi }{ 2 } \right\} \) है।
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 4
(iv) cosec-1 (-1)
cosec-1 की मुख्य मान शाखा \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) – {0} है।
माना cosec-1 (-1) = x
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 5
(v) cot-1 \(\left( -\frac { 1 }{ \sqrt { 3 } } \right) \)
cot-1 का मुख्य मान शाखा [0, π] है।
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 6a
(vi) tan-1 \(\left( \frac { 1 }{ \sqrt { 3 } } \right) \)
tan-1 की मुख्य मान शाखा \(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \) है।
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 7

प्रश्न 2.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 8
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 9
इति सिद्धम्।

प्रश्न 3.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 10
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 11
इति सिद्धम्।

प्रश्न 4.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 12
हल :
LHS
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 13
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 14
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 15

इति सिद्धम्।

प्रश्न 5.
sec² (tan-12) + cosec² (cot-13) = 15
हल :
माना tan-12 = θ ⇒ tan θ = 2
∴ sec² θ = 1 + tan² θ
= 1 + (2)² = 1 + 4 = 5
∴ sec²(tan-12) = 5 …(i)
माना cot-13 = Φ
⇒ cot Φ = 3
∴ cosec² Φ = 1 + cot² Φ
= 1 + (3)² = 1 + 9 = 10
∴ cosec²(cot-13) = 10 …(ii)
(i) और (ii) को जोड़ने पर
sec²(tan-12) + cosec² (cot-13) = 5 + 10
sec² (tan-12) + cosec² (cot-13) = 15.
इति सिद्धम्

प्रश्न 6.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 16
हल :
(i) 2 tan-1x = sin-1 \(\frac { 2x }{ 1+{ x }^{ 2 } } \)
माना tan-1 x = θ
∴ x = tan θ
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 17
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 18
इति सिद्धम्।

प्रश्न 7.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 19
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 20
= tan-1 (0)
= tan-1 (π)
= π
= RHS
इति सिद्धम्।

प्रश्न 8.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 21
हल :
माना tan-1 x = θ
x = tan θ
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 22
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 23

प्रश्न 9.
यदि cos-1x + cos-1y + cos-1z = π, तो सिद्ध कीजिए कि x² + y² + x² + 2xyz = 1.
हल :
∵ cos-1x + cos-1y + cos-1z = π
⇒ cos-1x + cos-1y = π – cos-1z
⇒ cos-1 [xy – √1 – x² √1 – y²] = cos-1(-z)
⇒ [∵ cos-1x + cos-1y = cos-1 [xy – √1 – x² √1 – x²] तथा (cos-1(-x) = π – cos-1x)
⇒ xy – √1 – x² √1 – x² = (- z)
⇒ xy + z = √1 – x² √1 – x²
⇒ (xy + z)² = (1 – x²)(1 – y²)
⇒ x²y² + z² + 2xyz = 1 – y² – x² + x²y²
⇒ z² + 2xyz = 1 – y² – x²
⇒ x² + y² + z² + 2xyz = 1
इति सिद्धम्।

प्रश्न 10.
यदि sin-1 + x + sin-1 y + sin-1 z = π, तो सिद्ध कीजिए कि \(x\sqrt { 1-{ x }^{ 2 } } +y\sqrt { 1-y^{ 2 } } +z\sqrt { 1-{ z }^{ 2 } } =2xyz\).
हल :
माना
sin-1 x = A ∴ x = sin A
sin-1 y = B ∴ y = sin B
sin-1 z = C ∴ z = sin C
∴ sin-1 x + sin-1 y + sin-1 z = π
A + B + C = π
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 24
⇒ \(\frac { 1 }{ 2 }\)[2 sin (A + B) cos (A – B) + 2 sin C cos C]
⇒ sin (A + B) cos (A – B) + sin C cos C
⇒ sin (π – C) cos (A – B) + sin cos C
⇒ sin C cos (A – B) + sin C cos C
⇒ sin ( [cos (A – B) + cos {π – (A + B)}]
⇒ sin ( [cos (A – B) – cos (A + B)]
⇒ sin ( [2 sin A sin B]
[∵ cos C – cos D = 2sin\(\frac { C+D }{ 2 }\)sin\(\frac { D-C }{ 2 }\)]
= 2 sin A sin B sin C
= 2 xyz
= R.H.S.
इति सिद्धम्

प्रश्न 11.
यदि tan-1x + tan-1y + tan-1z = \(\frac { \pi }{ 2 } \), तो सिद्ध कीजिए कि xy + yz + zx = 1.
हल :
प्रश्नानुसार
tan-1x + tan-1y + tan-1z = \(\frac { \pi }{ 2 } \)
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 25
⇒ 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1
इति सिद्धम्

प्रश्न 12.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 26
हल :
माना x = tanA, y = tanB, z = tanC
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 27 RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 28
⇒ x + y = -2(1 – xy) = -z + xyz
x + y + z = xyz.
इति सिद्धम्

प्रश्न 13.
यदि
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 29
तो सिद्ध कीजिए कि x + y + z = xyz.
हल :
प्रश्नानुसार
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 30
x + y + z – xyz = 0
x + y + z = xyz
इति सिद्धम्।।

प्रश्न 14.
सिद्ध कीजिए कि
tan-1 x + cot-1 (x + 1) = tan-1 (x² + x + 1)
हल :
LHS
= tan-1 x + cot-1 (x + 1)
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 31
= tan-1 (x² + x + 1)
= RHS
इति सिद्धम्।।

प्रश्न 15.
यदि tan-1x, tan-1y, tan-1z, समान्तर श्रेढ़ी में हो, तो सिद्ध कीजिए कि y² (x + z) + 2y(1 – xz) – x – z = 0
हल :
tan-1x, tan-1y, tan-1z, समान्तर श्रेढ़ी में हैं, अतः
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 32
⇒ (z + x) (1 – y²) = 2y (1 – zx)
⇒ z + x – y²(x + z) = 2y (1 – xz)
⇒ y²(x + 2) + 2y (1 – xz) – x – z = 0
इति सिद्धम्।।

प्रश्न 16.
यदि x3 + px2 + qx + p = 0 के मूल α, β, γ हो, तो सिद्ध कीजिए कि एक विशेष परिस्थिति के अलावा tan-1 α + tan-1 β + tan-1 γ = nπ और वह विशेष स्थिति भी ज्ञात कीजिए जब ऐसा नहीं होता है।
हल :
दिया है :
α, β, γ समीकरण : x3 + px2 + qx + p = 0 के मूल हैं; अत:
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 33
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 34
= tan-1 (0) [∵ α + β + γ = αβγ = -p]
= nπ
= RHS

प्रश्न 17.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 35
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 36
⇒   x² = a²b²
⇒   x = ± ab.

प्रश्न 18.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 37
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 38

प्रश्न 19.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 39
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 40
⇒ x(3x² – 7x – 6) =0
⇒ x(3x² – 9x + 2x – 6) = 0
⇒ x[3x(x – 3) + 2(x – 3) = 0
⇒ x(x – 3)(3x + 2) = 0
⇒ x = 0, x = 3, x = \(-\frac { 2 }{ 3 }\)

प्रश्न 20.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 41
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 42
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 43

प्रश्न 21.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 44
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 45
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 46
⇒ 2x – 1 = 2 + x
⇒ 2x – x = 2 + 1
⇒ x = 3

प्रश्न 22.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 47
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 48
⇒ 3 + x = 3x – 1
⇒ 2x = 4
⇒ x = 2

प्रश्न 23.
sin 2 (cos-1 {cot (2 tan-1 x)} = 0
हल :
दी गई समीकरण है
sin 2 [cos-1 {cot (2 tan-1 x)}] = 0
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 50
⇒ \(\sqrt { { 6x }^{ 2 }-1-{ x }^{ 4 } } \sqrt { { -2x }^{ 2 }+{ x }^{ 4 }+1 } =0\)
⇒ (6x² – 1 – x4) (x4 – 2x² + 1) = 0
⇒ 6x² – 1 – x4 = 0 ….(1)
⇒ x4 – 2x² + 1 = 0 …(2)
समीकरण (1) से,
⇒ 6x² – 1 – x4 = 0
⇒ x4 – 6x² + 1 = 0
⇒ x4 – 2 x 3x² + (3)² – 8 = 0
⇒ (x² – 3)² = 8
⇒ x² – 3 = ± 2√2
⇒ x² = 3 ± 2√2
⇒ x² = 1 + 2 ± 2√2
⇒ x² = (1)² + (√2)² ± 2√2
⇒ x² = (1±√2)²
⇒ x = ± (1+√2)
समीकरण (2) से
⇒ x4 – 2x² + 1 = 0
⇒ (x²)² – 2x² + (1)² = 0
⇒ (x² – 1)² = 0
⇒ x² = 1
⇒ x = ± 1

प्रश्न 24.
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 51
हल :
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 52
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 53

प्रश्न 25.
sin-1x – sin-1y = \(\frac { 2\pi }{ 3 } \); cos-1x – cos-1y = \(\frac { \pi }{ 3 } \)
हल :
प्रश्नानुसार,
sin-1x – sin-1y = \(\frac { 2\pi }{ 3 } \) …(i)
cos-1x – cos-1y = \(\frac { \pi }{ 3 } \) ….(ii)
RBSE Solutions for Class 12 Maths Chapter 2 Ex 2.1 54

We hope the given RBSE Solutions for Class 12 Maths Chapter 2 प्रतिलोम वृत्तीय फलन Ex 2.1 will help you. If you have any query regarding RBSE Solutions for Class 12 Maths Chapter 2 प्रतिलोम वृत्तीय फलन Ex 2.1, drop a comment below and we will get back to you at the earliest.

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