4. Solved In-Text Think & Reflect Proofs
4. हल किए गए In-Text Think & Reflect Proofs
Think & Reflect · p.71 (1) What can you say about $a$ and $b$ if $(a+b)^2 < a^2+b^2$? (2) What if $(a+b)^2 > a^2+b^2$? (3) When will $(a+b)^2 = a^2+b^2$? [NCERT p.71]
Think & Reflect · p.71 (1) यदि $(a+b)^2 < a^2+b^2$ हो तो $a$ और $b$ के बारे में क्या कह सकते हैं? (2) यदि $(a+b)^2 > a^2+b^2$ हो तो? (3) $(a+b)^2 = a^2+b^2$ कब होगा? [NCERT p.71]
1. Algebraic Relation $(a+b)^2 = a^2 + b^2 + 2ab \implies (a+b)^2 - (a^2+b^2) = 2ab$. The sign of the cross-term $2ab$ decides which is larger.
2. Part (1) $(a+b)^2 < a^2+b^2$ This occurs when $2ab < 0 \implies ab < 0$. That means $a$ and $b$ have opposite signs (one is positive, one is negative).
3. Part (2) $(a+b)^2 > a^2+b^2$ This occurs when $2ab > 0 \implies ab > 0$. That means $a$ and $b$ have the same sign (both positive or both negative).
4. Part (3) $(a+b)^2 = a^2+b^2$ This occurs when $2ab = 0 \implies$ at least one of $a$ or $b$ is zero ($a = 0$ or $b = 0$).
🎯 Conclusion: (1) $a, b$ opposite signs; (2) $a, b$ same signs; (3) $a=0$ or $b=0$.
चरण 1: Relation की पहचान $(a+b)^2 = a^2 + b^2 + 2ab$ होता है, इसलिए अंतर $(a+b)^2 - (a^2+b^2) = 2ab$ है। $2ab$ का चिन्ह ही तय करता है कि कौन बड़ा है।
चरण 2: भाग (1) का हल $(a+b)^2 < a^2+b^2$ तब होगा जब $2ab < 0$ हो, अर्थात् $a$ और $b$ के opposite signs हों (एक धनात्मक, एक ऋणात्मक)।
चरण 3: भाग (2) का हल $(a+b)^2 > a^2+b^2$ तब होगा जब $2ab > 0$ हो, अर्थात् $a$ और $b$ के समान चिन्ह हों (दोनों positive या दोनों negative)।
चरण 4: भाग (3) का हल $(a+b)^2 = a^2+b^2$ तब होगा जब $2ab = 0$ हो, अर्थात् $a$ या $b$ में से कम से कम एक zero हो ($a = 0$ या $b = 0$)।
🎯 उत्तर: (1) $a, b$ opposite signs; (2) $a, b$ समान चिन्ह; (3) $a = 0$ या $b = 0$।
Think & Reflect · p.73 Explain geometrically how $(a - b)^2 = a^2 - 2ab + b^2$ is derived by area subtraction. Why is $b^2$ added back? [NCERT p.73]
Think & Reflect · p.73 Geometrically समझाएं कि area घटाकर $(a - b)^2 = a^2 - 2ab + b^2$ कैसे प्राप्त होता है? $b^2$ को वापस क्यों जोड़ते हैं? [NCERT p.73]
1. Base Region Start with a square of side $a$ and area $a^2$.
2. Subtract Strips Subtract two rectangular strips of dimensions $a \times b$.
3. Overlap Corner The corner region $b \times b = b^2$ is removed twice, so adding $+b^2$ once restores exact geometric balance: $\text{Area} = a^2 - 2ab + b^2$.
🎯 Formula: $\mathbf{(a - b)^2 = a^2 - 2ab + b^2}$.
चरण 1: मूल वर्ग भुजा $a$ वाले बड़े square का कुल area $a^2$ लेते हैं।
चरण 2: पट्टियाँ घटाना $a \times b$ dimensions वाली दो rectangular पट्टियों को घटाते हैं।
चरण 3: दोहरे घटाव की भरपाई कोने का $b \times b = b^2$ square दो बार घट जाता है, इसलिए $+b^2$ को एक बार जोड़कर संतुलन बनाते हैं।
🎯 सूत्र: $\mathbf{(a - b)^2 = a^2 - 2ab + b^2}$ प्राप्त होता है।
Think & Reflect · p.76 Label the squares and rectangles in Fig. 4.4 so that it represents the identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. [NCERT p.76]
Think & Reflect · p.76 Fig. 4.4 के squares और rectangles को label करें जिससे यह $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ को दर्शाए। [NCERT p.76]
1. Grid Partitions A square of side $(a+b+c)$ is split into 9 sub-regions.
2. Three Squares Diagonal squares have areas $a^2, b^2, c^2$.
3. Six Rectangles Off-diagonal rectangles form symmetric pairs: 2 of area $ab$, 2 of area $bc$, and 2 of area $ca$.
🎯 Total Area: $\mathbf{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}$.
चरण 1: कुल 9 भाग भुजा $(a+b+c)$ वाले square को 9 भागों में बांटा जाता है।
चरण 2: तीन squares main diagonal पर स्थित squares का area $a^2, b^2, c^2$ है।
चरण 3: छह rectangles pairs में बने rectangles: $ab$ area के 2, $bc$ area के 2 और $ca$ area के 2 rectangles हैं।
🎯 कुल area: $\mathbf{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}$ प्राप्त होता है।
Think & Reflect · p.85 Is $(x - y)$ always a factor of $x^n - y^n$ for all positive integers $n$? Check for $n = 1, 2, 3, 4$. [NCERT p.85]
Think & Reflect · p.85 क्या $(x - y)$ हर positive integer $n$ के लिए $x^n - y^n$ का factor होता है? $n = 1, 2, 3, 4$ के लिए check करें। [NCERT p.85]
1. Factor Theorem Setting $x = y$ in $P(x) = x^n - y^n$ yields $P(y) = y^n - y^n = 0$. Hence $(x - y)$ is always a factor.
2. Expansions
• $n=1: (x - y)(1)$
• $n=2: (x - y)(x + y)$
• $n=3: (x - y)(x^2 + xy + y^2)$
• $n=4: (x - y)(x^3 + x^2y + xy^2 + y^3)$
🎯 Answer: YES, $(x - y)$ is a factor of $x^n - y^n$ for all $n \in \mathbb{N}$.
चरण 1: Factor Theorem $P(x) = x^n - y^n$ में $x = y$ रखने पर $P(y) = y^n - y^n = 0$ आता है, अतः $(x - y)$ surely factor है।
चरण 2: प्रसार
• $n=1: (x - y)(1)$
• $n=2: (x - y)(x + y)$
• $n=3: (x - y)(x^2 + xy + y^2)$
• $n=4: (x - y)(x^3 + x^2y + xy^2 + y^3)$
🎯 उत्तर: हाँ, $(x - y)$ हर positive integer $n$ के लिए $x^n - y^n$ का factor होता है।
Think & Reflect · p.87 Simplify the rational expression: $\frac{36s^2 - 12st + t^2 - 6st}{24s^2 + 8st - t^2}$ by factoring numerator and denominator. [NCERT p.87]
Think & Reflect · p.87 Numerator और denominator को factorise करके rational expression को simplify करें: $\frac{36s^2 - 18st + t^2}{24s^2 + 8st - t^2}$। [NCERT p.87]
1. Simplify Terms
Numerator: $36s^2 - 18st + t^2 = (6s - t)^2$.
Denominator: $24s^2 + 8st - t^2 = -(t^2 - 8st - 24s^2) = -(t - 12s)(t + 2s) = (12s - t)(2s + t)$.
2. Rational Form
$$\frac{(6s - t)^2}{(12s - t)(2s + t)}$$
🎯 Simplified Form: $\frac{(6s - t)^2}{(12s - t)(2s + t)}$.
चरण 1: पदों का factorisation
Numerator: $36s^2 - 18st + t^2 = (6s - t)^2$।
Denominator: $24s^2 + 8st - t^2 = (12s - t)(2s + t)$।
चरण 2: Simplified form में लिखना
$$\frac{(6s - t)^2}{(12s - t)(2s + t)}$$
🎯 उत्तर: $\frac{(6s - t)^2}{(12s - t)(2s + t)}$ प्राप्त होता है।
NCERT
7. Exercise Set 4.3 — Trinomial & Numerical Squares (All Questions Q1 to Q4)
7. Exercise Set 4.3 — Trinomial और Numerical Squares (सभी प्रश्न Q1 से Q4)
Ex 4.3 · Q1 Find the following squares using one of the identities (determine which identity makes calculation easier): (i) $117^2$, (ii) $78^2$, (iii) $198^2$, (iv) $214^2$, (v) $1104^2$, (vi) $1120^2$. [NCERT p.76]
Ex 4.3 · Q1 उपयुक्त identities का use करके squares की value find करें: (i) $117^2$, (ii) $78^2$, (iii) $198^2$, (iv) $214^2$, (v) $1104^2$, (vi) $1120^2$। [NCERT p.76]
1. Parts (i–iii)
(i) $(100 + 17)^2 = 10000 + 3400 + 289 = \mathbf{13,689}$.
(ii) $(80 - 2)^2 = 6400 - 320 + 4 = \mathbf{6,084}$.
(iii) $(200 - 2)^2 = 40000 - 800 + 4 = \mathbf{39,204}$.
2. Parts (iv–vi)
(iv) $(200 + 14)^2 = 40000 + 5600 + 196 = \mathbf{45,796}$.
(v) $(1000 + 100 + 4)^2 = 10^6 + 10^4 + 16 + 2(10^5) + 2(400) + 2(4000) = \mathbf{1,218,816}$.
(vi) $(1000 + 120)^2 = 1000000 + 240000 + 14400 = \mathbf{1,254,400}$.
🎯 Final Answers: (i) $13,689$, (ii) $6,084$, (iii) $39,204$, (iv) $45,796$, (v) $1,218,816$, (vi) $1,254,400$.
भाग (i–iii)
(i) $\mathbf{13,689}$ | (ii) $\mathbf{6,084}$ | (iii) $\mathbf{39,204}$।
भाग (iv–vi)
(iv) $\mathbf{45,796}$ | (v) $(1000 + 100 + 4)^2 = \mathbf{1,218,816}$ | (vi) $(1000 + 120)^2 = \mathbf{1,254,400}$।
🎯 उत्तर: सभी 6 numerical squares के exact values प्राप्त होते हैं।
Ex 4.3 · Q2 Factor using suitable identities: (i) $16y^2 - 24y + 9$, (ii) $\frac{9}{4}s^2 + 6st + 4t^2$, (iii) $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$, (iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$, (v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$. [NCERT p.76]
Ex 4.3 · Q2 Suitable identities से factorise करें: (i) $16y^2 - 24y + 9$, (ii) $\frac{9}{4}s^2 + 6st + 4t^2$, (iii) $\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2$, (iv) $\frac{p^2}{16} - 2 + \frac{16}{p^2}$, (v) $9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc$। [NCERT p.76]
1. Parts (i–ii)
(i) $(4y)^2 - 2(4y)(3) + 3^2 = \mathbf{(4y - 3)^2}$.
(ii) $\left(\frac{3}{2}s\right)^2 + 2\left(\frac{3}{2}s\right)(2t) + (2t)^2 = \mathbf{\left(\frac{3}{2}s + 2t\right)^2}$.
2. Part (iii) Trinomial Square
$$\left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + 2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right) + 2\left(\frac{k}{2}\right)(3n) + 2(3n)\left(\frac{m}{3}\right) = \mathbf{\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2}$$
3. Parts (iv–v)
(iv) $\left(\frac{p}{4}\right)^2 - 2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right) + \left(\frac{4}{p}\right)^2 = \mathbf{\left(\frac{p}{4} - \frac{4}{p}\right)^2}$.
(v) $(-3a)^2 + (2b)^2 + (-c)^2$ or $(3a - 2b + c)^2$: $(-12ab = 2(3a)(-2b), 6ac = 2(3a)(c), -4bc = 2(-2b)(c)) \implies \mathbf{(3a - 2b + c)^2}$.
🎯 Final Answers: (i) $(4y-3)^2$, (ii) $\left(\frac{3}{2}s+2t\right)^2$, (iii) $\left(\frac{m}{3}+\frac{k}{2}+3n\right)^2$, (iv) $\left(\frac{p}{4}-\frac{4}{p}\right)^2$, (v) $(3a-2b+c)^2$.
भाग (i–ii)
(i) $\mathbf{(4y - 3)^2}$ | (ii) $\mathbf{\left(\frac{3}{2}s + 2t\right)^2}$।
भाग (iii) Trinomial Square
$\left(\frac{m}{3}\right)^2 + \left(\frac{k}{2}\right)^2 + (3n)^2 + \dots = \mathbf{\left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2}$।
भाग (iv–v)
(iv) $\mathbf{\left(\frac{p}{4} - \frac{4}{p}\right)^2}$ | (v) Signs check करके: $\mathbf{(3a - 2b + c)^2}$।
🎯 उत्तर: पाँचों expressions के perfect square factorisations प्राप्त होते हैं।
Ex 4.3 · Q3 Expand using $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$: (i) $(p + 3q + 7r)^2$, (ii) $(3x - 2y + 4z)^2$. [NCERT p.76]
Ex 4.3 · Q3 Trinomial identity से expand करें: (i) $(p + 3q + 7r)^2$, (ii) $(3x - 2y + 4z)^2$। [NCERT p.76]
1. Part (i) $(p + 3q + 7r)^2$
$$p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p) = \mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp}$$
2. Part (ii) $(3x - 2y + 4z)^2$
$$(3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x) = \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx}$$
🎯 Final Answers: (i) $p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp$, (ii) $9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx$.
भाग (i) $(p + 3q + 7r)^2$
$$\mathbf{p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14rp}$$
भाग (ii) $(3x - 2y + 4z)^2$
$$\mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx}$$
🎯 उत्तर: दोनों trinomials के complete expansions प्राप्त होते हैं।
Ex 4.3 · Q4 Is this an identity? $(a+b-c)^2 + (a-b+c)^2 + (a-b-c)^2 = 2a^2 + 2b^2 + 2c^2$. [NCERT p.76]
Ex 4.3 · Q4 क्या यह एक identity है? $(a+b-c)^2 + (a-b+c)^2 + (a-b-c)^2 = 2a^2 + 2b^2 + 2c^2$। [NCERT p.76]
1. Expand Left-Hand Side (LHS)
• $(a+b-c)^2 = a^2+b^2+c^2 + 2ab - 2bc - 2ca$
• $(a-b+c)^2 = a^2+b^2+c^2 - 2ab - 2bc + 2ca$
• $(a-b-c)^2 = a^2+b^2+c^2 - 2ab + 2bc + 2ca$
2. Sum the Three Expansions
$$\text{LHS} = 3(a^2 + b^2 + c^2) + (2ab - 2ab - 2ab) + (-2bc - 2bc + 2bc) + (-2ca + 2ca + 2ca)$$
$$\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc + 2ca \neq 2a^2 + 2b^2 + 2c^2$$
🎯 Conclusion: NO, it is not an identity (LHS $\neq$ RHS for general values of $a, b, c$).
चरण 1: LHS के तीनों पदों का विस्तार
तीनों trinomial squares को expand करके जोड़ने पर:
चरण 2: पदों का योग
$$\text{LHS} = 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc + 2ca \neq 2a^2 + 2b^2 + 2c^2$$
🎯 उत्तर: नहीं, यह एक identity नहीं है क्योंकि LHS और RHS बराबर नहीं हैं।
NCERT
8. Exercise Set 4.4 — Factorisation & Products (All Questions Q1 to Q3)
8. Exercise Set 4.4 — Factorisation और Products (सभी प्रश्न Q1 से Q3)
Ex 4.4 · Q1 Fill in the blanks to complete the following identities: (i) $s^2 - 11s + 24 = (\_\_\_\_)(\_\_\_\_)$, (ii) $(\_\_\_\_)(x + 1) = 3x^2 - 4x - 7$, (iii) $10x^2 - 11x - 6 = (2x - \_\_\_)(\_\_\_ + 2)$, (iv) $6x^2 + 7x + 2 = (\_\_\_\_)(\_\_\_\_)$. [NCERT p.81]
Ex 4.4 · Q1 निम्नलिखित identities को पूरा करने के लिए रिक्त स्थान भरें: (i) $s^2 - 11s + 24 = (\_\_\_\_)(\_\_\_\_)$, (ii) $(\_\_\_\_)(x + 1) = 3x^2 - 4x - 7$, (iii) $10x^2 - 11x - 6 = (2x - \_\_\_)(\_\_\_ + 2)$, (iv) $6x^2 + 7x + 2 = (\_\_\_\_)(\_\_\_\_)$। [NCERT p.81]
1. Part (i) $a+b = -11, ab = 24 \implies a = -8, b = -3$. Blank: $\mathbf{(s - 8)(s - 3)}$.
2. Part (ii) Divide $3x^2 - 4x - 7$ by $(x + 1) = 3x(x+1) - 7(x+1) = (3x - 7)(x + 1)$. Blank: $\mathbf{(3x - 7)}$.
3. Part (iii) $10x^2 - 11x - 6 = (2x - 3)(5x + 2)$. Blanks: $\mathbf{3}$ and $\mathbf{5x}$.
4. Part (iv) $6x^2 + 7x + 2 = 6x^2 + 4x + 3x + 2 = (2x + 1)(3x + 2)$. Blank: $\mathbf{(2x + 1)(3x + 2)}$.
🎯 Final Blanks: (i) $(s - 8)(s - 3)$, (ii) $(3x - 7)$, (iii) $3$ and $5x$, (iv) $(2x + 1)(3x + 2)$.
भाग (i) का हल $s^2 - 11s + 24 = \mathbf{(s - 8)(s - 3)}$।
भाग (ii) का हल $3x^2 - 4x - 7 = \mathbf{(3x - 7)}(x + 1)$।
भाग (iii) का हल $10x^2 - 11x - 6 = (2x - \mathbf{3})(\mathbf{5x} + 2)$।
भाग (iv) का हल $6x^2 + 7x + 2 = \mathbf{(2x + 1)(3x + 2)}$।
🎯 उत्तर: चारों blanks सही factors से भरे गए हैं।
Ex 4.4 · Q2 Find the following products without multiplying directly: (i) $41^2$, (ii) $27^2$, (iii) $23 \times 17$, (iv) $135^2$, (v) $97^2$, (vi) $18 \times 29$, (vii) $34 \times 43$, (viii) $205^2$. [NCERT p.81]
Ex 4.4 · Q2 सीधे गुणा किए बिना products calculate करें: (i) $41^2$, (ii) $27^2$, (iii) $23 \times 17$, (iv) $135^2$, (v) $97^2$, (vi) $18 \times 29$, (vii) $34 \times 43$, (viii) $205^2$। [NCERT p.81]
1. Parts (i–iv)
(i) $(40 + 1)^2 = 1600 + 80 + 1 = \mathbf{1,681}$.
(ii) $(30 - 3)^2 = 900 - 180 + 9 = \mathbf{729}$.
(iii) $(20 + 3)(20 - 3) = 400 - 9 = \mathbf{391}$.
(iv) $(100 + 35)^2 = 10000 + 7000 + 1225 = \mathbf{18,225}$.
2. Parts (v–viii)
(v) $(100 - 3)^2 = 10000 - 600 + 9 = \mathbf{9,409}$.
(vi) $(20 - 2)(30 - 1) = 600 - 20 - 60 + 2 = \mathbf{522}$.
(vii) $(30 + 4)(40 + 3) = 1200 + 90 + 160 + 12 = \mathbf{1,462}$.
(viii) $(200 + 5)^2 = 40000 + 2000 + 25 = \mathbf{42,025}$.
🎯 Final Answers: (i) 1,681, (ii) 729, (iii) 391, (iv) 18,225, (v) 9,409, (vi) 522, (vii) 1,462, (viii) 42,025.
भाग (i–iv)
(i) $\mathbf{1,681}$ | (ii) $\mathbf{729}$ | (iii) $\mathbf{391}$ | (iv) $\mathbf{18,225}$।
भाग (v–viii)
(v) $\mathbf{9,409}$ | (vi) $\mathbf{522}$ | (vii) $\mathbf{1,462}$ | (viii) $\mathbf{42,025}$।
🎯 उत्तर: सभी 8 numerical products के मान प्राप्त होते हैं।
Ex 4.4 · Q3 Factor the following: (i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$, (ii) $16s^2 + 25t^2 - 40st$, (iii) $r^2 - r - 42$, (iv) $49g^2 + 14gh + h^2$, (v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$. [NCERT p.81]
Ex 4.4 · Q3 निम्नलिखित को factorise करें: (i) $9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc$, (ii) $16s^2 + 25t^2 - 40st$, (iii) $r^2 - r - 42$, (iv) $49g^2 + 14gh + h^2$, (v) $64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw$। [NCERT p.81]
1. Parts (i–ii)
(i) $(3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(2c)(3a) = \mathbf{(3a - b + 2c)^2}$.
(ii) $(4s)^2 - 2(4s)(5t) + (5t)^2 = \mathbf{(4s - 5t)^2}$.
2. Parts (iii–iv)
(iii) Split middle term: product $=-42$, sum $=-1 \implies -7, 6 \implies \mathbf{(r - 7)(r + 6)}$.
(iv) $(7g)^2 + 2(7g)(h) + h^2 = \mathbf{(7g + h)^2}$.
3. Part (v)
$$(8u)^2 + (-11v)^2 + (-2w)^2 + 2(8u)(-11v) + 2(-11v)(-2w) + 2(-2w)(8u) = \mathbf{(8u - 11v - 2w)^2}$$
🎯 Final Answers: (i) $(3a-b+2c)^2$, (ii) $(4s-5t)^2$, (iii) $(r-7)(r+6)$, (iv) $(7g+h)^2$, (v) $(8u-11v-2w)^2$.
भाग (i–ii)
(i) $\mathbf{(3a - b + 2c)^2}$ | (ii) $\mathbf{(4s - 5t)^2}$।
भाग (iii–iv)
(iii) $\mathbf{(r - 7)(r + 6)}$ | (iv) $\mathbf{(7g + h)^2}$।
भाग (v)
Trinomial square: $\mathbf{(8u - 11v - 2w)^2}$।
🎯 उत्तर: पाँचों algebraic expressions के factorisations प्राप्त होते हैं।
NCERT
10. End-of-Chapter Exercises — Complete Solutions (Q1 to Q13)
10. End-of-Chapter Exercises — सम्पूर्ण हल (Q1 से Q13)
End Ex · Q1 Use suitable identities to find the following products: (i) $(-3x + 4)^2$, (ii) $(2s + 7)(2s - 7)$, (iii) $\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)$, (iv) $(2n + 7)(2n - 7)$, (v) $(s - 2t)(s^2 + 2st + 4t^2)$, (vi) $\left(\frac{1}{2r} - 4r\right)^2$, (vii) $(-3m + 4k - l)^2$, (viii) $\left(x - \frac{1}{3}y\right)^3$, (ix) $\left(\frac{7}{2}k - \frac{2}{3}m\right)^3$. [NCERT p.88]
End Ex · Q1 उपयुक्त identities का use करके products calculate करें: (i) $(-3x + 4)^2$, (ii) $(2s + 7)(2s - 7)$, (iii) $\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)$, (iv) $(2n + 7)(2n - 7)$, (v) $(s - 2t)(s^2 + 2st + 4t^2)$, (vi) $\left(\frac{1}{2r} - 4r\right)^2$, (vii) $(-3m + 4k - l)^2$, (viii) $\left(x - \frac{1}{3}y\right)^3$, (ix) $\left(\frac{7}{2}k - \frac{2}{3}m\right)^3$। [NCERT p.88]
1. Parts (i–iii)
(i) $(-3x)^2 + 2(-3x)(4) + 4^2 = \mathbf{9x^2 - 24x + 16}$.
(ii) $(2s)^2 - 7^2 = \mathbf{4s^2 - 49}$.
(iii) $(p^2)^2 - \left(\frac{1}{2}\right)^2 = \mathbf{p^4 - \frac{1}{4}}$.
2. Parts (iv–vi)
(iv) $(2n)^2 - 7^2 = \mathbf{4n^2 - 49}$.
(v) Difference of cubes: $(s)^3 - (2t)^3 = \mathbf{s^3 - 8t^3}$.
(vi) $\left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2 = \mathbf{\frac{1}{4r^2} - 4 + 16r^2}$.
3. Parts (vii–ix)
(vii) Trinomial: $(-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m) = \mathbf{9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm}$.
(viii) Cubic: $x^3 - 3(x)^2\left(\frac{1}{3}y\right) + 3(x)\left(\frac{1}{9}y^2\right) - \frac{1}{27}y^3 = \mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}$.
(ix) Cubic: $\left(\frac{7}{2}k\right)^3 - 3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2 - \left(\frac{2}{3}m\right)^3 = \mathbf{\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3}$.
🎯 Final Answers: All 9 subparts expanded using standard identities.
भाग (i–iii)
(i) $\mathbf{9x^2 - 24x + 16}$ | (ii) $\mathbf{4s^2 - 49}$ | (iii) $\mathbf{p^4 - \frac{1}{4}}$।
भाग (iv–vi)
(iv) $\mathbf{4n^2 - 49}$ | (v) Difference of cubes: $\mathbf{s^3 - 8t^3}$ | (vi) $\mathbf{\frac{1}{4r^2} - 4 + 16r^2}$।
भाग (vii–ix)
(vii) Trinomial square: $\mathbf{9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm}$।
(viii) Cube expansion: $\mathbf{x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3}$।
(ix) Cube expansion: $\mathbf{\frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3}$।
🎯 उत्तर: सभी 9 sub-parts के expansions प्राप्त होते हैं।
End Ex · Q2 Find the values using suitable identities: (i) $17 \times 21$, (ii) $104 \times 96$, (iii) $24 \times 16$, (iv) $147^3$, (v) $199^3$, (vi) $127^3$, (vii) $(-107)^3$, (viii) $(-299)^3$. [NCERT p.89]
End Ex · Q2 Suitable identities का use करके values find करें: (i) $17 \times 21$, (ii) $104 \times 96$, (iii) $24 \times 16$, (iv) $147^3$, (v) $199^3$, (vi) $127^3$, (vii) $(-107)^3$, (viii) $(-299)^3$। [NCERT p.89]
1. Parts (i–iii)
(i) $(20 - 3)(20 + 1) = 400 - 40 - 3 = \mathbf{357}$.
(ii) $(100 + 4)(100 - 4) = 100^2 - 4^2 = 10000 - 16 = \mathbf{9,984}$.
(iii) $(20 + 4)(20 - 4) = 20^2 - 4^2 = 400 - 16 = \mathbf{384}$.
2. Parts (iv–v)
(iv) $(150 - 3)^3 = 150^3 - 3(150)^2(3) + 3(150)(3)^2 - 3^3 = 3375000 - 202500 + 4050 - 27 = \mathbf{3,176,523}$.
(v) $(200 - 1)^3 = 200^3 - 3(200)^2(1) + 3(200)(1) - 1 = 8000000 - 120000 + 600 - 1 = \mathbf{7,880,599}$.
3. Parts (vi–viii)
(vi) $(130 - 3)^3 = 2197000 - 152100 + 3510 - 27 = \mathbf{2,048,383}$.
(vii) $-[(100 + 7)^3] = -(1000000 + 210000 + 14700 + 343) = \mathbf{-1,225,043}$.
(viii) $-[(300 - 1)^3] = -(27000000 - 270000 + 900 - 1) = \mathbf{-26,730,899}$.
🎯 Final Answers: (i) 357, (ii) 9,984, (iii) 384, (iv) 3,176,523, (v) 7,880,599, (vi) 2,048,383, (vii) -1,225,043, (viii) -26,730,899.
भाग (i–iii)
(i) $(20 - 3)(20 + 1) = \mathbf{357}$ | (ii) $(100 + 4)(100 - 4) = \mathbf{9,984}$ | (iii) $(20 + 4)(20 - 4) = \mathbf{384}$।
भाग (iv–v)
(iv) $(150 - 3)^3 = \mathbf{3,176,523}$ | (v) $(200 - 1)^3 = \mathbf{7,880,599}$।
भाग (vi–viii)
(vi) $(130 - 3)^3 = \mathbf{2,048,383}$ | (vii) $-(100 + 7)^3 = \mathbf{-1,225,043}$ | (viii) $-(300 - 1)^3 = \mathbf{-26,730,899}$।
🎯 उत्तर: सभी 8 numerical calculations के exact values प्राप्त होते हैं।
End Ex · Q3 Factor the following algebraic expressions: (i) $4y^2 + 1 + \frac{1}{16y^2}$, (ii) $9m^2 - \frac{1}{25n^2}$, (iii) $27b^3 - \frac{1}{64b^3}$, (iv) $x^2 + \frac{5x}{6} + \frac{1}{6}$, (v) $27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}$, (vi) $64y^3 + \frac{1}{125}z^3$, (vii) $p^3 + 27q^3 + r^3 - 9pqr$, (viii) $9m^2 - 12m + 4$, (ix) $9x^3 - \frac{8}{3}y^3 + \frac{1}{3}z^3 + 6xyz$, (x) $4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy$, (xi) $27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}$. [NCERT p.89]
End Ex · Q3 Algebraic expressions को factorise करें: (i) $4y^2 + 1 + \frac{1}{16y^2}$, (ii) $9m^2 - \frac{1}{25n^2}$, (iii) $27b^3 - \frac{1}{64b^3}$, (iv) $x^2 + \frac{5x}{6} + \frac{1}{6}$, (v) $27u^3 - \frac{1}{125} - \frac{27u^2}{5} + \frac{9u}{25}$, (vi) $64y^3 + \frac{1}{125}z^3$, (vii) $p^3 + 27q^3 + r^3 - 9pqr$, (viii) $9m^2 - 12m + 4$, (ix) $9x^3 - \frac{8}{3}y^3 + \frac{1}{3}z^3 + 6xyz$, (x) $4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy$, (xi) $27u^3 - \frac{1}{216} - \frac{9u^2}{2} + \frac{u}{4}$। [NCERT p.89]
1. Parts (i–iv)
(i) $(2y)^2 + 2(2y)\left(\frac{1}{4y}\right) + \left(\frac{1}{4y}\right)^2 = \mathbf{\left(2y + \frac{1}{4y}\right)^2}$.
(ii) $(3m)^2 - \left(\frac{1}{5n}\right)^2 = \mathbf{\left(3m + \frac{1}{5n}\right)\left(3m - \frac{1}{5n}\right)}$.
(iii) $(3b)^3 - \left(\frac{1}{4b}\right)^3 = \mathbf{\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)}$.
(iv) $x^2 + \frac{1}{2}x + \frac{1}{3}x + \frac{1}{6} = \mathbf{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)}$.
2. Parts (v–viii)
(v) $(3u)^3 - \left(\frac{1}{5}\right)^3 - 3(3u)^2\left(\frac{1}{5}\right) + 3(3u)\left(\frac{1}{25}\right) = \mathbf{\left(3u - \frac{1}{5}\right)^3}$.
(vi) $(4y)^3 + \left(\frac{1}{5}z\right)^3 = \mathbf{\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2\right)}$.
(vii) $p^3 + (3q)^3 + r^3 - 3(p)(3q)(r) = \mathbf{(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - rp)}$.
(viii) $(3m)^2 - 2(3m)(2) + 2^2 = \mathbf{(3m - 2)^2}$.
3. Parts (ix–xi)
(ix) Factor out $\frac{1}{3}$: $\frac{1}{3}(27x^3 - 8y^3 + z^3 + 18xyz) = \mathbf{\frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx)}$.
(x) $(2x)^2 + (3y)^2 + (6z)^2 + 2(2x)(3y) + 2(3y)(6z) + 2(6z)(2x) = \mathbf{(2x + 3y + 6z)^2}$.
(xi) $(3u)^3 - \left(\frac{1}{6}\right)^3 - 3(3u)^2\left(\frac{1}{6}\right) + 3(3u)\left(\frac{1}{36}\right) = \mathbf{\left(3u - \frac{1}{6}\right)^3}$.
🎯 Final Answers: All 11 expressions factorised cleanly.
भाग (i–iv)
(i) $\mathbf{\left(2y + \frac{1}{4y}\right)^2}$ | (ii) $\mathbf{\left(3m + \frac{1}{5n}\right)\left(3m - \frac{1}{5n}\right)}$ | (iii) $\mathbf{\left(3b - \frac{1}{4b}\right)\left(9b^2 + \frac{3}{4} + \frac{1}{16b^2}\right)}$ | (iv) $\mathbf{\left(x + \frac{1}{2}\right)\left(x + \frac{1}{3}\right)}$।
भाग (v–viii)
(v) $\mathbf{\left(3u - \frac{1}{5}\right)^3}$ | (vi) $\mathbf{\left(4y + \frac{1}{5}z\right)\left(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2\right)}$ | (vii) $\mathbf{(p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - rp)}$ | (viii) $\mathbf{(3m - 2)^2}$।
भाग (ix–xi)
(ix) $\mathbf{\frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3zx)}$ | (x) $\mathbf{(2x + 3y + 6z)^2}$ | (xi) $\mathbf{\left(3u - \frac{1}{6}\right)^3}$।
🎯 उत्तर: सभी 11 expressions के complete factorisations प्राप्त होते हैं।
End Ex · Q4 Simplify: (i) $\frac{4x^2 + 4x + 1}{4x^2 - 1}$, (ii) $\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$, (iii) $\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$. [NCERT p.89]
End Ex · Q4 Simplify करें: (i) $\frac{4x^2 + 4x + 1}{4x^2 - 1}$, (ii) $\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}$, (iii) $\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}$। [NCERT p.89]
1. Part (i)
$$\frac{(2x + 1)^2}{(2x - 1)(2x + 1)} = \mathbf{\frac{2x + 1}{2x - 1}}$$
2. Part (ii)
$$\frac{27(a^3 - 8b^3)}{9(a^2 - 4b^2)} = \frac{3(a - 2b)(a^2 + 2ab + 4b^2)}{(a - 2b)(a + 2b)} = \mathbf{\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}}$$
3. Part (iii)
$$\frac{(s + 5t)(s^2 - 5st + 25t^2)}{(s - 7t)(s + 5t)} = \mathbf{\frac{s^2 - 5st + 25t^2}{s - 7t}}$$
🎯 Final Answers: (i) $\frac{2x+1}{2x-1}$, (ii) $\frac{3(a^2+2ab+4b^2)}{a+2b}$, (iii) $\frac{s^2 - 5st + 25t^2}{s - 7t}$.
भाग (i) का हल $\frac{(2x + 1)^2}{(2x - 1)(2x + 1)} = \mathbf{\frac{2x + 1}{2x - 1}}$।
भाग (ii) का हल $(a - 2b)$ factor cancel करने पर: $\mathbf{\frac{3(a^2 + 2ab + 4b^2)}{a + 2b}}$।
भाग (iii) का हल $(s + 5t)$ factor cancel करने पर: $\mathbf{\frac{s^2 - 5st + 25t^2}{s - 7t}}$।
🎯 उत्तर: तीनों rational expressions के simplified forms प्राप्त होते हैं।
End Ex · Q5 Find possible expressions for the length and breadth of rectangles whose areas are: (i) $25a^2 - 30ab + 9b^2$, (ii) $36s^2 - 49t^2$. [NCERT p.89]
End Ex · Q5 Rectangles की length और breadth के possible expressions find करें जिनका area है: (i) $25a^2 - 30ab + 9b^2$, (ii) $36s^2 - 49t^2$। [NCERT p.89]
1. Part (i) Area $= 25a^2 - 30ab + 9b^2$
$$\text{Area} = (5a)^2 - 2(5a)(3b) + (3b)^2 = (5a - 3b)(5a - 3b)$$
$\text{Length} = (5a - 3b), \quad \text{Breadth} = (5a - 3b)$ (A square rectangle).
2. Part (ii) Area $= 36s^2 - 49t^2$
$$\text{Area} = (6s)^2 - (7t)^2 = (6s + 7t)(6s - 7t)$$
$\text{Length} = (6s + 7t), \quad \text{Breadth} = (6s - 7t)$.
🎯 Final Answers: (i) $(5a - 3b)$ and $(5a - 3b)$; (ii) $(6s + 7t)$ and $(6s - 7t)$.
भाग (i) Area $= 25a^2 - 30ab + 9b^2$
$\text{Area} = (5a - 3b)^2 = (5a - 3b)(5a - 3b)$। Length $= (5a - 3b)$, Breadth $= (5a - 3b)$।
भाग (ii) Area $= 36s^2 - 49t^2$
$\text{Area} = (6s + 7t)(6s - 7t)$। Length $= (6s + 7t)$, Breadth $= (6s - 7t)$।
🎯 उत्तर: (i) $(5a - 3b)$ और $(5a - 3b)$; (ii) $(6s + 7t)$ और $(6s - 7t)$ प्राप्त होते हैं।
End Ex · Q6 Find possible expressions for the length, breadth, and heights of cuboids whose volumes are: (i) $6a^2 - 24b^2$, (ii) $3ps^2 - 15ps + 12p$. [NCERT p.89]
End Ex · Q6 Cuboids की dimensions (length, breadth, height) के possible expressions find करें जिनका volume है: (i) $6a^2 - 24b^2$, (ii) $3ps^2 - 15ps + 12p$। [NCERT p.89]
1. Part (i) Volume $= 6a^2 - 24b^2$
$$V = 6(a^2 - 4b^2) = 6(a + 2b)(a - 2b)$$
$\text{Dimensions} = 6, \quad (a + 2b), \quad (a - 2b)\text{ units}$.
2. Part (ii) Volume $= 3ps^2 - 15ps + 12p$
$$V = 3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)$$
$\text{Dimensions} = 3p, \quad (s - 1), \quad (s - 4)\text{ units}$.
🎯 Final Answers: (i) $6, (a+2b), (a-2b)$; (ii) $3p, (s-1), (s-4)$.
भाग (i) Volume $= 6a^2 - 24b^2$
$V = 6(a^2 - 4b^2) = 6(a + 2b)(a - 2b)$। Dimensions: $6, (a + 2b), (a - 2b)$ units।
भाग (ii) Volume $= 3ps^2 - 15ps + 12p$
$V = 3p(s^2 - 5s + 4) = 3p(s - 1)(s - 4)$। Dimensions: $3p, (s - 1), (s - 4)$ units।
🎯 उत्तर: (i) $6, (a+2b), (a-2b)$; (ii) $3p, (s-1), (s-4)$ प्राप्त होते हैं।
End Ex · Q7 The village playground is a square of side $40\text{ m}$. A path of width $s\text{ m}$ is created around the playground for walking. Find an expression for the area of the path in terms of $s$. [NCERT p.90]
End Ex · Q7 गाँव का playground $40\text{ m}$ भुजा का एक square है। उसके चारों ओर $s\text{ m}$ चौड़ा रास्ता (path) बनाया गया है। Path के area का algebraic expression $s$ के पदों में find करें। [NCERT p.90]
1. Inner & Outer Dimensions
Inner square side $= 40\text{ m} \implies \text{Area}_{\text{inner}} = 40^2 = 1600\text{ m}^2$.
Outer square side $= 40 + 2s\text{ m} \implies \text{Area}_{\text{outer}} = (40 + 2s)^2\text{ m}^2$.
2. Area of Path
$$\text{Area}_{\text{path}} = (40 + 2s)^2 - 40^2 = 40^2 + 2(40)(2s) + (2s)^2 - 40^2 = \mathbf{4s^2 + 160s}\text{ sq metres}$$
🎯 Final Answer: $\text{Area} = 4s^2 + 160s = 4s(s + 40)\text{ m}^2$.
चरण 1: Dimensions
Inner square भुजा $= 40\text{ m} \implies \text{Area}_{\text{inner}} = 1600\text{ m}^2$।
Outer square भुजा $= 40 + 2s\text{ m} \implies \text{Area}_{\text{outer}} = (40 + 2s)^2\text{ m}^2$।
चरण 2: Path का Area
$$\text{Area} = (40 + 2s)^2 - 40^2 = 1600 + 160s + 4s^2 - 1600 = \mathbf{4s^2 + 160s}\text{ m}^2$$
🎯 उत्तर: Path का area $4s^2 + 160s\text{ m}^2$ प्राप्त होता है।
End Ex · Q8 If a number plus its reciprocal equals $\frac{10}{3}$, find the number. [NCERT p.90]
End Ex · Q8 यदि किसी संख्या और उसके व्युत्क्रम (reciprocal) का योग $\frac{10}{3}$ है, तो वह संख्या find करें। [NCERT p.90]
1. Form Equation Let the number be $x$.
$$x + \frac{1}{x} = \frac{10}{3} \implies \frac{x^2 + 1}{x} = \frac{10}{3} \implies 3x^2 - 10x + 3 = 0$$
2. Factorise Quadratic
$$3x^2 - 9x - x + 3 = 0 \implies 3x(x - 3) - 1(x - 3) = (3x - 1)(x - 3) = 0$$
$$x = 3 \quad \text{or} \quad x = \frac{1}{3}$$
🎯 Final Answer: The number is $3$ or $\frac{1}{3}$.
चरण 1: Equation बनाना संख्या $x$ मान लेते हैं:
$$x + \frac{1}{x} = \frac{10}{3} \implies 3x^2 - 10x + 3 = 0$$
चरण 2: Factorise करना
$$3x(x - 3) - 1(x - 3) = (3x - 1)(x - 3) = 0 \implies x = 3 \text{ या } x = \frac{1}{3}$$
🎯 उत्तर: वह संख्या $3$ या $\frac{1}{3}$ है।
End Ex · Q9 A rectangular pool has area $2x^2 + 7x + 3$ square hastas. If its width is $2x + 1$ hastas, find its length. [NCERT p.90]
End Ex · Q9 एक rectangular pool का area $2x^2 + 7x + 3\text{ hastas}^2$ है। यदि उसकी breadth $2x + 1\text{ hastas}$ है, तो उसकी length find करें। [NCERT p.90]
1. Factorise Area Polynomial
$$\text{Area} = 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$$
2. Compute Length
$$\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{(2x + 1)(x + 3)}{2x + 1} = \mathbf{x + 3}\text{ hastas}$$
🎯 Final Answer: Length $= (x + 3)\text{ hastas}$.
चरण 1: Area का factorisation
$$\text{Area} = 2x^2 + 6x + x + 3 = (2x + 1)(x + 3)$$
चरण 2: Length निकालना
$$\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{(2x + 1)(x + 3)}{2x + 1} = \mathbf{x + 3}\text{ hastas}$$
🎯 उत्तर: Pool की length $(x + 3)\text{ hastas}$ होगी।
End Ex · Q10 (Advanced) If both $(x - 2)$ and $\left(x - \frac{1}{2}\right)$ are factors of $px^2 + 5x + r$, show that $p = r$. [NCERT p.90]
End Ex · Q10 (Advanced) यदि $(x - 2)$ और $\left(x - \frac{1}{2}\right)$ दोनों $px^2 + 5x + r$ के factors हैं, तो prove करें कि $p = r$ होगा। [NCERT p.90]
1. Factor Theorem for $x = 2$
$$P(2) = p(2)^2 + 5(2) + r = 4p + 10 + r = 0 \implies 4p + r = -10 \quad \text{--- (1)}$$
2. Factor Theorem for $x = 1/2$
$$P\left(\frac{1}{2}\right) = p\left(\frac{1}{4}\right) + 5\left(\frac{1}{2}\right) + r = 0 \implies \frac{p}{4} + \frac{5}{2} + r = 0 \implies p + 4r = -10 \quad \text{--- (2)}$$
3. Equate Equations
$$4p + r = p + 4r \implies 3p = 3r \implies \mathbf{p = r}$$
🎯 Conclusion: $p = r$ (Proved).
चरण 1: $x = 2$ रखने पर
$$P(2) = 4p + 10 + r = 0 \implies 4p + r = -10 \quad \text{--- (1)}$$
चरण 2: $x = 1/2$ रखने पर
$$P\left(\frac{1}{2}\right) = \frac{p}{4} + \frac{5}{2} + r = 0 \implies p + 4r = -10 \quad \text{--- (2)}$$
चरण 3: दोनों equations की तुलना
$$4p + r = p + 4r \implies 3p = 3r \implies \mathbf{p = r}$$
🎯 Result: $p = r$ proved होता है।
End Ex · Q11 (Advanced) If $a + b + c = 5$ and $ab + bc + ca = 10$, prove that $a^3 + b^3 + c^3 - 3abc = -25$. [NCERT p.90]
End Ex · Q11 (Advanced) यदि $a + b + c = 5$ और $ab + bc + ca = 10$ है, तो prove करें कि $a^3 + b^3 + c^3 - 3abc = -25$ होगा। [NCERT p.90]
1. Find $a^2 + b^2 + c^2$
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$
$$5^2 = a^2 + b^2 + c^2 + 2(10) \implies a^2 + b^2 + c^2 = 25 - 20 = 5$$
2. Apply Master Identity
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)\left[(a^2 + b^2 + c^2) - (ab + bc + ca)\right]$$
$$= 5 \times (5 - 10) = 5 \times (-5) = \mathbf{-25}$$
🎯 Conclusion: $a^3 + b^3 + c^3 - 3abc = -25$ (Proved).
चरण 1: $a^2 + b^2 + c^2$ का मान
$$5^2 = a^2 + b^2 + c^2 + 2(10) \implies a^2 + b^2 + c^2 = 25 - 20 = 5$$
चरण 2: Identity में मान रखना
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)\left[(a^2 + b^2 + c^2) - (ab + bc + ca)\right]$$
$$= 5 \times (5 - 10) = 5 \times (-5) = \mathbf{-25}$$
🎯 Result: $a^3 + b^3 + c^3 - 3abc = -25$ proved होता है।
End Ex · Q12 (Advanced) By factoring the expression, check that $n^3 - n$ is always divisible by 6 for all natural numbers $n$. Give reasons. [NCERT p.90]
End Ex · Q12 (Advanced) Expression को factorise करके जांचें कि $n^3 - n$ हर natural numbers $n$ के लिए 6 से हमेशा divisible होता है। कारण सहित बताएं। [NCERT p.90]
1. Factorise Expression
$$n^3 - n = n(n^2 - 1) = (n - 1) \cdot n \cdot (n + 1)$$
2. Product of Three Consecutive Integers
$(n - 1), n, (n + 1)$ are three consecutive integers.
• In any three consecutive integers, exactly one is divisible by 3.
• At least one is even (divisible by 2).
Since $\gcd(2, 3) = 1$, their product is always divisible by $2 \times 3 = 6$.
🎯 Conclusion: $n^3 - n$ is always divisible by 6 for all $n \in \mathbb{N}$.
चरण 1: Factorise करना
$$n^3 - n = n(n^2 - 1) = (n - 1) \cdot n \cdot (n + 1)$$
चरण 2: तीन लगातार संख्याएं
$(n - 1), n, (n + 1)$ तीन लगातार natural numbers हैं।
• तीन लगातार संख्याओं में से एक संख्या हमेशा 3 से divisible होती है।
• कम से कम एक संख्या सम (2 से divisible) होती है।
अतः इनका गुणनफल हमेशा $2 \times 3 = 6$ से divisible होगा।
🎯 Result: $n^3 - n$ हर natural numbers $n$ के लिए 6 से divisible होता है।
End Ex · Q13 (Advanced) Find the value of: (i) $x^3 + y^3 - 12xy + 64$, when $x + y = -4$; (ii) $x^3 - 8y^3 - 36xy - 216$, when $x = 2y + 6$. [NCERT p.90]
End Ex · Q13 (Advanced) मान find करें: (i) $x^3 + y^3 - 12xy + 64$, जब $x + y = -4$ हो; (ii) $x^3 - 8y^3 - 36xy - 216$, जब $x = 2y + 6$ हो। [NCERT p.90]
1. Part (i) When $x + y = -4$
Rewrite as $x + y + 4 = 0$. Let $a = x, b = y, c = 4$. Since $a + b + c = 0$:
$$x^3 + y^3 + 4^3 - 3(x)(y)(4) = x^3 + y^3 + 64 - 12xy = \mathbf{0}$$
2. Part (ii) When $x = 2y + 6$
Rewrite as $x + (-2y) + (-6) = 0$. Let $a = x, b = -2y, c = -6$. Since $a + b + c = 0$:
$$x^3 + (-2y)^3 + (-6)^3 - 3(x)(-2y)(-6) = x^3 - 8y^3 - 216 - 36xy = \mathbf{0}$$
🎯 Final Answers: (i) $0$, (ii) $0$.
भाग (i) जब $x + y = -4$
$x + y + 4 = 0$ लिखने पर: चूँकि तीनों का योग $0$ है, अतः $x^3 + y^3 + 4^3 - 3(x)(y)(4) = \mathbf{0}$ होगा।
भाग (ii) जब $x = 2y + 6$
$x - 2y - 6 = 0$ लिखने पर: $a = x, b = -2y, c = -6$ का योग $0$ है, अतः $x^3 - 8y^3 - 216 - 36xy = \mathbf{0}$ होगा।
🎯 उत्तर: (i) $0$, (ii) $0$ प्राप्त होते हैं।