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UP Board Class 11 Maths (8. द्विपद प्रमेय) solution PDF

UP Board Class 11 Maths 8. द्विपद प्रमेय is a Hindi Medium Solution which is prescribed by Uttar Pradesh Board for their students. These Solutions is completely prepared considering the latest syllabus and it covers every single topis, so that every student get organised and conceptual learning of the concepts. Class 11 Students of UP Board who have selected hindi medium as their study medium they can use these Hindi medium textSolutions to prepare themselves for exam and learn the concept with ease.

UP Board Class 11 Maths (8. द्विपद प्रमेय) solution

UP Board Class 11 Maths 8. द्विपद प्रमेय Hindi Medium Solutions - PDF

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अध्याय 8: द्विपद प्रमेय (Binomial Theorem)

प्रश्नावली 8.1

निर्देश (प्रश्न संख्या 1 से 5): प्रत्येक व्यंजक का प्रसार कीजिए।

दिए गए व्यंजक का प्रसार करने के लिए, हम सूत्र (x + y)n = nC0 xn + nC1 xn-1 y + ... + nCn yn तथा (x - y)n = nC0 xn - nC1 xn-1 y + ... + (-1)n nCn yn का प्रयोग करते हैं।

प्रश्न 1. (1 - 2x)5 का प्रसार कीजिए।

हल:
(1 - 2x)5 = 5C0 (1)5 - 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 - 5C3 (1)2 (2x)3 + 5C4 (1) (2x)4 - 5C5 (2x)5
= 1 - 5(2x) + 10(4x2) - 10(8x3) + 5(16x4) - 1(32x5)
= 1 - 10x + 40x2 - 80x3 + 80x4 - 32x5

प्रश्न 2. \(\left(\frac{2}{x} - \frac{x}{2}\right)^5\) का प्रसार कीजिए।

हल:
\(\left(\frac{2}{x} - \frac{x}{2}\right)^5\) = 5C0 \(\left(\frac{2}{x}\right)^5\) - 5C1 \(\left(\frac{2}{x}\right)^4\)\(\left(\frac{x}{2}\right)\) + 5C2 \(\left(\frac{2}{x}\right)^3\)\(\left(\frac{x}{2}\right)^2\) - 5C3 \(\left(\frac{2}{x}\right)^2\)\(\left(\frac{x}{2}\right)^3\) + 5C4 \(\left(\frac{2}{x}\right)\)\(\left(\frac{x}{2}\right)^4\) - 5C5 \(\left(\frac{x}{2}\right)^5\)
= \(\frac{32}{x^5} - 5 \cdot \frac{16}{x^4} \cdot \frac{x}{2} + 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4} - 10 \cdot \frac{4}{x^2} \cdot \frac{x^3}{8} + 5 \cdot \frac{2}{x} \cdot \frac{x^4}{16} - \frac{x^5}{32}\)
= \(\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}\)

प्रश्न 3. (2x - 3)6 का प्रसार कीजिए।

हल:
(2x - 3)6 = 6C0 (2x)6 - 6C1 (2x)5(3) + 6C2 (2x)4(3)2 - 6C3 (2x)3(3)3 + 6C4 (2x)2(3)4 - 6C5 (2x)(3)5 + 6C6 (3)6
= 64x6 - 6(32x5)(3) + 15(16x4)(9) - 20(8x3)(27) + 15(4x2)(81) - 6(2x)(243) + 729
= 64x6 - 576x5 + 2160x4 - 4320x3 + 4860x2 - 2916x + 729

प्रश्न 4. \(\left(\frac{x}{3} + \frac{1}{x}\right)^5\) का प्रसार कीजिए।

हल:
\(\left(\frac{x}{3} + \frac{1}{x}\right)^5\) = 5C0 \(\left(\frac{x}{3}\right)^5\) + 5C1 \(\left(\frac{x}{3}\right)^4\)\(\left(\frac{1}{x}\right)\) + 5C2 \(\left(\frac{x}{3}\right)^3\)\(\left(\frac{1}{x}\right)^2\) + 5C3 \(\left(\frac{x}{3}\right)^2\)\(\left(\frac{1}{x}\right)^3\) + 5C4 \(\left(\frac{x}{3}\right)\)\(\left(\frac{1}{x}\right)^4\) + 5C5 \(\left(\frac{1}{x}\right)^5\)
= \(\frac{x^5}{243} + 5 \cdot \frac{x^4}{81} \cdot \frac{1}{x} + 10 \cdot \frac{x^3}{27} \cdot \frac{1}{x^2} + 10 \cdot \frac{x^2}{9} \cdot \frac{1}{x^3} + 5 \cdot \frac{x}{3} \cdot \frac{1}{x^4} + \frac{1}{x^5}\)
= \(\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}\)

प्रश्न 5. \(\left(x + \frac{1}{x}\right)^6\) का प्रसार कीजिए।

हल:
\(\left(x + \frac{1}{x}\right)^6\) = 6C0 x6 + 6C1 x5}\)\(\left(\frac{1}{x}\right)\) + 6C2 x4}\)\(\left(\frac{1}{x}\right)^2\) + 6C3 x3}\)\(\left(\frac{1}{x}\right)^3\) + 6C4 x2}\)\(\left(\frac{1}{x}\right)^4\) + 6C5 x\(\left(\frac{1}{x}\right)^5\) + 6C6 \(\left(\frac{1}{x}\right)^6\)
= x6 + 6x4 + 15x2 + 20 + \(\frac{15}{x^2}\) + \(\frac{6}{x^4}\) + \(\frac{1}{x^6}\)

निर्देश (प्रश्न संख्या 6 से 9):

द्विपद प्रमेय का प्रयोग करके निम्नलिखित का मान ज्ञात कीजिए।
टिप्पणी: किसी संख्या की घात का मान ज्ञात करने के लिए, हम संख्या को दो भागों में इस प्रकार तोड़ते हैं कि द्विपद प्रमेय लागू हो सके, जैसे 96 = 100 - 4, 102 = 100 + 2, आदि।

प्रश्न 6. (96)3

हल:
(96)3 = (100 - 4)3 = 3C0 (100)3 - 3C1 (100)2(4) + 3C2 (100)(4)2 - 3C3 (4)3
= 1000000 - 3(10000)(4) + 3(100)(16) - 64
= 1000000 - 120000 + 4800 - 64
= 884736

प्रश्न 7. (102)5

हल:
(102)5 = (100 + 2)5
= 5C0 (100)5 + 5C1 (100)4(2) + 5C2 (100)3(2)2 + 5C3 (100)2(2)3 + 5C4 (100)(2)4 + 5C5 (2)5
= 10000000000 + 5(100000000)(2) + 10(1000000)(4) + 10(10000)(8) + 5(100)(16) + 32
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

प्रश्न 8. (101)4

हल:
(101)4 = (100 + 1)4
= 4C0 (100)4 + 4C1 (100)3(1) + 4C2 (100)2(1)2 + 4C3 (100)(1)3 + 4C4 (1)4
= 100000000 + 4(1000000) + 6(10000) + 4(100) + 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401

प्रश्न 9. (99)5

हल:
(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)4 - 5C5 (1)5
= 10000000000 - 5(100000000) + 10(1000000) - 10(10000) + 5(100) - 1
= 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1
= 9509900499

प्रश्न 10. द्विपद प्रमेय का प्रयोग करते हुए बताइए कौन-सी संख्या बड़ी है: (1.1)10000 या 1000?

हल:
(1.1)10000 = (1 + 0.1)10000
= 1 + 10000(0.1) + अन्य धनात्मक पद
= 1 + 1000 + धनात्मक संख्या > 1000
अतः (1.1)10000 > 1000 है।

प्रश्नावली 8.2

निर्देश (प्रश्न संख्या 1 से 2): निम्नलिखित प्रसार में गुणांक ज्ञात कीजिए।

प्रश्न 1. (x + 3)8 में x5 का गुणांक

हल:
व्यापक पद Tr+1 = 8Cr x8-r (3)r
x5 के लिए, 8 - r = 5 ⇒ r = 3
T4 = 8C3 x5 (3)3 = 56 × 27 × x5
अतः गुणांक = 1512

प्रश्न 2. (a - 2b)12 में a5b7 का गुणांक

हल:
व्यापक पद Tr+1 = 12Cr a12-r (-2b)r
a5b7 के लिए, 12 - r = 5 ⇒ r = 7
T8 = 12C7 a5 (-2b)7 = 12C5 a5 (-128) b7
12C5 = 792
गुणांक = 792 × (-128) = -101376

निर्देश (प्रश्न संख्या 3 से 4): निम्नलिखित प्रश्नों के प्रसार में व्यापक पद लिखिए।

प्रश्न 3. (x2 - y)6

हल:
व्यापक पद Tr+1 = 6Cr (x2)6-r (-y)r = 6Cr x12-2r (-1)r yr

प्रश्न 4. (x2 - yx)12, x ≠ 0

हल:
व्यापक पद Tr+1 = 12Cr (x2)12-r (-yx)r = 12Cr x24-2r (-1)r yr xr = 12Cr (-1)r yr x24-r

प्रश्न 5. (x - 2y)12 के प्रसार में चौथा पद ज्ञात कीजिए।

हल:
चौथा पद T4 = T3+1 = 12C3 x9 (-2y)3
= 220 × x9 × (-8y3)
= -1760 x9 y3

प्रश्न 6. \(\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}\), x ≠ 0 के प्रसार में 13वाँ पद ज्ञात कीजिए।

हल:
13वाँ पद T13 = T12+1 = 18C12 (9x)6 \(\left(-\frac{1}{3\sqrt{x}}\right)^{12}\)
= 18C6 (96 x6) \(\left(\frac{1}{3^{12} x^{6}}\right)\)
= 18564 × (531441 x6) × \(\left(\frac{1}{531441 x^{6}}\right)\)
= 18564

विविध प्रश्नावली

प्रश्न 1. यदि (a + b)n के प्रसार में प्रथम तीन पद क्रमशः 729, 7290 तथा 30375 हों, तो a, b और n ज्ञात कीजिए।

हल:
दिया है: nC0 an = 729 ...(i)
nC1 an-1 b = 7290 ...(ii)
nC2 an-2 b2 = 30375 ...(iii)
(ii) को (i) से भाग देने पर: \(\frac{n b}{a} = 10\) ...(iv)
(iii) को (ii) से भाग देने पर: \(\frac{(n-1) b}{2a} = \frac{25}{6}\) ...(v)
(iv) और (v) को हल करने पर: n = 6, a = 3, b = 5
उत्तर: a = 3, b = 5, n = 6

प्रश्न 2. यदि (3 + ax)9 के प्रसार में x2 तथा x3 के गुणांक समान हों, तो a का मान ज्ञात कीजिए।

हल:
x2 का गुणांक: 9C2 37 a2
x3 का गुणांक: 9C3 36 a3
दिया है: 9C2 37 a2 = 9C3 36 a3
⇒ 36 × 37 a2 = 84 × 36 a3
⇒ 108 a2 = 84 a3
⇒ a = \(\frac{108}{84} = \frac{9}{7}\)
उत्तर: a = \(\frac{9}{7}\)

प्रश्न 3. गुणनफल (1 + 2x)6 (1 - x)7 में x5 का गुणांक ज्ञात कीजिए।

हल:
(1 + 2x)6 = 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6
(1 - x)7 = 1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7
गुणनफल में x5 का गुणांक ज्ञात करने पर: 171

प्रश्न 7. (0.99)5 के प्रसार के पहले तीन पदों का प्रयोग करते हुए इसका निकटतम मान ज्ञात कीजिए।

हल:
(0.99)5 = (1 - 0.01)5
≈ 1 - 5(0.01) + 10(0.01)2
= 1 - 0.05 + 0.001
= 0.951

--- अध्याय समाप्त ---

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Other Chapters of Class 11 Maths
1. सम्मुच्य
2. संबंध एवं फलन
3. त्रिकोणमितीय फलन
4. गणितीय आगमन का सिद्धांत
5. सम्मिश्र संख्याएँ और द्विघातीय समीकरण
6. रैखिक असमिकाएँ
7. क्रमचय एवं संचय
8. द्विपद प्रमेय
9. अनुक्रम तथा श्रेणी
10. सरल रेखाएँ
11. शांकव परिच्छेदन
12. त्रिविमीय ज्यामिति का परिचय
13. सीमा और अवकलज
14. गणितीय विवेचन
15. सांख्यिकी
16. प्रायिकता
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