Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2 शेषफल प्रमेय तथा गुणनखण्ड प्रमेय

Ex 6.2 Remainder Theorem and Factor Theorem अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

प्रश्न 1.
बहुपद (25x2 – 1) + (1 – 5x) का एक गुणनखण्ड ज्ञात कीजिए।
हलः
(25x2 – 1) + (1 – 5x)2 = 25x2 – 1 + 1 + 25x2 – 10x = 50x2 – 10x = 10x(5x – 1)
x बहुपद का एक गुणनखण्ड है।

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 2.
बहुपद x3 – 6x2 + 11x – 6 के गुणनखण्ड ज्ञात कीजिए।
हल:
x3 – 6x2 + 11x – 6 में x = 1 रखने पर
शेषफल = (1)3 – 6(1)2 + 11(1) – 6 = 1 – 6 + 11 – 6 = 0
अतः (x – 1) इसका एक गुणनखण्ड है।
इसी प्रकार (x – 2) व (x – 3) भी इसके गुणनखण्ड हैं।

प्रश्न 3.
यदि (x + 1) बहुपद f (x) = 2x2 + kx, का एक गुणनखण्ड है तो k का मान ज्ञात कीजिए। (NCERT Exemplar)
हलः
यदि (x + 1), f(x) = 2x2 + kx का एक गुणनखण्ड है तो x + 1 = 0
∴ x = 0 – 1 = -1 रखने पर f(-1) = 0
f(-1) = 2(-1)2 + k(-1)
0 = 2 – k ⇒ k = 2

प्रश्न 4.
यदि (x – 2) बहुपद 4x3 + 3x2 – 4x + k का एक गुणनखण्ड है तो k का मान ज्ञात कीजिए।
हलः
यदि (x – 2), f(x) = 4x3 + 3x2 – 4x + k का एक गुणनखण्ड है तो x – 2 = 0 या x = 2 रखने पर
∴ f(2) = 0
4(2)3 + 3(2)2 – 4(2) + k = 0
32 + 12 – 8 + k=0
36 + k = 0 ⇒ k = -36

Ex 6.2 Remainder Theorem and Factor Theorem लघु उत्तरीय प्रश्न – I (Short Answer Type Questions – I)

प्रश्न 5.
a का मान ज्ञात कीजिए यदि (x + 1) बहुपद 2x3 – ax2 -(2a – 3)x + 2 का एक गुणनखण्ड है।
हलः
यदि (x + 1) बहुपद 2x3– ax2 – (2a – 3)x + 2 का एक गुणनखण्ड है तो
x + 1 = 0 या x = -1 रखने पर।
शेषफल = 0
2(-1)3 – a(-1)2 – (2a – 3)(-1) + 2 = 0
-2 – a + 2a – 3 + 2 = 0
a – 3 = 0 ⇒ a = 3

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 6.
k का मान ज्ञात कीजिए, यदि (x – 3) बहुपद k2x2 – kx – 2 का गुणनखण्ड है।
हलः
यदि (x – 3), k2x2 – kx – 2 का एक गुणनखण्ड है तो x – 3 = 0 या x = 3 रखने पर
शेषफल = 0
k2. (3)2 – k. 3 – 2 = 0
9k2 – 3k – 2 = 0
9k2 –(6 – 3)k – 2 = 0
9k2 – 6k + 3k – 2 = 0
3k(3k – 2) + 1(k – 2) = 0
(3k – 2)(3k + 1) = 0
यदि 3k – 2 = 0 ∴ k = [latex]\frac{2}{3}[/latex]
यदि 3k + 1 = 0 ∴ k = [latex]\frac{1}{3}[/latex]

Ex 6.2 Remainder Theorem and Factor Theorem लघु उत्तरीय प्रश्न – II (Short Answer Type Questions – II)

Remainder Theorem Calculator is a free online tool that displays the quotient and remainder of division for the given polynomial expressions.

प्रश्न 7.
यदि (x – 1) बहुपद x4 – 3x3 + bx2 + 8x – 4 का एक गुणनखण्ड है, तो b का मान ज्ञात कीजिए।
हलः
यदि (x – 1), x4 – 3x3 + bx2 + 8x – 4 का एक गुणनखण्ड है तो x – 1 = 0 या x = 1 रखने पर
शेषफल = 0
(1)4 – 3(1)3 + b(1)2 + 8(1) – 4 = 0
1 – 3 + b + 8 – 4 = 0
b + 2 = 0
b = 0 – 2 ⇒ b = -2

प्रश्न 8.
सिद्ध कीजिए कि (x – 3) व (x + 4) बहुपद x2 + x – 12 के गुणनखण्ड हैं।
हलः
बहुपद x2 + x – 12 के गुणनखण्ड (x – 3) तथा (x + 4) होंगे।
यदि x – 3 = 0 या x = 3 रखने पर शेषफल = (3)2 + 3 – 12 = 9 + 3 – 12 = 0
यदि x + 4 = 0 या x = -4 रखने पर शेषफल = (-4)2 – 4 – 12 = 16 – 16 = 0
∴ (x – 3) व (x + 4) बहुपद x2 + x – 12 के गुणनखण्ड हैं।

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 9.
गुणनखण्ड प्रमेय का प्रयोग करके जाँचिये कि g(x), बहुपद f(x) का गुणनखण्ड है या नहीं।
(i) f(x) = x3 – 6x2 -19x + 84 तथा g(x) = x – 7
(ii) f(x) = x3 – 3x2 +4x – 4 तथा g(x) = x – 2
(iii) f(x) = 3x4 + 17x3 + 9x2 – 7x – 10 तथा g(x) = x + 5
(iv) f(x) = 2x3 + 4x + 6 तथा g(x) = x + 1
हल:
(i) f(x) = x3 – 6x2 – 19x + 84 तथा g(x) = x – 7
g(x) = x – 7 = 0 या x = 7 का मान f(x) में रखने पर
f(7) = (7)3 – 6(7)2 – 19(7) + 84
= 343 – 294 – 133 + 84
=427 – 427 = 0
अतः g(x), f(x) का एक गुणनखण्ड है।

(ii) f(x) = x3 – 3x2 + 4x – 4 तथा g(x) = x – 2
g(x) = 0 या x – 2 = 0 या x = 2 रखने पर
f(2) = (2)3 – 3(2)2 + 4(2) – 4
= 8 – 12 + 8 – 4
= 16 – 16 = 0
अतः g(x), f(x) का एक गुणनखण्ड है।

(iii) f(x) = 3x4 + 17x3 + 9x2 – 7x – 10 तथा g(x) = x + 5
g(x) = 0 या x + 5 = 0 या x = -5 रखने पर
f(-5) = 3(-5)4 + 17(-5)3 + 9(-5)2 – 7(-5) – 10
= 3 × 625 – 17 × 125 + 9 × 25 + 35 – 10
= 1875 – 2125 + 225 + 25
= 2125 – 2125 = 0
अतः g(x), f(x) का एक गुणनखण्ड है।

(iv) f(x) = 2x3 + 4x + 6 तथा g(x) = x +1
g(x) = 0 या x + 1 = 0 या x = -1 रखने पर
f(-1) = 2(-1)3 + 4(-1) + 6
= -2 – 4 + 6 = 0
अतः g(x), f (x) का एक गुणनखण्ड है।

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 10.
सिद्ध कीजिए कि 2x4 – 6x3 + 3x2 + 3x – 2; x2 – 3x + 2 से पूर्णतया विभाजित है।
हलः
2x4 – 6x3 + 3x2 + 3x – 2 को x2 – 3x + 2 से भाग करने पर
∵ x2 – 3x + 2 = (x – 2)(x -1) यदि x – 2 = 0 या x = 2 रखने पर
शेषफल = 2(2)4 – 6(2)3 + 3(2)2 + 3(2) – 2
= 32 – 48 + 12 + 6 – 2
= 50 – 50 = 0
∴ (x – 2) से पूर्णतया विभाजित है।
यदि x – 1 = 0 या x = 1 रखने पर
शेषफल = 2(1)4 – 6(1)3 + 3(1)2 + 3(1) – 2
= 2 – 6 + 3 + 3 – 2
= 8 – 8 = 0
∴ (x – 1) से पूर्णतया विभाजित है।

प्रश्न 11.
सिद्ध कीजिए कि (x – 1), बहुपद x10 – 1 तथा x11 – 1 का गुणनखण्ड है।
हलः
(x – 1), बहुपद x10 – 1 का गुणनखण्ड होगा। यदि x – 1 = 0 या x = 1 रखने पर
x10 – 1 का शेषफल = (1)10 – 1 = 1 – 1 = 0
∴ (x – 1), x10 – 1 का गुणनखण्ड है।
x11 – 1 का शेषफल = (1)11 – 1 = 1 – 1 = 0
∴ (x – 1), x11 – 1 का गुणनखण्ड है।

प्रश्न 12.
बहुपद 4x3 + 16x2 – x + 5 से क्या घटाया जाये कि ऐसा बहुपद प्राप्त हो जो (x + 5) से पूर्णतया विभाजित हो?
हलः
यदि (x + 5) से 4x3 + 16x2 – x + 5 को पूर्णतया विभाजित किया जाए तो
x + 5 = 0 या x = 0 – 5 = -5 रखने पर
शेषफल = 4(-5)3 + 16(-5)2 – (-5) + 5
= 4(-125) + 16(25) + 5 + 5
= -500 + 400 + 10 = -90

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 13.
गुणनखण्ड प्रमेय के प्रयोग से k का मान ज्ञात कीजिए यदि (x + 2), बहुपद (x + 1)7 + (2x + k)3 का एक गुणनखण्ड है।
हलः
यदि (x + 2), बहुपद (x + 1)7 + (2x + k)3 का एक गुणनखण्ड है तो
x + 2 = 0 या x = 0 – 2 = -2 रखने पर
शेषफल = 0
(-2 + 1)7 + (2 × -2 + k)3 = 0
(-1)7 + (-4 + k)3 = 0
-1 + (-4 + k)3 = 0
(-4 + k)3 = 1
(-4 + k)3 = (1)3
-4 + k = 1
k = 1 + 4 = 5

प्रश्न 14.
m व n के मान ज्ञात कीजिए यदि (x – 1) तथा (x + 2), बहुपद 2x3 + mx2 + nx – 14 के गुणनखण्ड हैं।
हलः
यदि (x – 1), बहुपद 2x3 + mx2 + nx – 14 का एक गुणनखण्ड है तो x – 1 = 0 या x = 1 रखने पर
2(1)3 + m(1)2 + n(1) – 14 = 0
2 + m + n – 14 = 0
m + n = 12 ………….(1)
यदि (x + 2), बहुपद 2x3 + mx2 + nx – 14 का एक गुणनखण्ड है तो x + 2 = 0 या x = -2 रखने पर
2(-2) + m(-2)2 + n(-2) – 14 = 0
-16 + 4m – 2n – 14 = 0
Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2
समीकरण (1) में m का मान रखने पर 9 + n = 12
n = 12 – 9 = 3

Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 15.
α व β के मान ज्ञात कीजिए यदि (x + 1) तथा (x + 2), बहुपद x3 + 3x2 – 2αx + β के गुणनखण्ड हैं।
हलः
यदि (x + 1), x3 + 3x2 – 2αx + β का एक गुणनखण्ड है तो
x + 1 = 0 या x = 0 – 1 = -1 रखने पर
शेषफल = (-1)3 + 3(-1)2 – 2α . (-1) + β = 0
-1 + 3 + 2α + β = 0
2 + 2α + β = 0
2α + β = -2 ………………….(1)
यदि (x + 2), x3 + 3x2 – 2α.x + β का एक गुणनखण्ड है। तो .
x + 2 = 0 या x = 0 – 2 = -2 रखने पर
शेषफल ⇒ (-2)3 + 3(-2)2 – 2α (-2) + β = 0
Balaji Class 9 Maths Solutions Chapter 6 Remainder Theorem and Factor Theorem Ex 6.2

प्रश्न 16.
गणनखण्ड प्रमेय का प्रयोग करके सिद्ध कीजिए कि a + b, b + c, c + a बहुपद (a + b + c)3 – (a3 + b3 + c3) के गुणनखण्ड हैं।
हल:
∵ a + b एक गुणनखण्ड होगा (a + b + c)3 – (a3 + b3 + c3) का
यदि a + b = 0 या a = – b रखने पर,
शेषफल = (-b + b + c)3 – (-b3 + b3 + c3) = c3 – c3 = 0
∴ (a + b) इसका एक गुणनखण्ड है। .
∵ b + c एक गुणनखण्ड है (a + b + c)3 – (a3 + b3 + c3) का
यदि b + c = 0 या b = -c रखने पर,
शेषफल = (a – c + c)3 – (a3 – c3 + c3) = a3 – a3 = 0
∴ (b + c) इसका एक गुणनखण्ड है।
∵ c + a एक गुणनखण्ड है (a + b + c)3 – (a3 + b3 + c3) का।
यदि c + a = 0 या c = -a रखने पर ,
शेषफल = (a + b – a)3 – (a3 + b3 – a) = b3 – b3 = 0
∴ c + a इसका एक गुणनखण्ड है।

Balaji Publications Mathematics Class 9 Solutions

UP Board Solutions for Class 9 Maths Chapter 2 Polynomials

UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद)

These Solutions are part of UP Board Solutions for Class 9 Maths. Here we have given UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद).

प्रश्नावली 2.1

प्रश्न 1.
निम्नलिखित व्यंजकों में कौन-कौन एक चर में बहुपद हैं और कौन-कौन नहीं हैं? कारण के साथ अपने उत्तर दीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-7

प्रश्न 2.
निम्नलिखित में से प्रत्येक में x² का गुणांक लिखिए :
(i) 2 + x² + x
(ii) 2 – x² + x3
(iii) [latex]\frac { \pi }{ 2 }[/latex] x² + x
(iv) √2 x – 1
हल :
(i) 2 + x² + x में x² का गुणांक = 1
(ii) 2 – x² + x3 में x² का गुणांक = -1
(iii) [latex]\frac { \pi }{ 2 }[/latex] x² + x में x² का गुणांक = [latex]\frac { \pi }{ 2 }[/latex]
(iv) √2 x – 1 अर्थात 0.x2 + √2 x – 1 में x² का गुणांक = 0

UP Board Solutions

प्रश्न 3.
35 घात के द्विपद का और 100 घात के एकपदी का एक-एक उदाहरण दीजिए।
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-8

प्रश्न 4.
निम्नलिखित बहुपदों में से प्रत्येक बहुपद की घात लिखिए :
(i) 5x3 + 4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
हल :
(i) 5x3 + 4x² + 7x में चर x की अधिकतम घात = 3
दिए हुए बहुपद की घात= 3
(ii) 4 – y² में चर y की अधिकतम घात = 2
दिए हुए बहुपद की घात = 2
(iii) 5t – √7 में चर है की अधिकतम घात = 1
दिए हुए बहुपद की घात = 1
(iv) 3 एक अचर पद है अर्थात 3.x0
दिए हुए बहुपद की घात = 0

Synthetic Division Calculator, Calculator will divide polynomial by binomial using synthetic divsion.

प्रश्न 5.
बताइए कि निम्नलिखित बहुपदों में कौन-कौन बहुपद रैखिक है, कौन-कौन द्विघाती हैं और कौन-कौन त्रिघाती हैं :
(i) x² + x
(ii) x – x3
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x3
हल :
(i) बहुपद x² + x में चर x की अधिकतम घात = 2
यह बहुपद द्विघाती है।
(ii) बहुपद x – x3 में चर x की अधिकतम घात = 3
यह बहुपद त्रिघाती है।
(iii) बहुपद y + y² + 4 में चर y की अधिकतम घात = 2
यह बहुपद द्विघाती है।
(iv) बहुपद 1 + x में चर x की अधिकतम घात 1 है।
यह बहुपद रैखिक है।
(v) बहुपद 3t में चर है की अधिकतम घात 1 है।
यह बहुपद रैखिक है।
(vi) बहुपद r² में चर r की अधिकतम घात 2 है।
यह बहुपद द्विघाती है।
(vii) बहुपद 7x3 में चर x की अधिकतम घात 3 है।
यह बहुपद त्रिघाती है।

प्रटनावली 2.2

प्रश्न 1.
निम्नलिखित पर बहुपद 5x – 4x² + 3 के मान ज्ञात कीजिए।
(i) x = 0
(ii) x = – 1
(iii) x = 2
हल :
माना बहुपद p (x) = 5 – 4x² + 3
(i) x = 0 पर बहुपद p (x) का मान
p(0)= 5 (0) – 4 (0)² + 3 = 3
(ii) x = -1 पर बहुपद p (x) का मान
p(-1) = 5 (-1) – 4 (-1)² + 3 = – 5 – 4 + 3 = -6
(iii) x = 2 पर बहुपद p (x) का मान
p(2) = 5 (2) – 4 (2)2 + 3 = 10 – 16 + 3 = -3

UP Board Solutions

प्रश्न 2.
निम्नलिखित बहुपदों में से प्रत्येक बहुपद के लिए p (0), p (1) और p (2) ज्ञात कीजिए :
(i) p(y) = y² – y + 1
(ii) p(t) = 2 + t + 2t² – t3
(iii) p(x) = x3
(iv) p(x) = (x – 1)(x + 1)
हल :
(i) p(y) = y² – y + 1
p (0) = 0² – 0 + 1 = 0 – 0 + 1 = 1
p (1) = 1² – 1 + 1 = 1 – 1 + 1 = 1
p(2) = 2² – 2 + 1 = 4 – 2 + 1 = 3
(ii) p(t) = 2 + t + 2t² – t3
p(0) = 2 + 0 + 2 (0)² – (0)3 = 2
p (1) = 2 + 1 + 2 (1)² – (1)3 = 2 + 1 + 2 – 1 = 4
p (2) = 2 + 2 + 2 (2)² – (2)3 = 2 + 2 + 8 – 8 = 4
(iii) p (x) = x3
p(0) = (0)3 = 0
p (1) = (1)3 = 1
p (2) = (2)3 = 8
(iv) p (x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1) = (-1) (1) = -1
p (1) = (1 – 1) (1 + 1) = (0) (2) = 0
p (3) = (2 – 1) (2 + 1) = (1) (3) = 3

प्रश्न 3.
सत्यापित कीजिए कि दिखाए गए मान निम्नलिखित स्थितियों में संगत बहुपद के शून्यक हैं :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-9
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-10
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-1

प्रश्न 4.
निम्नलिखित स्थितियों में से प्रत्येक स्थिति में बहुपद को शून्यक ज्ञात कीजिए :
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax; a ≠ 0
(vii) p (x) = cx + d; c ≠ 0, c, d वास्तविक संख्याएँ हैं।
हल :
(i) बहुपद p (x) = x + 5 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(x) = 0
⇒ x + 5 = 0
⇒ x = – 5
p(3) को शून्यक = – 5
(ii) बहुपद p (x) = x – 5 को शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p (x) = 0
⇒ x – 5 = 0
⇒ x = 5
p(x) का शून्यक = 5
(iii) बहुपद p (x) = 2x + 5 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(3) = 0
⇒ 2x + 5 = 0
⇒ 2x = – 5
⇒ x = [latex]\frac { -5 }{ 2 }[/latex]
p (x) का शून्यके = [latex]\frac { -5 }{ 2 }[/latex]
(iv) बहुपद p (x) = 3x – 2 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p (5) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = [latex]\frac { 2 }{ 3 }[/latex]
p (x) का शून्यक = [latex]\frac { 2 }{ 3 }[/latex]
(v) बहुपद p (x) = 3x का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p (x) = 0
⇒ 3x = 0
⇒ x = 0
p (x) का शून्यक = 0
(vi) बहुपद p(x) = ax; a ≠ 0 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(x) = 0
⇒ ax = 0
⇒ x = 0 (a ≠ 0)
p(x) का शून्यक = 0
(vii) बहुपद p (x) = cx + d, c ≠ 0 का शून्यक ज्ञात करने के लिए इसे शून्य के बराबर रखते हैं।
p(x) = 0
cx + d = 0
cx = -d
x = [latex]\frac { -d }{ c }[/latex] (c ≠ 0)
p (x) का शून्यक = [latex]\frac { -d }{ c }[/latex]

UP Board Solutions

प्रश्नावली 2.3

प्रश्न 1.
x3 + 3x² + 3x + 1 को निम्नलिखित से भाग देने पर शेषफल ज्ञात कीजिए :
(i) x + 1
(ii) x – [latex]\frac { 1 }{ 2 }[/latex]
(iii) x
(iv) x + π
(v) 5 + 2x
हल :
माना p (x) = x3 + 3x² + 3x + 1
(i) माना x + 1 = 0 ⇒ x = -1
p (x) को + 1 से भाग देने पर शेषफल
p(- 1) = (-1)3 + 3(-1)² + 3(-1) + 1 = -1 + 3 – 3 + 1 = 0
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-11
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-12

प्रश्न 2.
x3 – ax² + 6x – a को x – a से भाग देने पर शेषफल ज्ञात कीजिए।
हल :
माना p (x) = x3 – ax² + 6x – a तथा x – a = 0
p (x) को x – a से भाग देने पर शेषफल = (a)3 – a(a)² + 6(a) – a = a3 – a3 + 6a – a = 5a

प्रश्न 3.
जाँच कीजिए कि 7 + 3x, 3x3 + 7x का एक गुणनखण्ड है या नहीं।
हल :
माना p (x) = 3x + 7x
यदि 7 + 3x, p (x) का एक गुणनखण्ड है तो p (x) को 7 + 3x से भाग देने पर शेषफल शून्य होना चाहिए।
माना 7 + 3x = 0 ⇒ 3x = – 7 ⇒ x = [latex]\frac { -3 }{ 7 }[/latex]
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-13

प्रश्नावली 2.4

प्रश्न 1.
बताइए कि निम्नलिखित बहुपदों में से किस बहुपद का एक गुणनखण्ड (x + 1) है।
(i) x3 + x2 + x + 1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 – x2 – (2 + √2) x + √2
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-14

UP Board Solutions

प्रश्न 2.
गुणनखण्ड प्रमेय लागू करके बताइए कि निम्नलिखित स्थितियों में से प्रत्येक स्थिति में g (x), p (x) का एक गुणनखण्ड है या नहीं :
(i) p(x) = 2x3 + x2 – 2x – 1, g (x) = x + 1
(ii) p(x) = x3 + 3x2 + 3x + 1, g (3) = x + 2
(iii) p(x) = x3 – 4x2 + x + 6, g (x) = x – 3
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-15

प्रश्न 3.
k का मान ज्ञात कीजिए जबकि निम्नलिखित स्थितियों में से प्रत्येक स्थिति में (x – 1), p (x) का एक गुणनखण्ड हो :
(i) p(3) = x2 + x + k
(ii) p(x) = 2x2 + kx + √2
(iii) p(x) = kx2 – √2 x + 1
(iv) p(x) = kx2 – 3x + k
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-16
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-2

प्रश्न 4.
गुणनखण्ड ज्ञात कीजिए :
(i) 12x2 – 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-17

प्रश्न 5.
गुणनखण्ड ज्ञात कीजिए :
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-18
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-19

प्रश्नावली 2.5

प्रश्न 1.
उपयुक्त सर्वसमिकाओं को प्रयोग करके निम्नलिखित गुणनफल ज्ञात कीजिए :
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y2 + [latex]\frac { 3 }{ 2 }[/latex]) (y2 – [latex]\frac { 3 }{ 2 }[/latex])
(v) (3 – 2x) (3 + 2x)
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-20
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-3

UP Board Solutions

प्रश्न 2.
सीधे गुणा किए बिना निम्नलिखित गुणनफलों के मान ज्ञात कीजिए :
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-21

प्रश्न 3.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित का गुणनखण्डन कीजिए :
(i) 9x2 + 6xy + y2
(ii) 4y2 – 4y + 1
(iii) x2 – [latex]\frac { { y }^{ 2 } }{ 100 }[/latex]
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-22

प्रश्न 4.
उपयुक्त सर्वसमिकाओं का प्रयोग करके निम्नलिखित में से प्रत्येक का प्रसार कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-4
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-23

प्रश्न 5.
गुणनखण्डन कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-24

प्रश्न 6.
निम्नलिखित घनों को प्रसारित रूप में लिखिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-25
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-26

प्रश्न 7.
उपयुक्त सर्वसमिकाएँ प्रयोग करके निम्नलिखित के मान ज्ञात कीजिए :
(i) (99)3
(ii) (102)3
(iii) (998)3
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-27

UP Board Solutions

प्रश्न 8.
निम्नलिखित में से प्रत्येक का गुणनखण्डन कीजिए।
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-5
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-28

प्रश्न 9.
सत्यापित कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-29

प्रश्न 10.
निम्नलिखित में से प्रत्येक का गुणनखण्डन कीजिए
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-30
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-6

प्रश्न 11.
गुणनखण्ड कीजिए : 27x3 + y3 + z3 – 9xyz
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-31

प्रश्न 12.
सत्यापित कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-32

प्रश्न 13.
यदि x + y + z = 0 हो तो दिखाइए कि x3 + y3 + z3 = 3xyz
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-33

प्रश्न 14.
घनों का परिकलन किए बिना निम्नलिखित में से प्रत्येक का मान ज्ञात कीजिए :
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-34

UP Board Solutions

प्रश्न 15.
नीचे दिए गए आयतों, जिनमें उनके क्षेत्रफल दिए गए हैं, में से प्रत्येक की लम्बाई और चौड़ाई के लिए सम्भव व्यंजक दीजिए।
(i) क्षेत्रफल : 25a2 – 35a + 12
(ii) क्षेत्रफल : 35y2 + 13y – 12
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-35

प्रश्न 16.
घनाभों (Cuboids), जिनके आयतन नीचे दिए गए हैं, की विमाओं के लिए सम्भव व्यंजक क्या हैं :
(i) आयतन : 3x2 – 12x
(ii) आयतन : 12ky2 + 8ky – 20k
UP Board Solutions for Class 9 Maths Chapter 2 Polynomials img-36

We hope the UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद) help you. If you have any query regarding UP Board Solutions for Class 9 Maths Chapter 2 Polynomials (बहुपद), drop a comment below and we will get back to you at the earliest.

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 परिमेयीकरण

Ex 3.2 Rationalisation अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

निम्न के मान ज्ञात कीजिए- (प्रश्न 1 – 6)

प्रश्न 1.
[latex]\frac{1}{\sqrt{4}-\sqrt{3}}[/latex]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 2.
[latex]\frac{1}{3+2 \sqrt{2}}[/latex]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 3.
[latex]\frac{1}{\sqrt{7}-\sqrt{6}}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 4.
[latex]\frac{1}{\sqrt{5}+\sqrt{2}}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

In Algebra, 1/32 to decimals are one of the types of numbers, which has a whole number and the fractional part separated by a decimal point.

प्रश्न 5.
[latex]\frac{1}{\sqrt{7}-2}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 6.
[latex]\frac{1}{\sqrt{7}}[/latex] [NCERT]
हलः
हर का परिमेयीकरण करने पर,
[latex]\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}[/latex]

Ex 3.2 Rationalisation लघु उत्तरीय प्रश्न (Short Answer Type Questions)

निम्न संख्याओं में हर का परिमेयीकरण कीजिए- (प्रश्न 7 – 10)

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 7.
[latex]\frac{1}{8+5 \sqrt{2}}[/latex]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 8.
[latex]\frac{6}{\sqrt{5}+\sqrt{2}}[/latex]
हलः
हर के संयुग्मी से गुणा व भाग करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Click here to get an answer to your question ✍️ How do you write 9/20 as a decimal?

प्रश्न 9.
[latex]\frac{2}{\sqrt{3}-\sqrt{5}}[/latex]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 10.
[latex]\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}[/latex] [NCERT Exemplar]
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 11.
यदि x = 2 + [latex]\sqrt{15}[/latex] तब x + [latex]\frac{1}{x}[/latex] का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 12.
यदि x = 2 + [latex]\sqrt{3}[/latex], तब x2 + [latex]\frac{1}{x^{2}}[/latex] का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 13.
यदि x = 7 + [latex]4 \sqrt{3}[/latex], तब x + [latex]\frac{1}{x}[/latex] का मान ज्ञात कीजिए। .
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 14.
निम्न में से प्रत्येक को परिमेय हर के रूप में व्यक्त कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 15

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 15.
हर का परिमेयीकरण कर, निम्न को सरल कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 16.
निम्न को सरल कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 19
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 17.
निम्न में से प्रत्येक से a तथा b के मान ज्ञात कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 24
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 18.
यदि x = [latex]\frac{\sqrt{5}-2}{\sqrt{5}+2}[/latex] तथा y = [latex]\frac{\sqrt{5}+2}{\sqrt{5}-2}[/latex], तब निम्न के मान ज्ञात कीजिए।
(i) x2
(ii) y2
(iii) xy
(iv) x2 + y2 + xy
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 19.
यदि x = [latex]5-2 \sqrt{6}[/latex], तब निम्न के मान ज्ञात कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 20.
यदि a = [latex]\frac{1}{3-\sqrt{8}}[/latex] , b = [latex]\frac{1}{3+\sqrt{8}}[/latex] तब निम्न के मान ज्ञात कीजिए।
(i) a2
(ii) b2
(iii) ab
(iv) 52 – 6ab + 3b2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 21.
निम्न में से प्रत्येक का मान दशमलव के तीन स्थानों तक ज्ञात कीजिए।
यदि दिया है: [latex]\sqrt{2}[/latex] = 1.4142, [latex]\sqrt{3}[/latex] = 1.732, [latex]\sqrt{5}[/latex] = 2.2360, [latex]\sqrt{6}[/latex] = 2.4445, [latex]\sqrt{10}[/latex] = 3.162
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 30
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 32

प्रश्न 22.
सरल कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 34
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 23.
यदि x = [latex]\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}[/latex] तथा y = [latex]\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}[/latex] तब 3x2 + 4xy-3y2 का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 24.
a तथा b का मान ज्ञात कीजिए।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 37
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 39

प्रश्न 25.
यदि [latex]\frac{\sqrt{3}+1}{\sqrt{3}-1}=a+b \sqrt{3}[/latex], तब a व b के मान ज्ञात कीजिए।
हलः
हर का परिमेयीकरण करने पर,
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAQ 41
दोनों पक्षों की तुलना से a = 2 व b = 1

प्रश्न 26.
यदि [latex]\frac{5-\sqrt{6}}{5+\sqrt{6}}=a-b \sqrt{6}[/latex], तब a व b के मान ज्ञात कीजिए।
हलः
हर का परिमेयीकरण करने पर
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 27.
यदि x =[latex]\frac{1}{3-2 \sqrt{2}}[/latex], y = [latex]\frac{1}{3+2 \sqrt{2}}[/latex], तब xy2 + x2y का मान ज्ञात कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 28.
यदि [latex]\sqrt{3}[/latex] = 1.732 व [latex]\sqrt{5}[/latex] = 2.236, तब [latex]\frac{6}{\sqrt{5}-\sqrt{3}}[/latex] का मान ज्ञात कीजिए।
हलः
हर का परिमेयीकरण करने पर
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Ex 3.2 Rationalisation बहुविकल्पीय (Multiple Choice Questions)

सही विकल्प का चयन कीजिए
प्रश्न 1.
यदि [latex]x+\sqrt{15}=4[/latex] तब [latex]x+\frac{1}{x}[/latex]
(a) 8
(b) 4
(c) 15
(d) इनमें से कोई नहीं
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 2.
[latex]3 \sqrt{5}-\sqrt{7}[/latex] का सरलतम परिमेय गुणनखण्ड है
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 MCQ 46
हल:
[latex]3 \sqrt{5}-\sqrt{7}[/latex] का सरलतम परिमेय गुणनखण्ड = [latex]3 \sqrt{5}-\sqrt{7}[/latex]
अतः विकल्प (c) सही है।

प्रश्न 3.
यदि x = 7 + [latex]4 \sqrt{3}[/latex] व xy = 1, तब [latex]\frac{1}{x^{2}}+\frac{1}{y^{2}}=[/latex] =
(a) 194
(b) 28
(c) 1915
(d) इनमें से कोई नहीं
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

प्रश्न 4.
यदि x = [latex]\sqrt[3]{2+\sqrt{3}}[/latex], तब [latex]x^{3}+\frac{1}{x^{3}}[/latex]
(a) 3
(b) 5
(c) 4
(d) 6
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (c) सही है।

प्रश्न 5.
[latex]\sqrt{3-2 \sqrt{2}}[/latex] का मान
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 MCQ 49
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (c) सही है।

प्रश्न 6.
यदि x = 4 – [latex]\sqrt{15}[/latex], तब x + [latex]\frac{\mathbf{1}}{\mathbf{x}}[/latex] =
(a) 8
(b) 4
(c) 15
(d) 12
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

प्रश्न 7.
7 + [latex]\sqrt{48}[/latex], का धनात्मक वर्गमूल है।
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 MCQ 52
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (a) सही है।

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 8.
यदि [latex]\sqrt{2}[/latex] = 1.414, तब [latex]\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=[/latex] (NCERT Exemplar)
(a) 1.414
(b) 2.07
(c) 0.414
(d) इनमें से कोई नहीं-
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2
अतः विकल्प (c) सही है।

Ex 3.2 Rationalisation स्वमूल्यांकन परीक्षण (Self Assessment Test)

प्रश्न 1.
[latex]\frac{5}{\sqrt{3}-\sqrt{5}}[/latex] = के हर का परिमेयीकरण कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 55

प्रश्न 2.
[latex]\frac{1}{7+3 \sqrt{2}}[/latex] के हर का परिमेयीकरण कीजिए।
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 3.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 57
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 4.
यदि a = 8 + [latex]3 \sqrt{7}[/latex] व b = [latex]\frac{1}{8+3 \sqrt{7}}[/latex], तब सिद्ध कीजिए कि a2 + b2 = 254
हलः
हर का परिमेयीकरण करने पर
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 5.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 60
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 6.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 62
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 7.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 64
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 8.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 66
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 9.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 68
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 10
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 70
हल:
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 11.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 72
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 12.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 74
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 13.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 76
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 14.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 78
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 15.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 80
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 81
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 16.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 83
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 17.
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 85
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

प्रश्न 18.
सिद्ध कीजिए कि
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2 SAT 87
हलः
Balaji Class 9 Maths Solutions Chapter 3 Rationalisation Ex 3.2

Balaji Publications Mathematics Class 9 Solutions

CBSE Sample Papers for Class 10 Maths Paper 11

These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 11.

CBSE Sample Papers for Class 10 Maths Paper 11

Board CBSE
Class X
Subject Maths
Sample Paper Set Paper 10
Category CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 11 of Solved CBSE Sample Paper for Class 10 Maths is given below with free pdf download solutions.

Time allowed: 3 Hours
Maximum Marks: 80 

 General Instructions

  • All questions are compulsory.
  • The question paper consists of 30 questions divided into four sections A, B, C and D.
  • Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
  • There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
  • Use of calculators is not permitted.

Section – A

Question 1.
Write whether the rational number [latex]\frac { 7 }{ 75 } [/latex] will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Question 2.
Find the value (s) of k, if the quadratic equation 3x2 – k √3x + 4 = 0 has equal roots.

Question 3.
Find the eleventh term from the last term of the AP: 27, 23, 19,… ,- 65.

How do you round 214.0822 to the nearest tenth? … Hint: We are given a decimal number.

Question 4.
Find the coordinates of the point on y- axis which is nearest to the point (-2, 5).

Question 5.
In given figure, ST || RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of ∆ PST to the area of ∆ PRQ.
CBSE Sample Papers for Class 10 Maths Paper 11 1

Question 6.
If cos A = [latex]\frac { 2 }{ 5 } [/latex] , find the value of 4 + 4 tan2 A.

Section – B

Question 7.
If two positive integers p and q are written as p = a2 b3 and q = a3 b; a, b are prime numbers, then verily: LCM (p, q) × HCF (p, q) = pq

Question 8.
The sum of first n terms of an AP is given by Sn = 2n2 + 3n. Find the sixteenth term of the A.P.

Question 9.
Find the value (s) of k for which the pair of linear equations kx + y = k2 and x + ky = 1 have infinitely many solutions.

Question 10.
If ( 1, [latex]\frac { p }{ 3 } [/latex] ) is the mid- point of the line segment joining the points (2,0) and ( 0, [latex]\frac { 2 }{ 9 } [/latex] ) then show that the line 5x + 3y + 2 = 0 passes through the point (-1 , 3p)

Question 11.
A box contains cards numbered 11 to 123. A card is drawn at random from the box. Find the probability that the number on the drawn card is

  1. a square number
  2. a multiple of 7.

To convert 10 11/16 to decimal you can use the long division method explained in our article fraction to decimal, which you can find in the header menu.

Question 12.
A box contains 12 balls of which some are red in colour. If 6 more red balls are put in the box and a ball is drawn at random, the probability of drawing a red ball doubles than what it was before. Find the number of red balls in the bag.

Section – C

Question 13.
Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Question 14.
Find all the zeroes of the polynomial 3x4 + 6x3 – 2x2 – 10x – 5 if two of its zeroes are √ [latex]\frac { 5 }{ 3 } [/latex] and – √ [latex]\frac { 5 }{ 3 } [/latex]

Question 15.
Seven times a two digit number is equal to four times the number obtained by reversing the order of its digits. Ifthe difference ofthe digits is 3, determine the number.

Question 16.
In what ratio does the x-axis divide the line segment joining the points (-4,-6) and (-1, 7)? Find the co-ordinates of the point of division.
OR
The points A (4, -2), B (7,2), C (0,9) and D (-3,5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.

Question 17.
In given figure ∠1 = ∠2 and ∆NSQ = ∆MTR, then prove that ∆ PTS ~ PRQ
CBSE Sample Papers for Class 10 Maths Paper 11 2
OR
In an equilateral triangle ABC, D is a point on the side BC such that BD = [latex]\frac { 1 }{ 3 } [/latex] BC. Prove that 9 AD2 = 7 AB2
CBSE Sample Papers for Class 10 Maths Paper 11 3

Question 18.
In given figure XY and X’ Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’ Y’ at B. Prove that ∠ AOB = 90°.
CBSE Sample Papers for Class 10 Maths Paper 11 4

Question 19.
CBSE Sample Papers for Class 10 Maths Paper 11 5

Question 20.
In given figure ABPC is a quadrant of a circle ofradius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region
CBSE Sample Papers for Class 10 Maths Paper 11 6

Question 21.
Water in a canal, 6 m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
OR
A cone of maximum size is carved out from a cube of edge 14 cm. Find the surface area of the remaining solid after the cone is carved out.

Question 22.
Find the mode of the following distribution of marks obtained by the students in an examination :

Marks obtained 0-20 20-40 40-60 60-80 80-100
Number of students 15 18 21 29 17

Given the mean of the above distribution is 53, using empirical relationship estimate the value of its median.

Question 23.
A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train.
OR
Check whether the equation 5x2 – 6x – 2 = 0 has real roots and if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.

Question 24.
An AP consists of 3 7 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Question 25.
Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
OR
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Question 26.
Draw a triangle ABC with side BC = 7 cm, ∠B=45°, ∠A= 105°. 1 Then, construct a triangle whose sides are [latex]\frac { 4 }{ 3 } [/latex] times the corresponding sides of ∆ABC.

Question 27.
CBSE Sample Papers for Class 10 Maths Paper 11 7

Question 28.
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Question 29.
Two dairy owners A and B sell flavoured milk filled to capacity in mugs of negligible thickness, which are cylindrical in shape with a raised hemispherical bottom. The mugs are 14 cm high and have diameter of 7 cm as shown in given figure. Both A and B sell flavoured milk at the rate of ₹ 80 per litre. The dairy owner A uses the formula nrh to find the volume of milk in the mug and charges ₹ 43.12 for it. The dairy owner B is of the view that the price of actual quantity of milk should be charged. What according to him should be the price of one mug of milk? Which value is exhibited by the dairy owner B ? [ use π = [latex]\frac { 22 }{ 7 } [/latex] )
CBSE Sample Papers for Class 10 Maths Paper 11 8

Question 30.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency k.

Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Number of children 3 6 9 13 k 5 4

The following frequency distribution shows the distance (in metres) thrown by 68 students in a Javelin throw competition.

Distance (in m) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Number of students 4 5 13 20 14 8 4

Draw a less than type O give for the given data and find the median distance thrown using this curve.

Solutions

Solution 1.
Non terminating repeating decimal expansion.

Solution 2.
k = ± 4

Solution 3.
a11 = -25

Solution 4.
(0,5)

Solution 5.
9 : 49

Solution 6.
4 + tan2 A = 4 sec2 A = 4 ( [latex]\frac { 5 }{ 2 } [/latex] ) 2 = 25

Solution 7.
LCM (p, q) = a3 b3
HC,F (p, q) = a2 b
LCM (p, q) × HCF (p, q) = a5 b4 = (a2 b3)
(a3 b) = pq

Solution 8.
Sn = 2n2 + 3n
S1 = 5 = a1
S2 = a1 + a2 = 14 ⇒ a2 = 9
d= a2 – a1 = 4
a16 = a1 + 15d = 5+ 15(4) = 65

Solution 9.
CBSE Sample Papers for Class 10 Maths Paper 11 9

Solution 10.
CBSE Sample Papers for Class 10 Maths Paper 11 10

Solution 11.

  1. P (square number) = [latex]\frac { 8 }{ 113 } [/latex]
  2. P (multiple of 7) = [latex]\frac { 16 }{ 113 } [/latex]

Solution 12.
Let number of red balls be = x
∴ P (red ball) = [latex]\frac { x }{ 12 } [/latex]
If 6 more red balls are added: The number of red balls = x + 6 P (red ball) = [latex]\frac { x+6 }{ 18 } [/latex]
Since, [latex]\frac { x+6 }{ 18 } [/latex] = 2 ( [latex]\frac { x }{ 12 } [/latex] ) ⇒ x = 3
There are 3 red balls in the bag.

Solution 13.
Let n = 3k, 3k + 1 or 3k+ 2.
(i) When n = 3k: n is divisible by 3.
n + 2 = 3k + 2 ⇒ n + 2 is not divisible by 3.
n + 4 = 3k + 4 = 3 (k + 1)+ 1 ⇒ n + 4 is not divisible by 3.

(ii) When n = 3k + 1: n is not divisible by 3.
n + 2 = (3k + 1) + 2 = 3k+ 3 = 3 (k+ 1) ⇒ n + 2 is divisible by 3.
n + 4 = (3k+ l) + 4 = 3k + 5 = 3(k+ 1) + 2 ⇒ n + 4isnotdivisibleby3.

(iii) When n = 3k + 2: n is not divisible by 3.
n + 2 = (3k+2) + 2 = 3k + 4 = 3(k+ 1)+ 1 ⇒ n + 2 is not divisible by 3.
n + 4 = (3k + 2)+ 4 = 3k+ 6 = 3(k + 2) ⇒ n + 4 is divisible by 3.
Hence exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

Solution 14.
CBSE Sample Papers for Class 10 Maths Paper 11 11
CBSE Sample Papers for Class 10 Maths Paper 11 12

Solution 15.
Let the ten’s and the units digit be y and x respectively.
So,the number is 10 y + x.
The number when digits are reversed is 10 x + y.
Now, 7(10y+ x) = 4(10x + y) ⇒ 2y = x …( 1)
Also x – y3 …(ii)
Solving (1) and (2),we get y = 3 and x = 6.
Hence the number is 36.

Solution 16.
CBSE Sample Papers for Class 10 Maths Paper 11 13

Solution 17.
∠SQN = ∠TRM (CPCT as NSQ ≅ MTR)
CBSE Sample Papers for Class 10 Maths Paper 11 14
Since, ∠P + ∠1 + ∠2 = ∠P + ∠PQR + ∠PRQ (Angle sum property)
⇒ ∠1 + ∠2 = ∠PQR + ∠PRQ
⇒ 2 ∠1 = 2 ∠PQR (as ∠1 = ∠2 and ∠PQR= ∠PRQ) ⇒ ∠1 = ∠PQR
Also ∠2 = ∠PRQ and ∠SPT = ∠QPR (common) ∆PTS ~ ∆PRQ (By AAA similarity criterion)
CBSE Sample Papers for Class 10 Maths Paper 11 15

Solution 18.
CBSE Sample Papers for Class 10 Maths Paper 11 16

Solution 19.
CBSE Sample Papers for Class 10 Maths Paper 11 17
CBSE Sample Papers for Class 10 Maths Paper 11 18

Solution 20.
CBSE Sample Papers for Class 10 Maths Paper 11 19

Solution 21.
CBSE Sample Papers for Class 10 Maths Paper 11 20
CBSE Sample Papers for Class 10 Maths Paper 11 21

Solution 22.
CBSE Sample Papers for Class 10 Maths Paper 11 22

Solution 23.
CBSE Sample Papers for Class 10 Maths Paper 11 23
CBSE Sample Papers for Class 10 Maths Paper 11 24

Solution 24.
Let the three middle most terms of the AP be a – d, a, a + d
We have, (a – d) +a + (a + d) = 225
⇒ 3 a = 225 ⇒ a = 75
Now, the AP is a – 18d,…,a – 2d, a – d, a, a + d,a + 2d,…,a + 18d
Sum of last three terms :
(a + 18d) + (a + 17d) + (a + 16d) = 429
⇒ 3a + 51 d = 429 ⇒ (a + 17d) = 143
⇒ 75 + 17d = 143 ⇒ d = 4
Now, first term = a – 18d = 75 – 18(4) = 3
∴ The AP is 3,7,11,…, 147.

Solution 25.
Given : A right triangle ABC right angled at B.
To prove : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof : In ∆ ADB and ∆ ABC
∠ADB = ∠ABC (each 90°)
∠BAD = ∠CAB (common)
CBSE Sample Papers for Class 10 Maths Paper 11 25
CBSE Sample Papers for Class 10 Maths Paper 11 26
CBSE Sample Papers for Class 10 Maths Paper 11 27
CBSE Sample Papers for Class 10 Maths Paper 11 28

Solution 26.
Draw ∆ABC in which BC = 7 cm, ∠B = 45°, ∠A = 105° and hence ∠C = 30°.
Construction of similar triangle A’ BC’ as shown below.
CBSE Sample Papers for Class 10 Maths Paper 11 29

Solution 27.
CBSE Sample Papers for Class 10 Maths Paper 11 30

Solution 28.
CBSE Sample Papers for Class 10 Maths Paper 11 31
CBSE Sample Papers for Class 10 Maths Paper 11 32

Solution 29.
CBSE Sample Papers for Class 10 Maths Paper 11 33
CBSE Sample Papers for Class 10 Maths Paper 11 34

Solution 30.
CBSE Sample Papers for Class 10 Maths Paper 11 35
CBSE Sample Papers for Class 10 Maths Paper 11 36
CBSE Sample Papers for Class 10 Maths Paper 11 37

We hope the CBSE Sample Papers for Class 10 Maths paper 11 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths paper 11, drop a comment below and we will get back to you at the earliest.

UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics

UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics

Discrete Mathematics Long Answer Type Questions (8 Marks)

Question 1.
How are characters created by Binary Numbers? What are its different codes? Explain with examples. (UP 2011)
Answer:
Computer Code: Computer codes are used to convert data into binary form to make the computer understand it. Apart from this, they are responsible for error-free signal flow in the computer. Three popular computer codes are:
BCD (Binary Coded Decimal): It is one of the earliest developed (UPBoardSolutions.com) memory codes. In this, every digit is converted into binary form separately:
e.g.,
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 1
But four-bit code can handle only 24 = 16 dIfferent characters that are why it is extended to 6-bit code and It can handle 26 = 64 different characters. It is 6-bit code and divided into two parts i.e., zone bit and character code.
Zone bit consists of Z bits and character zone consists of 4 bits.
To understand more look at the table given below:

UP Board Solutions
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 2
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 3
EBCDIC (Extended Binary Coded Decimal Interchange Code): BCD can convert only 64 characters but we use more than 64 characters in computer to represent data. To overcome this problem, 2 more bits have been added to the zone bit to develop new 8-bit code and, that is why it is known as extended binary coded decimal interchange code. EBCDIC is 8- bit code which can encode 28 = 256 different characters. It is similar to BCD in working but it has 4 bits in bit zone. To understand more table is given below:

UP Board Solutions
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 4
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 5
ASCII (American Standard Code for Information Interchange): This code is the most popular and widely accepted computer code. It is the standard code for computers, developed by the American National Standards Institute in the year 1963 for (UPBoardSolutions.com) encoding different characters in the computer. It is used by almost every manufacturing company. ASCII codes are of two types:
(a) 7-bit Code: To encode 27 = 128 characters with 3 bits in zone bit and four in character zone.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 6
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 7
(b) 8-bit code: To encode 28 = 256 characters with 4 bits in bit zone and 4 bits in character zone.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 8
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 9

UP Board Solutions

What is 5/8 as a decimal you ask? Converting the fraction 5/8 into a decimal is very easy.

Question 2.
Define character representation in computers. (UP 2008)
Or
What is meant by “character representation”? Explain one such code in detail. (UP 2009)
Or
What is meant by character coding? Explain one coding methods in detail. (UP 2016)
Or
What is Character Representation? (UP 2018)
Answer:
Character Representation: Physical devices used to store and process data in computers are two-state devices. A switch, for example, is a two-state device. It can be either ON or OFF. Electronic devices such as transistors used in computers must function reliably when operated as switches. Thus, all data to be stored and processed in computers are transformed or coded as strings of two symbols, one symbol to represent each state.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 10
Coding of characters has been standardized to facilitate the exchange of recorded data between computers. The most popular standard is known as ASCII. Each letter is a unique combination of Binary Digits (BITS). That is, each letter (UPBoardSolutions.com) is a group of charged and uncharged transistors and it is grouped in such a way that a particular combination represents a specific character. A group of 8 BITS which is used to represent a character is called a byte. The length of 1 word is called word length which ranges from 1 byte to 64 bytes.
The internal code representation of string HARSH is:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 11

UP Board Solutions

Question 3.
Explain the Number System.
Answer:
Number System: Number systems are very important to understand because the design and organization of a computer system depend on it.
Number systems are basically of two types:
1. Non-positional Number System: In this number system, each symbol represents the same value regardless of its position in the number and the symbols are simply added to find out the value of a particular number. Since it is very difficult to perform arithmetical operations with such a number system, positional number systems have been developed.

2. Positional Number System: In a positional number system, there are only a few symbols (UPBoardSolutions.com) called digits, and these symbols represent different values depending on the position they occupy in the number.

The value of each digit in such a number is determined by three considerations :

  1. The digit itself
    Face Value: The face value of a digit always remains the same regardless of its position in the number, e.g., the face value of 4 in 554, 40567 etc. is 4.
  2. The position of the digit in the number.
    Place Value: Place value of a digit changes due to change in its position, e.g., place value of 2 in 4210 is 2 hundred, in 32,450 it is 2 thousand, etc.
  3. The base of the number system (where the base is defined as the total number of digits available in the number system), e.g., Decimal number system has base 10 since it includes only 10 digits 0, 1, 2, ….., 9, to represent any number.

UP Board Solutions

The various positional systems in use are:

  1. Binary number system
  2. Octal number system
  3. Decimal number system
  4. Hexadecimal number system.

Convert fraction 6 and 1/2 to decimal. What is 6 1/2 as a decimal? Answer: 6.5.

Question 4.
What are the logical operators? What are their different types? Explain operators making their Truth Table. (UP 2007, 08, 10)
Answer:
Logical Operators: AND, OR, and NOT are logical operators. Since these operators are operated on logical values 0 and 1, that is why these operators are called logical operators.

AND Operator: An AND operator is represented by the symbol ‘.’. Basically AND operator is used to performing logical multiplication. A, B, and C are three logical variables, where A, B, are input variables and C is the output variable. We can define the AND operator by listing all possible combinations of A and B and the resulting value of C in the operation A.B = C.

It may be noted that since the variables A and B can have only two possible values (0 or 1) so only four (22) combinations of inputs are possible as shown in the following table. The resulting output values for each of the four input combinations (UPBoardSolutions.com) are given in the table. Such a table is known as the truth table. Thus, the table is the truth table for the logical AND operator.

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UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 12
As we can observe from the truth table that in AND operation, the output will be 1 when all inputs are 1 else output will be 0.
OR Operator: An OR operator is represented by the symbol V. Basically an OR operator is used to perform logical addition. As in the previous example, A and B are input variables and C its output variable. We can define the OR operator by listing all possible combinations of A and B and the resulting value of C in the equation A + B = C. The truth table for Logical OR operator is shown in the following Table:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 13
As we can observe from the truth table that in logical addition, the output will be 1 when any one input is 1. It means if all inputs are 0 then the output will be 0.

NOT Operator: The two operators (AND and OR) are binary operators because they operate on two variables. NOT operator denoted by is a Unary operator because it operates on a single variable. NOT operator is also known as complementation operator or inverse operator.
Thus, complement of A is [latex]\bar { A } [/latex]. Complement of (A + B) is [latex]\bar { (A+B } )[/latex]. If value
of [latex]\bar { A } [/latex] is 0 then value of A is 1 and if value of A is 1 then value of [latex]\bar { A } [/latex] is 0. (UPBoardSolutions.com) The truth table for logical NOT operator is shown in table.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 14

Question 5.
Write about the postulates of Boolean Algebra.
Answer:
Postulates of Boolean Algebra: Boolean Algebra is an algebraic structure defined on a set of elements B together with two binary operators + and . provided the following postulates are satisfied:
(1) (a) Closure with respect to the operator +
(b) Closure with respect to the operator.

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(2) (a) An identity element with respect to +, designated by
0 : X + 0 = 0 + X = X.
(b) An identify element with respect to designated by
1 : X . 1 = 1 . X = X.

(3) (a) Commutative with respect to + : X + Y = Y + X
(b) Commutative with respect to . : X . Y = Y . X

(4) (a) . is distributive over : : X . (Y + Z) = (X . Y) + (X . Z)
(b) + is distributive over . : X + (Y . Z) = (X + Y) . (X + Z)

(5) For every element X ∈ B, there exists an element [latex]\bar { X } [/latex] ∈ B such that:
(a) X × [latex]\bar { X } [/latex] = 1
(b) X . [latex]\bar { X } [/latex] = 0
The postulates listed above are called Huntington (1904) Postulates and need (UPBoardSolutions.com) no proof. They are used to prove the theorems of Boolean Algebra.

Question 6.
What are the different Gates in Boolean Algebra? (UP 2004, 05, 07)
Or
What is ‘Truth Table’? How is it helpful in understanding GATES? (UP 2006)
Or
Explain the working of a NAND Gate. Give its two application. (UP 2009, 10)
Or
How can you show that NAND is a Universal Gate? Explain with diagrams and truth tables. (UP 2011, 19)
Answer:
Truth Table: A table that shows all the input-output possibilities of a logic circuit is called a truth table.
There are several types of the truth table. AND, OR, NOT, NAND, NOR, XOR, XNOR Gates are described below:
(a) AND Gate: In English language, Input A is ANDed with Input B to get output Y.

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UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 15
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 16
The truth table illustrates four ways to express the logical ANDing of A and B.
The AND Gate works on the principle that output will be high when all the inputs (UPBoardSolutions.com) are high otherwise output will be low.

(b) OR Gate: In OR Gate, input A is ORed with input B to get output Y.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 17
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 18
The OR Gate works on the principle that, if anyone input is high, the output will be high. Thus, the only case when output will be low is when all inputs are low i.e., 0.

(c) NOT Gate: The NOT Gate is an electronic circuit that generates an output signal which is the reverse of the input signal. A NOT gate is also known as an inverter because it inverts the input.

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UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 19
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 20

(d) NAND Gate: A NAND Gate is a complemented AND gate. That is, the output of NAND Gate will be 1 if anyone of the inputs is 0 and will be 0 when all the inputs are 1.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 21
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 22

(e) NOR Gate: A NOR Gate is a complemented OR Gate. That is, the output of a NOR Gate will be 1 only when all inputs are 0 and will be 0 if any input represents a 1.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 23
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 24

(f) XOR Gate (Exclusive-OR Gate): XOR Gate is a combination of AND, OR, and NOT (UPBoardSolutions.com) Gates. symbol denotes XOR operation. This Gate works on the principle that if an odd number of inputs are 1, the output will be 1 otherwise output will be 0.

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UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 25
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 26
As observed from the truth table, the output is 1 when odd numbers of inputs are 1.

(g) XNOR Gate (Exclusive-NOR Gate): Similarly, XNOR gate is also formed with a combination of AND, OR, and NOT gates. symbol denotes XNOR operation. Since this gate is the inverse of XOR gate, the output will be 0 when odd numbers of inputs are 1 otherwise output will be 1.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 27
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 28

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Question 7.
Explain the basic features of ASCII Code. (UP 2005, 08, 09, 10)
Or
Explain in detail the features of the ASCII character code. (UP 2007)
Answer:
ASCII: Binary numbers are coded to represent characters in the computer memory. Several codes are used for this purpose. One most commonly used code is the American Standard Code for Information Interchange (ASCII). ASCII has been adopted by several American computer manufacturers as their computer’s internal code. This code is popular in data communications, is used (UPBoardSolutions.com) almost exclusively to represent data internally in microcomputers, and is frequently found in the larger computers produced by some vendors.

ASCII is of two types: ASCII-7 and ASCII-8. ASCII-7 is a 7-bit code that represents 128 (27) different characters.
ASCII-8 is an extended version of ASCII-7. It is an 8-bit code that represents 256 (28) different characters rather than 128.
e.g. (i) A is given ASCII code 65.

Now if we convert 65 into Binary form we get 01000001 → 1 byte
In the same way, every character has its own ASCII value after converting into binary code stored on the computer.

Question 8.
Describe various Binary Arithmetic Operations. (UP 2008, 09, 11)
Or
What is binary arithmetic? Explain with suitable example. (UP 2017)
Answer:
Four basic arithmetic operations are performed inside a computer using binary numbers. These are addition, subtraction, multiplication, and division. Since binary numbers are made up of 0’s and 1’s, results of arithmetic operations are also in 0’s and 1’s only.

Binary Addition: Binary addition is performed in the same manner as decimal addition. However the binary system has only two digits, the addition table for binary arithmetic is very simple, consisting of only four entries. The complete table for binary addition is as follows:
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

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Plus a carry of 1 to next higher column. Carryovers are performed in the same manner as in decimal arithmetic. Since 1 is the largest digit in the binary system, any sum greater than 1 requires that a digit be carried over.
Example:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 29
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 30
Binary Subtraction: From the following table, it is clear that the lower digit is subtracted (UPBoardSolutions.com) from the upper digit. If the lower digit is larger than the upper digit, it is necessary to borrow from the column to the left which equals to 2 (10).
0 – 0 = 0
1 – 0 = 1
1 – 1 = 0
0 – 1 = 1
with borrow from the next column. Thus, the only case in which it is necessary to borrow is when 1 is subtracted from 0.
Example
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 31
Binary Multiplication: Multiplication in the binary system also follows the same general rules as decimal multiplication. The table for binary multiplication is as follows:
0 × 0 = 0
0 × 1 = 0
1 × 0 = 0
1 × 1 = 1
Example:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 32

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Binary Division: Binary division is, again, very simple. As in the decimal system (or in any other system), division by zero is meaningless, here too. Hence, the complete table for the binary division is as follows:
0/1 = 0
1/1 = 1
The division process is performed in a manner similar to the decimal division.
Example:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 33

Discrete Mathematics Short Answer Type Questions (4 Marks)

Question 1.
What is the order of precedence in Boolean Algebra? (UP 2007, 09, 19)
Answer:
In a Boolean expression, many operators are used. The order in which they are operated is known as precedence. The precedence of Boolean operators is as follows:

  1. The expression is scanned from left to right.
  2. Expressions enclosed within parentheses are evaluated first.
  3. All complement (NOT) operations are performed next.
  4. All ‘.’ (AND) operations are performed after that.
  5. Finally, all ‘+’ (OR) operations are performed in the end.

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Question 2.
Define the Principal of Quality. (UP 2016)
Answer:
The Huntington Postulates have been listed in two parts: (a) and (b). One part may be obtained from the other if ‘+’ is interchanged ‘+’ with ‘.’ and ‘0’ in interchanged with ‘1’ and vice-versa. This important property of Boolean Algebra (UPBoardSolutions.com) is called Principle of Quality. This principle ensures that, if a theorem is proved using the postulates, then a dual theorem obtained interchanging ‘+’ with ‘.’ and ‘0’ with ‘1’ automatically holds and need not be proved separately.
The table below lists theorems and their corresponding dual theorems.
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 34

Question 3.
Write a note on De Morgan’s Theorems to prove it.
Answer:
Theorem (a): De Morgan’s Theorems: (x + y)’ = x’. y’
Proof: The truth table for proving this theorem is given below:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 35
From the truth table, it is clear that both sides of the theorem are equal. Hence, the theorem is proved.
Theorem (b): (x + y)’ = x’ + y’
Proof: The truth table for proving this theorem is given below:
UP Board Solutions for Class 10 Computer Science Chapter 4 Discrete Mathematics 36

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From the truth table, It is clear that both sides of the theorem are equal. Hence, the theorem is proved.
Theorems 6(a) and 6(b) are very important and useful. They are known as De Morgan’s theorems. They can be extended to n variables as given below:
(X1 + X2 + X3 + ………. + Xn)’ = X1‘ . X2‘ . X3‘ …….. Xn
(X1 . X2 . X3. …… Xn)’ = X1‘ + X’2 + X3‘ + ……….. + Xn

Question 4.
Write AND and OR LAWS of Discrete mathematics.
Answer:
AND LAWS: AND LAWS are the laws which work on logical multiplication. They are:
1. x . 1 = x
2. x . x’ = 0
3. x . x = x
4. x . 0 = 0
“The tabular representations of truth values of a compound statement based on the truth values of the prime connective ness of statements is called TRUTH TABLE.”
Truth table consists of horizontal lines (rows) and vertical lines (columns). If a compound statement consists of N statements, the number of rows will be 2^N. The number of columns in a truth table depends upon the number of relationships between these statements.

Discrete Mathematics Very Short Answer Type Questions (2 Marks)

Question 1.
Discuss De-Morgan’s Theorem. (UP 2014)
Answer:
First Theorem: This theorem states that the complement of a sum of the binary variable is equal to the product of the complement of the binary variables.
Second Theorem: The theorem states that the complement of a product of binary (UPBoardSolutions.com) variable is equal to the sum of the complement of the binary variable

Question 2.
What is the full form of ASCII? (UP 2014)
Answer:
The full form of ASGII is American Standard Code for Information Interchange.

Question 3.
If A = 0 and B = 1, then find the value of y from the following expression:
Y = (A . B)
Answer:
Y = [latex]\bar { (0.1 } )[/latex] = [latex]\bar { (0 } )[/latex] = 1

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Question 4.
Give the name of the Boolean operators. (UP 2014)
Answer:
AND Operator, Or operator and NOT operator.

Question 5.
Write a full form of EBCDIC. (UP 2017)
Answer:
Extended Binay Coded Decimal Interchange Code.

Discrete Mathematics Objective Type Questions (1 Marks)

There are four alternative answers for each part of the questions. Select the (UPBoardSolutions.com) correct one and write in your answer book:

Question 1.
Each letter is a unique combination of:
(a) Bits
(b) Bytes
(c) Word length
(d) Binary.
Answer:
(a) Bits

Question 2.
A group of 8 bits which is used to represent a character is called :
(a) Bits
(b) Bytes
(c) Integer
(d) None of these.
Answer:
(b) Bytes

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Question 3.
The most popular standard is known as:
(a) BCD
(b) ABC
(c) ASC
(d) ASCII
Answer:
(d) ASCII

Question 4.
In the hexadecimal number system, the base :
(a) 8
(b) 10
(c) 16
(d) None of these.
Answer:
(c) 16

Question 5.
The binay equivalent of the number (15)10. (UP 2014)
(a) (1101)
(b) (1110)2
(c) (1111)2
(d) (1000)2.
Answer:
(a) (1101)

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Question 6.
Which logic gate has only one input and one output? (UP 2015)
(a) NOT
(b) NOR
(c) OR
(d) AND.
Answer:
(a) NOT

Question 7.
The value of the binary number (1010)2 would be?
(a) (14)10
(b) (12)10
(c) (10)10
(d) (11)10
Answer:
(c) (10)10

Question 8.
What is binary equivalent of [31]10. (UP 2017)
(a) 10000
(b) 11111
(c) 100000
(d) 11110.
Answer:
(b) 11111

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Question 9.
Which of the following is a single input logic gate? (UP 2018)
(a) NAND
(b) AND
(c) NOT
(d) NOR.
Answer:
(c) NOT

UP Board Solutions for Class 10 Computer Science