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RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3

June 10, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3

प्रश्न 1.
निम्न सम्मिश्र संख्याओं के वर्गमूल ज्ञात कीजिए
(i) – 5 + 12i
(ii) 8 – 6i
(iii) – i
हल-
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3

प्रश्न 2.
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3
का मान ज्ञात कीजिए।
हल-
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3
माना 4 + 3√20i = (x + iy)²
4 + 3√20i = x² – y² + 2xyi
वास्तविक एवं काल्पनिक भागों को अलग-अलग करने पर
x² – y² = 4 ………(1)
एवं 2xy = 3√20 …..(2)
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3

प्रश्न 3.
निम्न के घनमूल ज्ञात कीजिए
(i) -216
(ii) -512
हल-
(i) माना (-216)1/3= x
∴ x = (-216 x 1)1/3
x = (-216)1/3 x (1)1/3
= – 6 x (1, w, w²)
∵ 1, w, w² इकाई के घनमूल हैं।
x = – 6, – 6w, – 6w²

(ii) माना (-512)1/3 = x
∴ x = (-512)1/3 = (-512 x 1)1/3
x = -8 x (1)1/3 = -8(1, w, w²)
x = -8, -8w, -8w²

प्रश्न 4.
सिद्ध कीजिए
(i) 1 + wn + w2n = 0, जबकि n = 2, 4
(ii) 1 + wn + w2n = 3, जबकि n, 3 की गुणज है।
हल-
(i) L.H.S.= 1 + wn +w2n
जब n = 2
तब L.H.S. = 1 + w2 + w4 = 1 + w2 + w3.w
L.H.S. = 1 + w2 + w ∵ w³ = 1
L.H.S. = 0 = R.H.S.
∵ 1 + w + w² = 0
जब n = 4
तब L.H.S. = 1 + w4 + w8
L.H.S. = 1 + w3.w + w3.w3.w3
L.H.S. = 1 +w3.w + (w3)2.w2
L.H.S. = 1 + w + w2 ∵ w3 = 1
L.H.S. = 0 = R.H.S.
∵ 1 + w + w2 = 0

(ii) L.H.S. = 1 + wn + w2n
जब n, 3 का गुणन है, तब
माना n = 3m
∴ L.H.S. = 1 + w3m + w2.3m
L.H.S. = 1 + (w3)m + (w3)2m
= 1 + (1) + (1)m ∵ w3 = 1
L.H.S. = 1 + 1 + 1 = 3 = R.H.S.

प्रश्न 5.
सिद्ध कीजिए
(i)
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3
(ii) (1 + 5w² + w)(1 + 5w + w²)(5 + w + w²) = 64
हल-
(i) माना
RBSE Solutions for Class 11 Maths Chapter 5 सम्मिश्र संख्याएँ Ex 5.3
= (w)29 + (w2)29
= w29 + w58
= (w3)9.w2 + (w3)19.w
= 1.w2 + 1.w ∵ w3 = 1
∵ 1 + w + w2 = 0
= w2 + w = -1 = R.H.S.

(ii) (1 + 5w2 + w)(1 + 5w + w2)(5 + w + w2) = 64
LHS = (1 + 5w2 + w)(1 + 5w + w2)(5 + w + w2)
LHS = (1 + 4w2 + w2 + w)(1 + 4w + w + w2) x (5 + w + w2)
∵ w2 + w + 1 = 0 ⇒ w2 + w = -1
LHS = (1 + 4w2 – 1)(1 + 4w – 1)(5 – 1)
LHS = 4w2 x 4w x 4 = 64w3 = 64 = R.H.S.

प्रश्न 6.
1, w, w² इकाई के घनमूल हो तो सिद्ध कीजिए
(1 + w)(1 + w²)(1 + w4)(1 + w8)……… 2n गुणनखण्ड = 1
हल-
LHS = (1 + w)(1 + w2)(1 + w4)(1 + w8) ……… 2n गुणनखण्ड
LHS = (1 + w)(1 + w2)(1 + w3.w)(1 + w3.w3.w2) ……… 2n गुणनखण्ड
LHS = (1 + w)(1 + w2)(1 + w)(1 + w2) ………. 2n गुणनखण्ड
LHS = (-w2)(-w)(-w2)(-w) ………. 2n गुणनखण्ड
LHS = w3.w3.w3………. n बार
LHS = 1.1.1.1.1.1 …….. n बार = 1

RBSE Solutions for Class 11 Maths

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