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RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

June 12, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

निम्न ( प्रश्न 1 से 5 तक) में प्रत्येक व्यंजक का प्रसार कीजिए।

प्रश्न 1.
(2 – x)³
हल-
हम जानते हैं,
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

प्रश्न 2.
\({ \left( \frac { 2 }{ x } -\frac { x }{ 2 } \right) }^{ 5 }\)
हल-
द्विपद प्रमेय से हम जानते हैं कि
(x + a)n = nC0 xn a0 + nC1 xn-1. a + nC2 xn-2 a2 + ……… + nCr xn-r ar + ……… + an
प्रश्नानुर
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

प्रश्न 3.
\({ \left( \frac { x }{ 3 } +\frac { 1 }{ x } \right) }^{ 6 }\)
हल-
द्विपद प्रमेय के आधार पर हम जानते हैं कि
(x + a)n = nC0 xn a0 + nC1 xn-1. a + nC2 xn-2 a2 + ……… + nCr xn-r ar + ……… + nCn an
प्रश्नानुसार
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

प्रश्न 4.
(3x + 2)4
हल-
द्विपद प्रमेय के आधार पर हम जानते हैं कि
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

प्रश्न 5.
\({ \left( \sqrt { \frac { x }{ a } } -\sqrt { \frac { a }{ x } } \right) }^{ 6 }\)
हल-
द्विपद प्रमेय के आधार पर हम जानते हैं कि
(x + a)n = xn – nC1 xn-1. a + nC2 xn-2 a2 + ……… +(-1)r nCr xn-r ar + ……… +(-1)n an
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.1

द्विपद प्रमेय का प्रयोग करके निम्नलिखित ( प्रश्न 6-9) का मान ज्ञात कीजिए

प्रश्न 6.
(96)³
हल-
96 = 100 – 4 लिखने पर
(96)³ = (100 – 4)³
= 3C0 (100)³ – 3C1 (100)2 (4) + 3C2 (100)1 (4)2 – 3C3 (4)3
= 1000000 – 3 (10000) (4) + 3 (100) 16 – 64
= 1000000 – 120000 + 4800 – 64
= 1004800 – 120064
= 884736

प्रश्न 7.
(101)4
हल-
101 = 100 + 1 लिखने पर
(100 + 1)4 = 4C0 (100)4 + 4C1 (100)3 (1) + 4C2 (100)2 (1)2 + 4C3 (100)1 (1)3 + 4C4 (1)4
= 100000000 + 4 (1000000) + 6 (10000) + 4 (100) + 1
= 100000000 + 4000000 + 60000 + 400 +1
= 104060401

प्रश्न 8.
(99)5
हल-
99 = 100 – 1 लिखने पर
(99)5 = (100 – 1)5
= 5C0 (100)5 – 5C1 (100)4 . 1 + 5C2 (100)3 . (1)2 – 5C3 (100)2 . (1)3 + 5C4 (100)1 (1)4 – 5C5 (1)5
= (100)5 – 5 x (100)4 + 10 x (100)3 – 10 x (100)2 + 5 x 100 – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 10010000500 – 500 100001
= 9509900499

प्रश्न 9.
(1.1)6
हल-
1.1 = 1 + 0.1 लिखने पर
(1.1)6 = (1 + 0.1)6
∴ (1 + 0.1)6 = 6C0 (1)6 + 6C1 (1)5.(0.1) + 6C2 (1)4.(0.1)2 + 6C3 (1)3.(0.1)3 + 6C4 (1)2.(0.1)4 + 6C5 (1)(0.1)5 + (0.1)6
= 1 + 6 x 1 x 0.1 + 15 x 1 x 0.01 + 20 x 1 x 0.001 + 15 x 1 x 0.0001 + 6 x 1 x 0.00001 + 0.000001
= 1 + 0.6 + 0.15 + 0.020 + 0.0015 + 0.00006 + 0.000001
= 1.771561

प्रश्न 10.
द्विपद प्रमेय का प्रयोग करते हुए बताइए कौनसी संख्या बड़ी है। (1.1)10000 या 1000.
हल-
द्विपद प्रमेय की सहायता से (1.1)10000 का विस्तार करने पर
(1.1)10000 = (1 + 0.1)10000
= 10000C0 + 10000C1 (0.1) + 10000C2 (0.1) + अन्य धनात्मक पद
= 1 + 10000 x (0.1) + अन्य धनात्मक पद्
= 1 + 1000 + अन्य धनात्मक पद
= 1001 + अन्य धनात्मक पद
जो कि 1000 से बड़ी संख्या प्राप्त होगी। अतः
(1.1)10000 > 1000

प्रश्न 11.
(a + b)4 – (a – b)4 का विस्तार कीजिए। इसका प्रयोग करके (√3 + √2)4 – (√3 – √2)4 का मान ज्ञात कीजिए।
हल-
प्रश्नानुसार (a + b)4 – (a – b)4
= [4C0 a4 + 4C1 a3b + 4C2 a2b2 + 4C1 ab3 + 4C4.b4] – [4C0 a4 – 4C1 a3b + 4C2 a2b2 – 4C3 ab3 + 4C4 b4]
= (a4 + 4a3b + 6a2b2 + 4ab3 + b4) – (a4 – 4a3b + 6a2b2 – 4ab3 + b4)
= 2 [4a3b – 4ab3]
= 8ab (a2 + b2)
इसमें प्रश्नानुसार a = √3 तथा b = √2 रखने पर
(√3 + √2) – (√3 – √2)
= 8.√3.√2 [(√3)² + (√2)²]
= 8√6 (3 + 2)
= 8√6 (5)
= 40 √6

RBSE Solutions for Class 11 Maths

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