1. What is a Polynomial? Terms, Variables, and Degrees
1. Polynomial क्या है? Terms, Variables, और Degree
Section Goal: Understand how algebraic expressions are built from variables and coefficients, classify polynomials by their degree, and recognize the structure of a linear polynomial $P(x) = ax + b$.
Section Goal: समझें कि algebraic expressions variables और coefficients से कैसे बनते हैं, polynomials को उनकी degree के आधार पर classify कैसे करते हैं, और linear polynomial $P(x) = ax + b$ की structure को identify करें।
🚕 The Taxi Fare Analogy: Constant + Variable Rate
🚕 Taxi Fare का Example: Fixed Charge + Per Kilometer Rate
When you hire an auto-rickshaw, there is a fixed boarding charge of ₹20, plus ₹10 for every kilometer you travel. If you travel $x\text{ km}$, your total fare is:
$$\text{Fare}(x) = 10x + 20$$
Here, $10x + 20$ is a linear polynomial! The number $10$ is the coefficient of $x$, $x$ is the variable, and $20$ is the constant term [NCERT p.16].
जब आप auto-rickshaw hire करते हैं, तो ₹20 का fixed boarding charge होता है, और फिर per kilometer ₹10 लगता है। अगर आप $x\text{ km}$ travel करते हैं, तो total fare होगा:
$$\text{Fare}(x) = 10x + 20$$
यहाँ $10x + 20$ एक linear polynomial है! $10$ $x$ का coefficient है, $x$ variable है, और $20$ constant term है [NCERT p.16]।
An algebraic expression is called a polynomial in one variable $x$ if the exponents of $x$ in all terms are non-negative integers ($0, 1, 2, 3, \dots$) [NCERT p.17].
एक variable $x$ वाला algebraic expression polynomial कहलाता है यदि सभी terms में $x$ के exponents non-negative integers ($0, 1, 2, 3, \dots$) हों [NCERT p.17]।
4. Solved In-Text Think & Reflect Proofs
4. हल किए गए In-Text Think & Reflect Proofs
Think & Reflect · p.17 A wire of length $20\text{ cm}$ is bent into a rectangle with length $x\text{ cm}$ and width $(10 - x)\text{ cm}$. Identify terms, variables, coefficients, and evaluate area at $x = 6\text{ cm}$. [NCERT p.17, 20]
Think & Reflect · p.17 $20\text{ cm}$ लंबी wire को मोड़कर length $x\text{ cm}$ और width $(10 - x)\text{ cm}$ वाला rectangle बनाया गया। इसके terms, variables, coefficients identify करें और $x = 6\text{ cm}$ पर area calculate करें। [NCERT p.17, 20]
1. Algebraic Expression Perimeter $= 2(l + w) = 20 \implies l + w = 10 \implies w = 10 - x$.
$$\text{Area } A(x) = x(10 - x) = 10x - x^2$$
2. Terms & Coefficients
• Terms: $10x$ and $-x^2$.
• Variable: $x$.
• Coefficient of $x = 10$, Coefficient of $x^2 = -1$.
• Degree $= 2$ (Quadratic function).
3. Evaluation at $x = 6\text{ cm}$
$$A(6) = 10(6) - (6)^2 = 60 - 36 = 24\text{ cm}^2$$
🎯 Final Answer: $A(x) = \mathbf{10x - x^2}$, Area at $x = 6\text{ cm}$ is $\mathbf{24\text{ cm}^2}$.
1. Expression बनाना Rectangle का perimeter $= 20\text{ cm}$ है $\implies l + w = 10 \implies w = 10 - x$।
$$\text{Area } A(x) = x(10 - x) = 10x - x^2$$
2. Terms और Coefficients
• Terms: $10x$ और $-x^2$ हैं।
• Variable: $x$ है।
• $x$ का coefficient $= 10$ और $x^2$ का coefficient $= -1$ है।
• Degree $= 2$ (Quadratic function) है।
3. $x = 6\text{ cm}$ पर Area
$$A(6) = 10(6) - (6)^2 = 60 - 36 = 24\text{ cm}^2$$
🎯 Final Answer: Area expression $A(x) = \mathbf{10x - x^2}$ है और $x = 6\text{ cm}$ पर area $\mathbf{24\text{ cm}^2}$ प्राप्त होता है।
Think & Reflect · p.19 A chess club charges a joining fee of ₹200 plus ₹50 for every match played ($C(m) = 200 + 50m$). If a player paid ₹750, how many matches did he play? [NCERT p.19]
Think & Reflect · p.19 Chess club ₹200 joining fee और ₹50 per match charge करता है ($C(m) = 200 + 50m$)। यदि एक player ने ₹750 pay किए, तो उसने कितने matches खेले? [NCERT p.19]
1. Formulate Linear Equation Let $m$ be the number of matches.
$$200 + 50m = 750$$
2. Solve for $m$
$$50m = 750 - 200 = 550 \implies m = \frac{550}{50} = 11$$
🎯 Final Answer: The player played 11 matches.
1. Equation बनाना मान लेते हैं कि खेले गए matches की count $m$ है।
$$200 + 50m = 750$$
2. $m$ की Value calculate करना
$$50m = 750 - 200 = 550 \implies m = \frac{550}{50} = 11$$
🎯 Final Answer: Player ने कुल 11 matches खेले।
Think & Reflect · p.22 For the square tile pattern $T_n = 2n - 1$: (i) Find tiles in 15th and 26th stages; (ii) Which stages contain 21 tiles and 47 tiles? [NCERT p.22]
Think & Reflect · p.22 Square tile pattern $T_n = 2n - 1$ के लिए: (i) 15th और 26th stage में tiles count करें; (ii) कौन से stages में 21 tiles और 47 tiles होंगे? [NCERT p.22]
1. Evaluation at $n = 15$ and $n = 26$
• $T_{15} = 2(15) - 1 = 30 - 1 = 29\text{ tiles}$.
• $T_{26} = 2(26) - 1 = 52 - 1 = 51\text{ tiles}$.
2. Find Stages for 21 and 47 Tiles
• $2n - 1 = 21 \implies 2n = 22 \implies n = 11$ (Stage 11).
• $2n - 1 = 47 \implies 2n = 48 \implies n = 24$ (Stage 24).
🎯 Final Answer: 15th Stage $= \mathbf{29}$, 26th Stage $= \mathbf{51}$; 21 tiles in Stage 11, 47 tiles in Stage 24.
1. Tiles की Calculation
• $T_{15} = 2(15) - 1 = 30 - 1 = 29\text{ tiles}$।
• $T_{26} = 2(26) - 1 = 52 - 1 = 51\text{ tiles}$।
2. Stages निकालना
• $2n - 1 = 21 \implies 2n = 22 \implies n = 11$ (Stage 11)।
• $2n - 1 = 47 \implies 2n = 48 \implies n = 24$ (Stage 24)।
🎯 Final Answer: 15th Stage $= \mathbf{29}$, 26th Stage $= \mathbf{51}$; 21 tiles Stage 11 में और 47 tiles Stage 24 में होंगे।
Think & Reflect · p.23 Bela has ₹100 and spends ₹5 every day ($B(n) = 100 - 5n$). (i) Amount left on 15th day; (ii) Days to spend entire amount. [NCERT p.23]
Think & Reflect · p.23 Bela के पास ₹100 pocket money है और वह ₹5 per day spend करती है ($B(n) = 100 - 5n$)। (i) 15th day पर कितना amount बचेगा? (ii) पूरा पैसा ख़त्म होने में कितने days लगेंगे? [NCERT p.23]
1. Amount on 15th Day $B(15) = 100 - 5(15) = 100 - 75 = \text{₹}25$.
2. Zero of Polynomial (Entire Amount Spent)
$$100 - 5n = 0 \implies 5n = 100 \implies n = 20\text{ days}$$
🎯 Final Answer: (i) ₹25 left on 15th day; (ii) Entire amount spent in 20 days.
1. 15th Day पर बचा Balance $B(15) = 100 - 5(15) = 100 - 75 = \text{₹}25$।
2. Zero of Polynomial
$$100 - 5n = 0 \implies 5n = 100 \implies n = 20\text{ days}$$
🎯 Final Answer: (i) 15th day पर ₹25 बचेंगे; (ii) पूरा amount 20 days में ख़त्म होगा।
Think & Reflect · p.24 Journey cost function is $C(d) = 100 + 60d$. (i) Find cost for $15\text{ km}$; (ii) Find distance for cost of ₹700. [NCERT p.24]
Think & Reflect · p.24 Journey cost function $C(d) = 100 + 60d$ है। (i) $15\text{ km}$ travel का cost निकालें; (ii) ₹700 cost के लिए distance calculate करें। [NCERT p.24]
1. Cost for $15\text{ km}$ $C(15) = 100 + 60(15) = 100 + 900 = \text{₹}1000$.
2. Distance for ₹700
$$100 + 60d = 700 \implies 60d = 600 \implies d = 10\text{ km}$$
🎯 Final Answer: (i) ₹1000 for $15\text{ km}$; (ii) $10\text{ km}$ for ₹700.
1. $15\text{ km}$ का Cost $C(15) = 100 + 60(15) = 100 + 900 = \text{₹}1000$।
2. ₹700 के लिए Distance
$$100 + 60d = 700 \implies 60d = 600 \implies d = 10\text{ km}$$
🎯 Final Answer: (i) $15\text{ km}$ का cost ₹1000 है; (ii) ₹700 में $10\text{ km}$ travel होगा।
Think & Reflect · p.27 In the linear relationship $y = 20x + 150$ for monthly internet bill, what do the numbers $20$ and $150$ represent in real life? [NCERT p.27]
Think & Reflect · p.27 Monthly internet bill के linear relation $y = 20x + 150$ में number $20$ और $150$ real life में क्या represent करते हैं? [NCERT p.27]
1. Meaning of Slope $a = 20$ $20$ represents the variable rate of ₹20 per GB of data consumed (the rate of change $\frac{\Delta y}{\Delta x}$).
2. Meaning of $y$-Intercept $b = 150$ $150$ represents the fixed monthly line rental / access charge of ₹150 incurred even when 0 GB data is used ($x = 0$).
🎯 Final Answer: $20 = \mathbf{\text{₹}20\text{ per GB (Rate of Change)}}$, $150 = \mathbf{\text{₹}150\text{ Fixed Monthly Fee (y-Intercept)}}$.
1. Slope $a = 20$ का मतलब $20$ प्रति GB data usage का variable rate (₹20 per GB) दर्शाता है (rate of change $\frac{\Delta y}{\Delta x}$)।
2. $y$-Intercept $b = 150$ का मतलब $150$ network का fixed monthly rental charge (₹150) दर्शाता है, जो $0\text{ GB}$ data use करने पर भी लगता है।
🎯 Final Answer: $20 = \mathbf{\text{₹}20\text{ per GB (Slope)}}$ और $150 = \mathbf{\text{₹}150\text{ Fixed Charge (y-Intercept)}}$ है।
Think & Reflect · p.33 Differentiate between the graphs of the linear equations $y = 3x + 1$ and $y = -3x + 1$. [NCERT p.33]
Think & Reflect · p.33 Linear equations $y = 3x + 1$ और $y = -3x + 1$ के graphs के बीच खास अंतर स्पष्ट करें। [NCERT p.33]
1. Similarities Both lines share the same $y$-intercept $(0, 1)$ since $b = 1$.
2. Differences in Slope & Direction
• $y = 3x + 1$: Slope $a = +3 > 0$. Represents linear growth; line rises from left to right with $x$-intercept $\left(-\frac{1}{3}, 0\right)$.
• $y = -3x + 1$: Slope $a = -3 < 0$. Represents linear decay; line falls from left to right with $x$-intercept $\left(\frac{1}{3}, 0\right)$.
🎯 Final Answer: Both meet at $(0, 1)$; $y = 3x+1$ has positive slope (growth), while $y = -3x+1$ has negative slope (decay).
1. समान बात दोनों lines का $y$-intercept समान $(0, 1)$ है क्योंकि दोनों में $b = 1$ है।
2. Slope और Direction का अंतर
• $y = 3x + 1$: Slope $a = +3 > 0$ है। यह linear growth दर्शाता है और graph ऊपर उठता है ($x$-intercept $= -1/3$) ।
• $y = -3x + 1$: Slope $a = -3 < 0$ है। यह linear decay दर्शाता है और graph नीचे झुकता है ($x$-intercept $= 1/3$) ।
🎯 Final Answer: दोनों $(0, 1)$ पर intersect करती हैं; $y = 3x+1$ का slope positive (growth) है तथा $y = -3x+1$ का slope negative (decay) है।
NCERT
5. Complete NCERT Exercise Solutions (Sets 2.1 to 2.6)
5. सम्पूर्ण NCERT Solutions (Exercise Sets 2.1 से 2.6)
NCERT
Exercise Set 2.1 — Degrees and Coefficients
Exercise Set 2.1 — Degrees और Coefficients
Ex 2.1 · Q1 Find the degrees of the following polynomials: (i) $2x^2 - 5x + 3$, (ii) $y^3 + 2y - 1$, (iii) $-9$, (iv) $4z - 3$. [NCERT p.18]
Ex 2.1 · Q1 नीचे दिए गए polynomials की degree identify करें: (i) $2x^2 - 5x + 3$, (ii) $y^3 + 2y - 1$, (iii) $-9$, (iv) $4z - 3$। [NCERT p.18]
1. Concept The degree of a polynomial is the highest power of the variable present in it.
2. Step-by-Step Evaluation
• (i) Highest exponent of $x$ is $2 \implies \text{Degree} = \mathbf{2}$ (Quadratic).
• (ii) Highest exponent of $y$ is $3 \implies \text{Degree} = \mathbf{3}$ (Cubic).
• (iii) Constant $-9 = -9x^0 \implies \text{Degree} = \mathbf{0}$ (Constant).
• (iv) Highest exponent of $z$ is $1 \implies \text{Degree} = \mathbf{1}$ (Linear).
🎯 Final Answer: (i) $\mathbf{2}$, (ii) $\mathbf{3}$, (iii) $\mathbf{0}$, (iv) $\mathbf{1}$.
1. Concept Polynomial की degree variable की highest power होती है।
2. Step-by-Step Check
• (i) $x$ की highest power $2$ है $\implies \text{Degree} = \mathbf{2}$।
• (ii) $y$ की highest power $3$ है $\implies \text{Degree} = \mathbf{3}$।
• (iii) Constant term $-9 = -9x^0 \implies \text{Degree} = \mathbf{0}$।
• (iv) $z$ की highest power $1$ है $\implies \text{Degree} = \mathbf{1}$।
🎯 Final Answer: (i) $\mathbf{2}$, (ii) $\mathbf{3}$, (iii) $\mathbf{0}$, (iv) $\mathbf{1}$।
Ex 2.1 · Q2 Write polynomials of degrees 1, 2 and 3. [NCERT p.18]
Ex 2.1 · Q2 Degrees 1, 2 और 3 वाले polynomials के examples लिखें। [NCERT p.18]
1. Definitions
• Degree 1 (Linear): $P(x) = ax + b$ ($a \neq 0$). Example: $3x + 5$.
• Degree 2 (Quadratic): $P(x) = ax^2 + bx + c$ ($a \neq 0$). Example: $2x^2 - 4x + 1$.
• Degree 3 (Cubic): $P(x) = ax^3 + bx^2 + cx + d$ ($a \neq 0$). Example: $x^3 + 7$.
🎯 Final Answer: Degree 1: $\mathbf{3x + 5}$; Degree 2: $\mathbf{2x^2 - 4x + 1}$; Degree 3: $\mathbf{x^3 + 7}$.
1. Definitions और Examples
• Degree 1 (Linear): $3x + 5$ (variable की highest power 1 है)।
• Degree 2 (Quadratic): $2x^2 - 4x + 1$ (variable की highest power 2 है)।
• Degree 3 (Cubic): $x^3 + 7$ (variable की highest power 3 है)।
🎯 Final Answer: Degree 1: $\mathbf{3x + 5}$; Degree 2: $\mathbf{2x^2 - 4x + 1}$; Degree 3: $\mathbf{x^3 + 7}$।
Ex 2.1 · Q3 What are the coefficients of $x^2$ and $x^3$ in the polynomial $x^4 - 3x^3 + 6x^2 - 2x + 7$? [NCERT p.18]
Ex 2.1 · Q3 Polynomial $x^4 - 3x^3 + 6x^2 - 2x + 7$ में $x^2$ और $x^3$ के coefficients क्या हैं? [NCERT p.18]
1. Inspection Identify the numerical multiplier before each power of $x$.
2. Coefficients
• Term containing $x^3$ is $-3x^3 \implies \text{Coefficient of } x^3 = \mathbf{-3}$.
• Term containing $x^2$ is $+6x^2 \implies \text{Coefficient of } x^2 = \mathbf{6}$.
🎯 Final Answer: Coefficient of $x^3 = \mathbf{-3}$, Coefficient of $x^2 = \mathbf{6}$.
1. Identification $x$ की हर power के आगे आने वाले numerical multiplier को identify करते हैं।
2. Coefficients निकालना
• $x^3$ वाला term $-3x^3$ है, इसलिए coefficient of $x^3 = -3$ होगा।
• $x^2$ वाला term $+6x^2$ है, इसलिए coefficient of $x^2 = 6$ होगा।
🎯 Final Answer: $x^3$ का coefficient $= \mathbf{-3}$ और $x^2$ का coefficient $= \mathbf{6}$ प्राप्त होता है।
Ex 2.1 · Q4 What is the coefficient of $z$ in the polynomial $4z^3 + 5z^2 - 11$? [NCERT p.19]
Ex 2.1 · Q4 Polynomial $4z^3 + 5z^2 - 11$ में $z$ का coefficient क्या है? [NCERT p.19]
1. Standard Form We can write the polynomial including missing powers as:
$$4z^3 + 5z^2 + 0z - 11$$
2. Coefficient The term with $z^1$ has multiplier $0$.
🎯 Final Answer: Coefficient of $z = \mathbf{0}$.
1. Standard Form Polynomial को missing terms के साथ लिख सकते हैं:
$$4z^3 + 5z^2 + 0z - 11$$
2. Coefficient चूंकि $z$ वाला term absent है, इसका multiplier $0$ है।
🎯 Final Answer: $z$ का coefficient $= \mathbf{0}$ है।
Ex 2.1 · Q5 What is the constant term of the polynomial $9x^3 + 5x^2 - 8x - 10$? [NCERT p.19]
Ex 2.1 · Q5 Polynomial $9x^3 + 5x^2 - 8x - 10$ का constant term क्या है? [NCERT p.19]
1. Definition The constant term is the term that does not contain the variable (or has exponent $x^0$).
2. Identification The term independent of $x$ is $-10$.
🎯 Final Answer: Constant term $= \mathbf{-10}$.
1. Definition Constant term वह term होता है जिसमें कोई variable नहीं होता ($x^0$ का term)।
2. Identification Variable $x$ से स्वतंत्र term $-10$ है।
🎯 Final Answer: Constant term $= \mathbf{-10}$ है।
NCERT
Exercise Set 2.2 — Polynomial Evaluation & Linear Word Problems
Exercise Set 2.2 — Polynomial Evaluation और Word Problems
Ex 2.2 · Q1 Find the value of the linear polynomial $P(x) = 5x - 3$ if: (i) $x = 0$, (ii) $x = -1$, (iii) $x = 2$. [NCERT p.21]
Ex 2.2 · Q1 Linear polynomial $P(x) = 5x - 3$ की value calculate करें यदि: (i) $x = 0$, (ii) $x = -1$, (iii) $x = 2$। [NCERT p.21]
1. Part (i) at $x = 0$ $P(0) = 5(0) - 3 = 0 - 3 = -3$.
2. Part (ii) at $x = -1$ $P(-1) = 5(-1) - 3 = -5 - 3 = -8$.
3. Part (iii) at $x = 2$ $P(2) = 5(2) - 3 = 10 - 3 = 7$.
🎯 Final Answer: (i) $\mathbf{-3}$, (ii) $\mathbf{-8}$, (iii) $\mathbf{7}$.
1. भाग (i) $x = 0$ पर $P(0) = 5(0) - 3 = 0 - 3 = -3$।
2. भाग (ii) $x = -1$ पर $P(-1) = 5(-1) - 3 = -5 - 3 = -8$।
3. भाग (iii) $x = 2$ पर $P(2) = 5(2) - 3 = 10 - 3 = 7$।
🎯 Final Answer: (i) $\mathbf{-3}$, (ii) $\mathbf{-8}$, (iii) $\mathbf{7}$ प्राप्त होता है।
Ex 2.2 · Q2 Find the value of the quadratic polynomial $Q(s) = 7s^2 - 4s + 6$ if: (i) $s = 0$, (ii) $s = -3$, (iii) $s = 4$. [NCERT p.21]
Ex 2.2 · Q2 Quadratic polynomial $Q(s) = 7s^2 - 4s + 6$ की value calculate करें यदि: (i) $s = 0$, (ii) $s = -3$, (iii) $s = 4$। [NCERT p.21]
1. Part (i) at $s = 0$ $Q(0) = 7(0)^2 - 4(0) + 6 = 6$.
2. Part (ii) at $s = -3$ $Q(-3) = 7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 18 = 81$.
3. Part (iii) at $s = 4$ $Q(4) = 7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 10 = 102$.
🎯 Final Answer: (i) $\mathbf{6}$, (ii) $\mathbf{81}$, (iii) $\mathbf{102}$.
1. भाग (i) $s = 0$ पर $Q(0) = 7(0)^2 - 4(0) + 6 = 6$।
2. भाग (ii) $s = -3$ पर $Q(-3) = 7(9) + 12 + 6 = 63 + 18 = 81$।
3. भाग (iii) $s = 4$ पर $Q(4) = 7(16) - 16 + 6 = 112 - 10 = 102$।
🎯 Final Answer: (i) $\mathbf{6}$, (ii) $\mathbf{81}$, (iii) $\mathbf{102}$ प्राप्त होता है।
Ex 2.2 · Q3 The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages. [NCERT p.21]
Ex 2.2 · Q3 सलिल की mother की present age सलिल की present age की 3 गुना है। 5 years बाद, उनकी ages का sum 70 years होगा। उनकी present ages find करें। [NCERT p.21]
1. Formulate Variables Let Salil's present age $= x\text{ years}$. Mother's present age $= 3x\text{ years}$.
2. After 5 Years Salil's age $= x + 5$; Mother's age $= 3x + 5$.
3. Linear Equation
$$(x + 5) + (3x + 5) = 70 \implies 4x + 10 = 70 \implies 4x = 60 \implies x = 15$$
🎯 Final Answer: Salil's present age $= \mathbf{15\text{ years}}$, Mother's present age $= 3(15) = \mathbf{45\text{ years}}$.
1. Variables मानना हम मान लेते हैं कि सलिल की present age $= x\text{ years}$ है, तो उसकी mother की present age $= 3x\text{ years}$ हो जाएगी।
2. 5 Years के बाद की आयु 5 years बीतने के बाद सलिल की age $= x + 5$ वर्ष और mother की age $= 3x + 5$ वर्ष हो जाएगी।
3. Equation solve करेंगे Question के according to दोनों की ages का sum 70 है:
$$(x + 5) + (3x + 5) = 70 \implies 4x + 10 = 70 \implies 4x = 60 \implies x = 15$$
🎯 Final Answer: अतः सलिल की present age $= 15\text{ years}$ है और उसकी mother की present age $= 45\text{ years}$ होगी।
Ex 2.2 · Q4 The difference between two positive integers is 63. The ratio of the two integers is $2:5$. Find the two integers. [NCERT p.21]
Ex 2.2 · Q4 दो positive integers का difference 63 है। दोनों integers का ratio $2:5$ है। दोनों integers find करें। [NCERT p.21]
1. Setup Ratio Variables Let the two integers be $2x$ and $5x$ where $x > 0$.
2. Linear Equation
$$5x - 2x = 63 \implies 3x = 63 \implies x = 21$$
3. Find Integers
• Smaller integer $= 2(21) = 42$.
• Larger integer $= 5(21) = 105$.
🎯 Final Answer: The two integers are 42 and 105.
1. Ratio Variables मानना मान लेते हैं कि दोनों integers $2x$ और $5x$ हैं।
2. Equation solve करना
$$5x - 2x = 63 \implies 3x = 63 \implies x = 21$$
3. Integers निकालना
• पहला integer $= 2(21) = 42$ है।
• दूसरा integer $= 5(21) = 105$ है।
🎯 Final Answer: दोनों integers 42 और 105 हैं।
Ex 2.2 · Q5 Ruby has 3 times as many ₹2 coins as she has ₹5 coins. If she has a total of ₹88, how many coins does she have of each type? [NCERT p.21]
Ex 2.2 · Q5 Ruby के पास ₹5 के coins से 3 गुना ज्यादा ₹2 के coins हैं। यदि उसके पास कुल ₹88 हैं, तो उसके पास हर एक प्रकार के कितने coins हैं? [NCERT p.21]
1. Define Variables Let the number of ₹5 coins $= x$. Then the number of ₹2 coins $= 3x$.
2. Total Value Equation
$$5(x) + 2(3x) = 88 \implies 5x + 6x = 88 \implies 11x = 88 \implies x = 8$$
3. Calculate Coin Counts
• ₹5 coins $= 8$.
• ₹2 coins $= 3(8) = 24$.
🎯 Final Answer: 8 coins of ₹5 and 24 coins of ₹2.
1. Variables मानना मान लेते हैं कि ₹5 के coins की count $= x$ है, तो ₹2 के coins की count $= 3x$ होगी।
2. Total Value Equation
$$5(x) + 2(3x) = 88 \implies 5x + 6x = 88 \implies 11x = 88 \implies x = 8$$
3. Coins की Count निकालना
• ₹5 के coins $= 8$ हैं।
• ₹2 के coins $= 3(8) = 24$ हैं।
🎯 Final Answer: Ruby के पास ₹5 के 8 coins और ₹2 के 24 coins हैं।
Ex 2.2 · Q6 A farmer cuts a $300\text{ feet}$ fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces? [NCERT p.21]
Ex 2.2 · Q6 एक किसान $300\text{ feet}$ लंबी fence को दो टुकड़ों में काटता है। बड़ा piece छोटे piece से 4 गुना लंबा है। दोनों pieces की length निकालें। [NCERT p.21]
1. Setup Variable Let the length of shorter piece $= x\text{ ft}$. Then longer piece $= 4x\text{ ft}$.
2. Equation
$$x + 4x = 300 \implies 5x = 300 \implies x = 60\text{ ft}$$
3. Longer Piece $4(60) = 240\text{ ft}$.
🎯 Final Answer: Shorter piece $= \mathbf{60\text{ ft}}$, Longer piece $= \mathbf{240\text{ ft}}$.
1. Variable मानना मान लेते हैं कि छोटे piece की length $= x\text{ ft}$ है, तो बड़े piece की length $= 4x\text{ ft}$ होगी।
2. Equation solve करना
$$x + 4x = 300 \implies 5x = 300 \implies x = 60\text{ ft}$$
3. बड़े piece की Length $4(60) = 240\text{ ft}$।
🎯 Final Answer: छोटा piece $60\text{ ft}$ और बड़ा piece $240\text{ ft}$ लंबा है।
Ex 2.2 · Q7 If the length of a rectangle is three more than twice its width and its perimeter is $24\text{ cm}$, what are the dimensions of the rectangle? [NCERT p.21]
Ex 2.2 · Q7 यदि एक rectangle की length उसकी width के दोगुने से $3\text{ cm}$ अधिक है और perimeter $24\text{ cm}$ है, तो rectangle की dimensions find करें। [NCERT p.21]
1. Variable Formulation Let width $= w\text{ cm}$. Length $l = 2w + 3\text{ cm}$.
2. Perimeter Formula
$$\text{Perimeter} = 2(l + w) = 24 \implies l + w = 12$$
$$(2w + 3) + w = 12 \implies 3w + 3 = 12 \implies 3w = 9 \implies w = 3\text{ cm}$$
3. Length $l = 2(3) + 3 = 9\text{ cm}$.
🎯 Final Answer: Width $= \mathbf{3\text{ cm}}$, Length $= \mathbf{9\text{ cm}}$.
1. Variables मानना मान लेते हैं कि width $= w\text{ cm}$ है। Length $l = 2w + 3\text{ cm}$ होगी।
2. Perimeter Equation
$$\text{Perimeter} = 2(l + w) = 24 \implies l + w = 12$$
$$(2w + 3) + w = 12 \implies 3w + 3 = 12 \implies 3w = 9 \implies w = 3\text{ cm}$$
3. Length calculate करना $l = 2(3) + 3 = 9\text{ cm}$।
🎯 Final Answer: Rectangle की Width $= \mathbf{3\text{ cm}}$ और Length $= \mathbf{9\text{ cm}}$ है।
NCERT
Exercise Set 2.3 — Linear Patterns & Sequences
Exercise Set 2.3 — Linear Patterns और Sequences
Ex 2.3 · Q1 A student has ₹500 in her savings bank account and gets ₹150 every month. How much money will she have at the end of each month from 2nd month onwards? Find a linear expression for the $n$-th month. [NCERT p.23]
Ex 2.3 · Q1 एक student के savings account में ₹500 हैं और उसे हर महीने ₹150 pocket money मिलती है। दूसरे महीने से हर महीने के अंत में उसके पास कितना पैसा होगा? $n$-th month के लिए linear expression निकालें। [NCERT p.23]
1. Sequence Pattern
• Initial: ₹500.
• Month 1: $500 + 150(1) = 650$.
• Month 2: $500 + 150(2) = 800$.
• Month 3: $500 + 150(3) = 950$.
2. Linear Expression for $n$-th Month
$$A(n) = 500 + 150n$$
🎯 Final Answer: Pattern: ₹800, ₹950, ₹1100...; Linear expression $= \mathbf{150n + 500}$ (or $150n + 350$ if counting $n=1$ as starting month).
1. Sequence बनाना
• Initial balance: ₹500.
• Month 1: $500 + 150 = 650$.
• Month 2: $650 + 150 = 800$.
• Month 3: $800 + 150 = 950$।
2. $n$-th Month का Linear Expression
$$A(n) = 500 + 150n$$
🎯 Final Answer: Amounts: ₹800, ₹950, ₹1100...; Linear model $= \mathbf{150n + 500}$ प्राप्त होता है।
Ex 2.3 · Q2 A rally starts with 120 members. Each hour, 9 members drop out. How many members remain after 1, 2, 3, … hours? Find a linear expression for the end of the $n$-th hour. [NCERT p.24]
Ex 2.3 · Q2 एक rally में शुरुआत में 120 members हैं। हर घंटे 9 members drop out होते हैं। 1, 2, 3... घंटे बाद कितने members बचेंगे? $n$-th hour के अंत में members के लिए linear expression निकालें। [NCERT p.24]
1. Sequence Calculation
• After 1 hour: $120 - 9(1) = 111$.
• After 2 hours: $120 - 9(2) = 102$.
• After 3 hours: $120 - 9(3) = 93$.
2. General $n$-th Hour Expression
$$M(n) = 120 - 9n$$
🎯 Final Answer: Members after 1, 2, 3 hours: 111, 102, 93; Linear expression $= \mathbf{120 - 9n}$.
1. Sequence की Calculation
• 1 घंटे बाद: $120 - 9(1) = 111$ members.
• 2 घंटे बाद: $120 - 9(2) = 102$ members.
• 3 घंटे बाद: $120 - 9(3) = 93$ members.
2. $n$-th Hour का Linear Expression
$$M(n) = 120 - 9n$$
🎯 Final Answer: 1, 2, 3 hours बाद members: 111, 102, 93; Linear expression $= \mathbf{120 - 9n}$ है।
Ex 2.3 · Q3 If the length of a rectangle is $13\text{ cm}$, find the area if breadth is: (i) $12\text{ cm}$, (ii) $10\text{ cm}$, (iii) $8\text{ cm}$. Find the linear pattern representing the area. [NCERT p.24]
Ex 2.3 · Q3 यदि rectangle की length $13\text{ cm}$ है, तो area find करें यदि breadth: (i) $12\text{ cm}$, (ii) $10\text{ cm}$, (iii) $8\text{ cm}$ हो। Area का linear pattern बताएं। [NCERT p.24]
1. Area Formula $\text{Area} = \text{length} \times \text{breadth} = 13b$.
2. Evaluations
• (i) $b = 12 \implies \text{Area} = 13 \times 12 = 156\text{ cm}^2$.
• (ii) $b = 10 \implies \text{Area} = 13 \times 10 = 130\text{ cm}^2$.
• (iii) $b = 8 \implies \text{Area} = 13 \times 8 = 104\text{ cm}^2$.
🎯 Final Answer: (i) $\mathbf{156\text{ cm}^2}$, (ii) $\mathbf{130\text{ cm}^2}$, (iii) $\mathbf{104\text{ cm}^2}$; Linear pattern $= \mathbf{13b}$.
1. Area Formula $\text{Area} = \text{length} \times \text{breadth} = 13b$।
2. Calculations
• (i) $b = 12 \implies \text{Area} = 13 \times 12 = 156\text{ cm}^2$।
• (ii) $b = 10 \implies \text{Area} = 13 \times 10 = 130\text{ cm}^2$।
• (iii) $b = 8 \implies \text{Area} = 13 \times 8 = 104\text{ cm}^2$।
🎯 Final Answer: (i) $\mathbf{156\text{ cm}^2}$, (ii) $\mathbf{130\text{ cm}^2}$, (iii) $\mathbf{104\text{ cm}^2}$; Linear pattern $= \mathbf{13b}$ है।
Ex 2.3 · Q4 A box has length $7\text{ cm}$ and breadth $11\text{ cm}$. Find volume if height is: (i) $5\text{ cm}$, (ii) $9\text{ cm}$, (iii) $13\text{ cm}$. Find the linear pattern for volume. [NCERT p.24]
Ex 2.3 · Q4 एक box की length $7\text{ cm}$ और breadth $11\text{ cm}$ है। Volume find करें यदि height: (i) $5\text{ cm}$, (ii) $9\text{ cm}$, (iii) $13\text{ cm}$ हो। Volume का linear pattern बताएं। [NCERT p.24]
1. Volume Formula $V = l \times b \times h = 7 \times 11 \times h = 77h$.
2. Calculations
• (i) $h = 5 \implies V = 77 \times 5 = 385\text{ cm}^3$.
• (ii) $h = 9 \implies V = 77 \times 9 = 693\text{ cm}^3$.
• (iii) $h = 13 \implies V = 77 \times 13 = 1001\text{ cm}^3$.
🎯 Final Answer: (i) $\mathbf{385\text{ cm}^3}$, (ii) $\mathbf{693\text{ cm}^3}$, (iii) $\mathbf{1001\text{ cm}^3}$; Linear pattern $= \mathbf{77h}$.
1. Volume Formula $V = l \times b \times h = 7 \times 11 \times h = 77h$।
2. Calculations
• (i) $h = 5 \implies V = 77 \times 5 = 385\text{ cm}^3$।
• (ii) $h = 9 \implies V = 77 \times 9 = 693\text{ cm}^3$।
• (iii) $h = 13 \implies V = 77 \times 13 = 1001\text{ cm}^3$।
🎯 Final Answer: (i) $\mathbf{385\text{ cm}^3}$, (ii) $\mathbf{693\text{ cm}^3}$, (iii) $\mathbf{1001\text{ cm}^3}$; Linear pattern $= \mathbf{77h}$ है।
Ex 2.3 · Q5 Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern. [NCERT p.24]
Ex 2.3 · Q5 सरिता 500 pages की book पढ़ रही है। वह रोज 20 pages पढ़ती है। 15 days बाद कितने pages बचेंगे? इसे linear pattern में express करें। [NCERT p.24]
1. Linear Pattern Formulation Let $d$ be the number of days.
$$P(d) = 500 - 20d$$
2. Evaluation at $d = 15$
$$P(15) = 500 - 20(15) = 500 - 300 = 200\text{ pages}$$
🎯 Final Answer: Linear pattern: $P(d) = \mathbf{500 - 20d}$; Pages left after 15 days $= \mathbf{200\text{ pages}}$.
1. Linear Pattern बनाना मान लेते हैं कि दिनों की count $d$ है।
$$P(d) = 500 - 20d$$
2. $d = 15$ पर Calculation
$$P(15) = 500 - 20(15) = 500 - 300 = 200\text{ pages}$$
🎯 Final Answer: Linear pattern $P(d) = \mathbf{500 - 20d}$ है और 15 days बाद 200 pages बचेंगे।
NCERT
Exercise Set 2.4 — Linear Growth & Linear Decay Models
Exercise Set 2.4 — Linear Growth और Linear Decay Models
Ex 2.4 · Q1 A plant is $1.75\text{ ft}$ tall and grows $0.5\text{ ft}$ each month. (i) Find height after 7 months; (ii) Table for $t = 0 \dots 10$; (iii) Linear model relating $h$ and $t$ and classify behavior. [NCERT p.25]
Ex 2.4 · Q1 एक पौधा $1.75\text{ ft}$ लंबा है और हर महीने $0.5\text{ ft}$ बढ़ता है। (i) 7 months बाद height find करें; (ii) $t = 0 \dots 10$ के लिए table बनाएं; (iii) Linear model लिखें और behavior explain करें। [NCERT p.25]
1. Part (i) Height after 7 Months $h(7) = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25\text{ ft}$.
2. Part (ii) Table of Values ($t = 0 \dots 5$)
• $t = 0 \implies 1.75\text{ ft}$, $t = 1 \implies 2.25\text{ ft}$, $t = 2 \implies 2.75\text{ ft}$, $t = 3 \implies 3.25\text{ ft}$, $t = 5 \implies 4.25\text{ ft}$, $t = 10 \implies 6.75\text{ ft}$.
3. Part (iii) Linear Model $h(t) = 0.5t + 1.75$. Since the slope $+0.5 > 0$ (increases by a constant amount each month), it represents linear growth.
🎯 Final Answer: (i) $\mathbf{5.25\text{ ft}}$; (iii) $h(t) = \mathbf{0.5t + 1.75}$ (Linear Growth).
1. भाग (i) 7 Months बाद Height $h(7) = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25\text{ ft}$।
2. भाग (ii) Table of Values
• $t = 0 \implies 1.75\text{ ft}$, $t = 1 \implies 2.25\text{ ft}$, $t = 2 \implies 2.75\text{ ft}$, $t = 3 \implies 3.25\text{ ft}$, $t = 7 \implies 5.25\text{ ft}$।
3. भाग (iii) Linear Model $h(t) = 0.5t + 1.75$। चूंकि slope $a = +0.5 > 0$ (positive) है, इसलिए यह linear growth को दर्शाता है।
🎯 Final Answer: (i) $\mathbf{5.25\text{ ft}}$; (iii) $h(t) = \mathbf{0.5t + 1.75}$ (Linear Growth)।
Ex 2.4 · Q2 A mobile phone bought for ₹10,000 depreciates by ₹800 every year. (i) Value after 3 years; (ii) Table for $t = 0 \dots 8$; (iii) Linear model relating $v$ and $t$ and explain decay. [NCERT p.25]
Ex 2.4 · Q2 ₹10,000 में खरीदा गया mobile phone हर साल ₹800 depreciate होता है। (i) 3 years बाद value निकालें; (ii) $t = 0 \dots 8$ के लिए table बनाएं; (iii) Linear model लिखें और decay explain करें। [NCERT p.25]
1. Part (i) Value after 3 Years $v(3) = 10000 - 800(3) = 10000 - 2400 = \text{₹}7600$.
2. Part (ii) Depreciation Table
• $t = 0 \implies \text{₹}10000$, $t = 1 \implies \text{₹}9200$, $t = 2 \implies \text{₹}8400$, $t = 3 \implies \text{₹}7600$, $t = 5 \implies \text{₹}6000$, $t = 8 \implies \text{₹}3600$.
3. Part (iii) Linear Decay Model $v(t) = 10000 - 800t$. Since value decreases by a constant ₹800 each year (slope $-800 < 0$), it represents linear decay.
🎯 Final Answer: (i) ₹7,600; (iii) $v(t) = \mathbf{10000 - 800t}$ (Linear Decay).
1. भाग (i) 3 Years बाद Phone की Value $v(3) = 10000 - 800(3) = 10000 - 2400 = \text{₹}7600$।
2. भाग (ii) Depreciation Table
• $t = 0 \implies \text{₹}10000$, $t = 1 \implies \text{₹}9200$, $t = 2 \implies \text{₹}8400$, $t = 3 \implies \text{₹}7600$।
3. भाग (iii) Linear Decay Model $v(t) = 10000 - 800t$। चूंकि value हर साल ₹800 घट रही है (slope $a = -800 < 0$), इसलिए यह linear decay है।
🎯 Final Answer: (i) ₹7,600; (iii) $v(t) = \mathbf{10000 - 800t}$ (Linear Decay)।
Ex 2.4 · Q3 Village population is initially 750, and 50 people move in every year. (i) Population after 6 years; (ii) Table for $t = 0 \dots 10$; (iii) Linear model relating $P$ and $t$. [NCERT p.25-26]
Ex 2.4 · Q3 एक village की initial population 750 है और हर साल 50 people city से आते हैं। (i) 6 years बाद population निकालें; (ii) $t = 0 \dots 10$ table बनाएं; (iii) Linear model लिखें। [NCERT p.25-26]
1. Part (i) Population after 6 Years $P(6) = 750 + 50(6) = 750 + 300 = 1050\text{ people}$.
2. Part (ii) Table $t = 0 \to 750, t = 1 \to 800, t = 2 \to 850, t = 6 \to 1050, t = 10 \to 1250$.
3. Part (iii) Linear Growth Model $P(t) = 750 + 50t$. Positive slope $+50$ indicates linear growth.
🎯 Final Answer: (i) 1,050 people; (iii) $P(t) = \mathbf{750 + 50t}$ (Linear Growth).
1. भाग (i) 6 Years बाद Population $P(6) = 750 + 50(6) = 750 + 300 = 1050$ people।
2. भाग (ii) Table $t = 0 \to 750, t = 1 \to 800, t = 2 \to 850, t = 6 \to 1050$।
3. भाग (iii) Linear Growth Model $P(t) = 750 + 50t$। Positive rate $+50$ की वजह से यह linear growth है।
🎯 Final Answer: (i) 1,050 people; (iii) $P(t) = \mathbf{750 + 50t}$ (Linear Growth)।
Ex 2.4 · Q4 A ₹600 prepaid recharge balance reduces by ₹15 each day. (i) Model balance $b(x)$ after $x$ days and classify; (ii) Days until balance runs out; (iii) Table for $x = 1 \dots 10$. [NCERT p.26]
Ex 2.4 · Q4 ₹600 का prepaid balance हर दिन ₹15 reduce होता है। (i) $x$ days बाद balance $b(x)$ का model लिखें; (ii) कितने days में balance ख़त्म होगा? (iii) $x = 1 \dots 10$ table बनाएं। [NCERT p.26]
1. Part (i) Linear Equation $b(x) = 600 - 15x$. Slope $-15 < 0 \implies$ Linear Decay.
2. Part (ii) Balance Depletion
$$600 - 15x = 0 \implies 15x = 600 \implies x = 40\text{ days}$$
3. Part (iii) Table Day 1: 585, Day 2: 570, Day 3: 555, Day 5: 525, Day 10: 450.
🎯 Final Answer: (i) $b(x) = \mathbf{600 - 15x}$ (Linear Decay); (ii) Balance runs out in 40 days.
1. भाग (i) Linear Model $b(x) = 600 - 15x$। Negative slope $-15 < 0$ की वजह से यह linear decay है।
2. भाग (ii) Balance ख़त्म होने के दिन
$$600 - 15x = 0 \implies 15x = 600 \implies x = 40\text{ days}$$
3. भाग (iii) Table Day 1: ₹585, Day 2: ₹570, Day 3: ₹555, Day 10: ₹450।
🎯 Final Answer: (i) $b(x) = \mathbf{600 - 15x}$ (Linear Decay); (ii) Balance 40 days में ख़त्म होगा।
NCERT
Exercise Set 2.5 — Determining Linear Relationships ($y = ax + b$)
Exercise Set 2.5 — Linear Relationships निकालना ($y = ax + b$)
Ex 2.5 · Q1 A platform charges a fixed fee plus a cost per module: for 10 modules bill is ₹400, for 14 modules bill is ₹500. Find $a$ and $b$ in $y = ax + b$. [NCERT p.27]
Ex 2.5 · Q1 एक learning platform fixed fee और per module cost charge करता है: 10 modules के लिए bill ₹400 है, 14 modules के लिए bill ₹500 है। $y = ax + b$ में $a$ और $b$ की values find करें। [NCERT p.27]
1. Formulate System of Equations
$$10a + b = 400 \quad \text{--- (1)}$$
$$14a + b = 500 \quad \text{--- (2)}$$
2. Subtract (1) from (2)
$$(14a + b) - (10a + b) = 500 - 400 \implies 4a = 100 \implies a = 25$$
3. Substitute $a = 25$ into (1)
$$10(25) + b = 400 \implies 250 + b = 400 \implies b = 150$$
🎯 Final Answer: $a = \mathbf{25}$, $b = \mathbf{150}$; Linear relation is $\mathbf{y = 25x + 150}$.
1. Equations बनाना
$$10a + b = 400 \quad \text{--- (1)}$$
$$14a + b = 500 \quad \text{--- (2)}$$
2. Equation (2) में से (1) घटाना
$$4a = 100 \implies a = 25$$
3. $a = 25$ Equation (1) में रखना
$$10(25) + b = 400 \implies 250 + b = 400 \implies b = 150$$
🎯 Final Answer: $a = \mathbf{25}$ और $b = \mathbf{150}$ है; Linear relation $\mathbf{y = 25x + 150}$ बनता है।
Ex 2.5 · Q2 A gym charges a fixed fee plus hourly court fee: for 10 hours bill is ₹800, for 15 hours bill is ₹1100. Find $a$ and $b$ in $y = ax + b$. [NCERT p.27]
Ex 2.5 · Q2 Gym fixed fee और per hour court fee charge करता है: 10 hours के लिए bill ₹800 है, 15 hours के लिए bill ₹1100 है। $y = ax + b$ में $a$ और $b$ find करें। [NCERT p.27]
1. Formulate System
$$10a + b = 800 \quad \text{--- (1)}$$
$$15a + b = 1100 \quad \text{--- (2)}$$
2. Solve for $a$
$$(15a + b) - (10a + b) = 1100 - 800 \implies 5a = 300 \implies a = 60$$
3. Solve for $b$
$$10(60) + b = 800 \implies 600 + b = 800 \implies b = 200$$
🎯 Final Answer: $a = \mathbf{60}$, $b = \mathbf{200}$; Linear relation is $\mathbf{y = 60x + 200}$.
1. Equations बनाना
$$10a + b = 800 \quad \text{--- (1)}$$
$$15a + b = 1100 \quad \text{--- (2)}$$
2. $a$ की Calculation
$$5a = 300 \implies a = 60$$
3. $b$ की Calculation
$$10(60) + b = 800 \implies b = 200$$
🎯 Final Answer: $a = \mathbf{60}$ और $b = \mathbf{200}$ प्राप्त होता है ($y = 60x + 200$)।
Ex 2.5 · Q3 The temperature conversion relation is ${}^\circ\text{C} = a\,{}^\circ\text{F} + b$. Given ice melts at $(32\,{}^\circ\text{F}, 0\,{}^\circ\text{C})$ and water boils at $(212\,{}^\circ\text{F}, 100\,{}^\circ\text{C})$, find $a, b$ and the relation. [NCERT p.27]
Ex 2.5 · Q3 Temperature conversion relation ${}^\circ\text{C} = a\,{}^\circ\text{F} + b$ है। Ice $32\,{}^\circ\text{F}$ ($0\,{}^\circ\text{C}$) पर melt होती है और water $212\,{}^\circ\text{F}$ ($100\,{}^\circ\text{C}$) पर boil होता है। $a, b$ और linear relation find करें। [NCERT p.27]
1. Formulate System
$$32a + b = 0 \implies b = -32a \quad \text{--- (1)}$$
$$212a + b = 100 \quad \text{--- (2)}$$
2. Solve for $a$
$$212a - 32a = 100 \implies 180a = 100 \implies a = \frac{100}{180} = \frac{5}{9}$$
3. Solve for $b$
$$b = -32\left(\frac{5}{9}\right) = -\frac{160}{9}$$
🎯 Final Answer: $a = \mathbf{\frac{5}{9}}$, $b = \mathbf{-\frac{160}{9}}$; Formula: $\mathbf{{}^\circ\text{C} = \frac{5}{9}({}^\circ\text{F} - 32)}$.
1. Equations बनाना
$$32a + b = 0 \implies b = -32a \quad \text{--- (1)}$$
$$212a + b = 100 \quad \text{--- (2)}$$
2. $a$ calculate करना
$$180a = 100 \implies a = \frac{5}{9}$$
3. $b$ calculate करना
$$b = -32\left(\frac{5}{9}\right) = -\frac{160}{9}$$
🎯 Final Answer: $a = \mathbf{\frac{5}{9}}$, $b = \mathbf{-\frac{160}{9}}$; Formula: $\mathbf{{}^\circ\text{C} = \frac{5}{9}({}^\circ\text{F} - 32)}$ है।
NCERT
Exercise Set 2.6 — Graphical Reflection on Slopes & Intercepts
Exercise Set 2.6 — Slopes और Intercepts का Graphical अध्ययन
Ex 2.6 · Q1 Reflect on the roles of slope $a$ and $y$-intercept $b$ for the following sets of lines: (i) $y = 4x, 2x, x$; (ii) $y = -6x, -3x, -x$; (iii) $y = 5x, -5x$; (iv) $y = 3x - 1, 3x, 3x + 1$; (v) $y = -2x - 3, -2x, 2x + 3$. [NCERT p.36]
Ex 2.6 · Q1 नीचे दिए गए lines के sets के लिए slope $a$ और $y$-intercept $b$ के प्रभाव का विश्लेषण करें: (i) $y = 4x, 2x, x$; (ii) $y = -6x, -3x, -x$; (iii) $y = 5x, -5x$; (iv) $y = 3x - 1, 3x, 3x + 1$; (v) $y = -2x - 3, -2x, 2x + 3$। [NCERT p.36]
1. Parts (i) & (ii) Origin Lines
• (i) $y = 4x, 2x, x$: All have $b = 0$ (pass through origin $(0, 0)$). As $a$ increases ($1 \to 2 \to 4$), the line becomes steeper towards the $y$-axis.
• (ii) $y = -6x, -3x, -x$: All have $b = 0$, but negative slopes ($a < 0$) fall from left to right (linear decay).
2. Part (iii) Symmetrical Lines
• $y = 5x$ and $y = -5x$ are reflections of each other across the $y$-axis (and $x$-axis) with equal steepness ($|a| = 5$).
3. Parts (iv) & (v) Parallel Lines & Shifts
• (iv) $y = 3x - 1, 3x, 3x + 1$: All have identical slope $a = 3$, so they are strictly parallel. Varying $b = -1, 0, 1$ shifts the line vertically without changing tilt.
• (v) $y = -2x - 3$ and $y = -2x$ are parallel ($a = -2$). $y = 2x + 3$ intersects them with opposite positive slope $a = +2$.
🎯 Core Geometric Principles: $a$ controls steepness/tilt; $b$ controls vertical shift $(0, b)$; equal slopes with different intercepts create parallel lines.
1. भाग (i) व (ii) Origin Lines
• (i) $y = 4x, 2x, x$: सभी में $b = 0$ है, इसलिए origin $(0, 0)$ से गुजरती हैं। $a$ बढ़ने पर line अधिक steep होती है।
• (ii) $y = -6x, -3x, -x$: Negative slope ($a < 0$) के कारण ये decay lines हैं।
2. भाग (iii) Symmetrical Lines
• $y = 5x$ और $y = -5x$ $y$-axis के सापेक्ष mirror images हैं।
3. भाग (iv) व (v) Parallel Lines
• (iv) $y = 3x - 1, 3x, 3x + 1$: Slope $a = 3$ समान होने के कारण ये parallel lines हैं। $b$ बदलने से line vertical shift होती है।
• (v) $y = -2x - 3$ और $y = -2x$ parallel हैं।
🎯 मुख्य निष्कर्ष: Slope $a$ line का steepness तय करता है; $b$ $y$-intercept $(0, b)$ पर vertical shift तय करता है; समान slope वाली lines parallel होती हैं।
NCERT
11. End-of-Chapter Comprehensive Exercises (Q1 to Q14)
11. अध्याय के अंत के विस्तृत अभ्यास प्रश्न (Q1 से Q14)
End Ex · Q1 Write a polynomial of degree 3 in the variable $x$, in which the coefficient of the $x^2$ term is $-7$. [NCERT p.36]
End Ex · Q1 Variable $x$ में degree 3 का एक polynomial लिखें जिसमें $x^2$ term का coefficient $-7$ हो। [NCERT p.36]
1. Conditions Degree must be 3 (highest power $x^3$, coefficient non-zero) and $x^2$ coefficient must be $-7$.
2. Example Formulation $P(x) = 2x^3 - 7x^2 + 4x - 1$.
🎯 Final Answer: $\mathbf{2x^3 - 7x^2 + 4x - 1}$ (Any valid cubic with $-7x^2$).
1. Conditions Degree 3 होनी चाहिए ($x^3$ term उपस्थित हो) और $x^2$ का multiplier $-7$ होना चाहिए।
2. Example Polynomial $P(x) = 2x^3 - 7x^2 + 4x - 1$।
🎯 Final Answer: $\mathbf{2x^3 - 7x^2 + 4x - 1}$।
End Ex · Q2 Find the values of: (i) $P(x) = 5x^2 - 3x + 7$ if $x = 1$; (ii) $Q(t) = 4t^3 - t^2 + 6$ if $t = a$. [NCERT p.36]
End Ex · Q2 Values calculate करें: (i) $P(x) = 5x^2 - 3x + 7$ यदि $x = 1$; (ii) $Q(t) = 4t^3 - t^2 + 6$ यदि $t = a$। [NCERT p.36]
1. Part (i) at $x = 1$ $P(1) = 5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = 9$.
2. Part (ii) at $t = a$ $Q(a) = 4a^3 - a^2 + 6$.
🎯 Final Answer: (i) $\mathbf{9}$, (ii) $\mathbf{4a^3 - a^2 + 6}$.
1. भाग (i) $x = 1$ पर $P(1) = 5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = 9$।
2. भाग (ii) $t = a$ पर $Q(a) = 4a^3 - a^2 + 6$।
🎯 Final Answer: (i) $\mathbf{9}$, (ii) $\mathbf{4a^3 - a^2 + 6}$ प्राप्त होता है।
End Ex · Q3 If we multiply a number by $\frac{5}{2}$ and add $\frac{2}{3}$ to the product, we get $-\frac{7}{12}$. Find the number. [NCERT p.37]
End Ex · Q3 यदि हम किसी number को $\frac{5}{2}$ से multiply करके product में $\frac{2}{3}$ जोड़ें, तो हमें $-\frac{7}{12}$ मिलता है। Number find करें। [NCERT p.37]
1. Linear Equation Let the number be $x$.
$$\frac{5}{2}x + \frac{2}{3} = -\frac{7}{12}$$
2. Transpose Constant Term
$$\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3} = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4}$$
3. Solve for $x$
$$x = -\frac{5}{4} \times \frac{2}{5} = -\frac{2}{4} = -\frac{1}{2}$$
🎯 Final Answer: The number is $\mathbf{-\frac{1}{2}}$.
1. Linear Equation बनाना मान लेते हैं कि number $x$ है।
$$\frac{5}{2}x + \frac{2}{3} = -\frac{7}{12}$$
2. Fractions solve करना
$$\frac{5}{2}x = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4}$$
3. $x$ calculate करना
$$x = -\frac{5}{4} \times \frac{2}{5} = -\frac{1}{2}$$
🎯 Final Answer: वह number $\mathbf{-\frac{1}{2}}$ है।
End Ex · Q4 A positive number is 5 times another number. If 21 is added to both numbers, one becomes twice the other. Find the numbers. [NCERT p.37]
End Ex · Q4 एक positive number दूसरे number का 5 गुना है। यदि दोनों में 21 जोड़ दिया जाए, तो एक नया number दूसरे का दोगुना हो जाता है। Numbers find करें। [NCERT p.37]
1. Define Variables Let the smaller number $= x$. Larger number $= 5x$.
2. After Adding 21 New numbers are $5x + 21$ and $x + 21$.
$$5x + 21 = 2(x + 21) \implies 5x + 21 = 2x + 42$$
3. Solve for $x$
$$5x - 2x = 42 - 21 \implies 3x = 21 \implies x = 7$$
Larger number $= 5(7) = 35$.
🎯 Final Answer: The numbers are 7 and 35.
1. Variables मानना मान लेते हैं कि पहला number $= x$ है, तो दूसरा number $= 5x$ होगा।
2. 21 जोड़ने पर Equation
$$5x + 21 = 2(x + 21) \implies 5x + 21 = 2x + 42$$
3. $x$ calculate करना
$$3x = 21 \implies x = 7$$
दूसरा number $= 5(7) = 35$।
🎯 Final Answer: दोनों numbers 7 और 35 हैं।
End Ex · Q5 If you have ₹800 and save ₹250 every month, find the amount after: (i) 6 months; (ii) 2 years. Express as a linear pattern. [NCERT p.37]
End Ex · Q5 यदि आपके पास ₹800 हैं और आप हर महीने ₹250 save करते हैं, तो amount calculate करें: (i) 6 months बाद; (ii) 2 years बाद। इसे linear pattern में लिखें। [NCERT p.37]
1. Linear Pattern Model $A(m) = 800 + 250m$, where $m$ is months.
2. Part (i) After 6 Months $A(6) = 800 + 250(6) = 800 + 1500 = \text{₹}2300$.
3. Part (ii) After 2 Years ($m = 24$) $A(24) = 800 + 250(24) = 800 + 6000 = \text{₹}6800$.
🎯 Final Answer: Pattern: $A(m) = \mathbf{800 + 250m}$; (i) ₹2,300, (ii) ₹6,800.
1. Linear Pattern बनाना $m$ months बाद total amount: $A(m) = 800 + 250m$।
2. भाग (i) 6 Months बाद $A(6) = 800 + 250(6) = 800 + 1500 = \text{₹}2300$।
3. भाग (ii) 2 Years ($24\text{ months}$) बाद $A(24) = 800 + 250(24) = 800 + 6000 = \text{₹}6800$।
🎯 Final Answer: Linear model: $A(m) = \mathbf{800 + 250m}$; (i) ₹2,300, (ii) ₹6,800 प्राप्त होता है।
End Ex · Q6 The digits of a two-digit number differ by 3. If the digits are interchanged and added to the original number, the sum is 143. Find both numbers. [NCERT p.37]
End Ex · Q6 दो digits की number के digits का difference 3 है। यदि digits को interchange करके original number में जोड़ा जाए, तो sum 143 मिलता है। दोनों numbers find करें। [NCERT p.37]
1. Number Structure Let the tens digit $= t$ and units digit $= u$. Original number $= 10t + u$. Reversed number $= 10u + t$.
2. Sum Equation
$$(10t + u) + (10u + t) = 143 \implies 11t + 11u = 143 \implies t + u = 13 \quad \text{--- (1)}$$
3. Two Cases for Difference $|t - u| = 3$
• Case 1 ($t - u = 3$): Adding with (1) gives $2t = 16 \implies t = 8, u = 5 \implies \text{Number } = \mathbf{85}$.
• Case 2 ($u - t = 3$): Adding with (1) gives $2u = 16 \implies u = 8, t = 5 \implies \text{Number } = \mathbf{58}$.
🎯 Final Answer: The numbers are 58 and 85.
1. Number की Structure मान लेते हैं tens digit $= t$ और units digit $= u$ है। Number $= 10t + u$, Reversed $= 10u + t$।
2. Sum Equation
$$(10t + u) + (10u + t) = 143 \implies 11(t + u) = 143 \implies t + u = 13 \quad \text{--- (1)}$$
3. दोनों Cases का हल
• Case 1 ($t - u = 3$): $2t = 16 \implies t = 8, u = 5 \implies \text{Number } = \mathbf{85}$।
• Case 2 ($u - t = 3$): $2u = 16 \implies u = 8, t = 5 \implies \text{Number } = \mathbf{58}$।
🎯 Final Answer: वे numbers 58 और 85 हैं।
End Ex · Q7 Identify slopes and $y$-intercepts for: (i) $y = -3x + 4$, (ii) $2y = 4x + 7$, (iii) $5y = 6x - 10$, (iv) $3y = 6x - 11$. Are any lines parallel? [NCERT p.37]
End Ex · Q7 Slopes और $y$-intercepts identify करें: (i) $y = -3x + 4$, (ii) $2y = 4x + 7$, (iii) $5y = 6x - 10$, (iv) $3y = 6x - 11$। क्या कोई lines parallel हैं? [NCERT p.37]
1. Convert to Standard Form $y = ax + b$
• (i) $y = -3x + 4 \implies \text{Slope } a = -3, y\text{-intercept } = (0, 4)$.
• (ii) $y = 2x + \frac{7}{2} \implies \text{Slope } a = 2, y\text{-intercept } = (0, 3.5)$.
• (iii) $y = \frac{6}{5}x - 2 \implies \text{Slope } a = 1.2, y\text{-intercept } = (0, -2)$.
• (iv) $y = 2x - \frac{11}{3} \implies \text{Slope } a = 2, y\text{-intercept } = \left(0, -\frac{11}{3}\right)$.
2. Parallel Lines Check Lines (ii) and (iv) both have slope $a = 2$ with different $y$-intercepts, so Lines (ii) and (iv) are parallel.
🎯 Final Answer: Slopes: $-3, 2, 1.2, 2$; Lines (ii) and (iv) are strictly parallel ($a = 2$).
1. $y = ax + b$ form में बदलना
• (i) $y = -3x + 4 \implies \text{Slope } a = -3, y\text{-intercept } = (0, 4)$।
• (ii) $y = 2x + 3.5 \implies \text{Slope } a = 2, y\text{-intercept } = (0, 3.5)$।
• (iii) $y = 1.2x - 2 \implies \text{Slope } a = 1.2, y\text{-intercept } = (0, -2)$।
• (iv) $y = 2x - \frac{11}{3} \implies \text{Slope } a = 2, y\text{-intercept } = \left(0, -\frac{11}{3}\right)$।
2. Parallel Lines Lines (ii) और (iv) दोनों का slope $a = 2$ समान है, इसलिए Lines (ii) और (iv) parallel हैं।
🎯 Final Answer: Slopes: $-3, 2, 1.2, 2$; Lines (ii) और (iv) parallel हैं ($a = 2$)।
End Ex · Q8 Liquid temperature relation is $y = \frac{9}{5}(x - 273) + 32$ ($x\text{ K}, y\,{}^\circ\text{F}$). (i) Find ${}^\circ\text{F}$ for $313\text{ K}$; (ii) Find $\text{K}$ for $158\,{}^\circ\text{F}$. [NCERT p.37]
End Ex · Q8 Liquid temperature relation $y = \frac{9}{5}(x - 273) + 32$ है। (i) $313\text{ K}$ के लिए ${}^\circ\text{F}$ में temperature निकालें; (ii) $158\,{}^\circ\text{F}$ के लिए Kelvin में value find करें। [NCERT p.37]
1. Part (i) at $x = 313\text{ K}$
$$y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32 = 72 + 32 = 104\,{}^\circ\text{F}$$
2. Part (ii) at $y = 158\,{}^\circ\text{F}$
$$158 - 32 = \frac{9}{5}(x - 273) \implies 126 = \frac{9}{5}(x - 273)$$
$$x - 273 = 126 \times \frac{5}{9} = 14 \times 5 = 70 \implies x = 273 + 70 = 343\text{ K}$$
🎯 Final Answer: (i) $\mathbf{104\,{}^\circ\text{F}}$, (ii) $\mathbf{343\text{ K}}$.
1. भाग (i) $x = 313\text{ K}$ पर
$$y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32 = 72 + 32 = 104\,{}^\circ\text{F}$$
2. भाग (ii) $y = 158\,{}^\circ\text{F}$ पर
$$126 = \frac{9}{5}(x - 273) \implies x - 273 = 70 \implies x = 343\text{ K}$$
🎯 Final Answer: (i) $\mathbf{104\,{}^\circ\text{F}}$, (ii) $\mathbf{343\text{ K}}$ प्राप्त होता है।
End Ex · Q9 Work done by a constant force of 3 units is $W = 3d$. Find work done when distance $d = 2\text{ units}$. [NCERT p.37-38]
End Ex · Q9 3 units के constant force द्वारा किया गया work $W = 3d$ है। जब distance $d = 2\text{ units}$ हो, तो work done find करें। [NCERT p.37-38]
1. Linear Equation $W(d) = 3d$, representing direct proportional growth through the origin $(0, 0)$.
2. Evaluation at $d = 2$ $W(2) = 3(2) = 6\text{ units}$.
🎯 Final Answer: Work done $= \mathbf{6\text{ units}}$.
1. Linear Equation $W(d) = 3d$ (यह origin $(0, 0)$ से गुजरने वाली direct proportionality line है)।
2. $d = 2$ पर Calculation $W(2) = 3(2) = 6\text{ units}$।
🎯 Final Answer: किया गया work done $= \mathbf{6\text{ units}}$ है।
End Ex · Q10 The graph of $p(x) = ax + b$ passes through $(1, 5)$ and $(3, 11)$. (i) Find $p(x)$; (ii) Find axes intersection points. [NCERT p.38]
End Ex · Q10 Linear polynomial $p(x) = ax + b$ का graph $(1, 5)$ और $(3, 11)$ से गुजरता है। (i) $p(x)$ निकालें; (ii) Axes पर intersection points find करें। [NCERT p.38]
1. Calculate Slope $a$
$$a = \frac{11 - 5}{3 - 1} = \frac{6}{2} = 3$$
2. Calculate $y$-Intercept $b$ Using point $(1, 5)$:
$$5 = 3(1) + b \implies b = 2 \implies p(x) = 3x + 2$$
3. Axes Intercepts
• $y$-axis cut ($x = 0$): $(0, 2)$.
• $x$-axis cut ($y = 0$): $3x + 2 = 0 \implies x = -2/3 \implies \left(-\frac{2}{3}, 0\right)$.
🎯 Final Answer: $p(x) = \mathbf{3x + 2}$; Cuts $y$-axis at $\mathbf{(0, 2)}$, $x$-axis at $\mathbf{\left(-\frac{2}{3}, 0\right)}$.
1. Slope $a$ calculate करना
$$a = \frac{11 - 5}{3 - 1} = \frac{6}{2} = 3$$
2. $y$-Intercept $b$ निकालना Point $(1, 5)$ put करने पर:
$$5 = 3(1) + b \implies b = 2 \implies p(x) = 3x + 2$$
3. Axes के Intercepts
• $y$-axis पर cut ($x = 0$): $(0, 2)$।
• $x$-axis पर cut ($y = 0$): $3x + 2 = 0 \implies x = -2/3 \implies \left(-\frac{2}{3}, 0\right)$।
🎯 Final Answer: Polynomial $p(x) = \mathbf{3x + 2}$ है; Axes cuts: $\mathbf{(0, 2)}$ और $\mathbf{\left(-\frac{2}{3}, 0\right)}$ हैं।
End Ex · Q11 Let $p(x) = ax + b$ and $q(x) = cx + d$ such that: (i) $p(0) = 5$, (ii) $p(x) - q(x)$ cuts $x$-axis at $(3, 0)$, (iii) $p(x) + q(x) = 6x + 4$ for all $x$. Find $p(x)$ and $q(x)$. [NCERT p.38]
End Ex · Q11 Linear polynomials $p(x) = ax + b$ और $q(x) = cx + d$ इस प्रकार हैं कि: (i) $p(0) = 5$, (ii) $p(x) - q(x)$ $x$-axis को $(3, 0)$ पर काटता है, (iii) $p(x) + q(x) = 6x + 4$। $p(x)$ और $q(x)$ find करें। [NCERT p.38]
1. Evaluate Constants $b$ and $d$
• $p(0) = b = 5$.
• $p(x) + q(x) = (a+c)x + (b+d) = 6x + 4 \implies a + c = 6$ and $b + d = 4 \implies 5 + d = 4 \implies d = -1$.
2. Difference Polynomial $x$-Intercept
$$p(x) - q(x) = (a - c)x + (5 - (-1)) = (a - c)x + 6$$
At $(3, 0)$: $(a - c)(3) + 6 = 0 \implies 3(a - c) = -6 \implies a - c = -2$.
3. Solve for $a$ and $c$
$$a + c = 6 \quad \text{and} \quad a - c = -2$$
Adding gives $2a = 4 \implies a = 2$. Then $c = 6 - 2 = 4$.
🎯 Final Answer: $\mathbf{p(x) = 2x + 5}$ and $\mathbf{q(x) = 4x - 1}$.
1. Constants $b$ और $d$
• $p(0) = b = 5$।
• $p(x) + q(x) = (a+c)x + (b+d) = 6x + 4 \implies a + c = 6$ और $b + d = 4 \implies 5 + d = 4 \implies d = -1$।
2. Difference Polynomial
$$p(x) - q(x) = (a - c)x + 6$$
$(3, 0)$ point put करने पर: $3(a - c) + 6 = 0 \implies a - c = -2$।
3. $a$ और $c$ calculate करना
$$a + c = 6 \quad \text{तथा} \quad a - c = -2$$
समीकरण जोड़ने पर $2a = 4 \implies a = 2$ और $c = 4$ प्राप्त होता है।
🎯 Final Answer: $\mathbf{p(x) = 2x + 5}$ और $\mathbf{q(x) = 4x - 1}$ है।
End Ex · Q12 A growing pattern of hexagons made with matchsticks adds 1 shared hexagon per stage: (i) Matchsticks for stages 4 and 5; (ii) Complete table; (iii) Rule for $n$-th stage; (iv) Matchsticks for 15th stage; (v) Can 200 matchsticks form a stage? [NCERT p.38-39]
End Ex · Q12 Matchsticks से बने growing hexagons pattern में: (i) Stage 4 और 5 में matchsticks count; (ii) Table पूरा करें; (iii) $n$-th stage का rule; (iv) 15th stage में matchsticks; (v) क्या 200 matchsticks एक stage बना सकती हैं? [NCERT p.38-39]
1. Sequence & Rule
• Stage 1: 6 matchsticks.
• Each subsequent hexagon adds 5 new matchsticks (shares 1 side).
• Rule for Stage $n$: $M(n) = 6 + 5(n - 1) = \mathbf{5n + 1}$.
2. Stages 4, 5, and 15
• Stage 4: $5(4) + 1 = 21$.
• Stage 5: $5(5) + 1 = 26$.
• Stage 15: $5(15) + 1 = 76\text{ matchsticks}$.
3. Check 200 Matchsticks
$$5n + 1 = 200 \implies 5n = 199 \implies n = \frac{199}{5} = 39.8$$
Since $n$ must be a positive integer, 200 matchsticks cannot form a stage.
🎯 Final Answer: Rule $= \mathbf{5n + 1}$; Stage 15 needs 76 matchsticks; 200 matchsticks cannot form a stage ($n = 39.8$).
1. Pattern और Rule
• Stage 1: 6 matchsticks.
• हर नए hexagon के लिए 5 नई तीलियां लगती हैं।
• Stage $n$ का rule: $M(n) = 6 + 5(n - 1) = \mathbf{5n + 1}$।
2. Stages 4, 5 और 15
• Stage 4: $5(4) + 1 = 21$ तीलियां।
• Stage 5: $5(5) + 1 = 26$ तीलियां।
• Stage 15: $5(15) + 1 = 76$ तीलियां।
3. 200 Matchsticks की जांच
$$5n + 1 = 200 \implies 5n = 199 \implies n = 39.8$$
चूंकि $n$ एक whole number होना चाहिए, इसलिए 200 तीलियां कोई stage नहीं बना सकतीं।
🎯 Final Answer: Rule $= \mathbf{5n + 1}$; Stage 15 में 76 तीलियां; 200 तीलियां कोई stage नहीं बना सकतीं।
End Ex · Q13 $p(x)$ passes through $(2, 3)$ and $(6, 11)$. $q(x)$ passes through $(4, -1)$ and is parallel to $p(x)$. Find $p(x), q(x)$, and their $x$-intercepts. [NCERT p.39]
End Ex · Q13 $p(x)$ $(2, 3)$ और $(6, 11)$ से गुजरता है। $q(x)$ $(4, -1)$ से गुजरता है और $p(x)$ के parallel है। $p(x), q(x)$ और उनके $x$-intercepts find करें। [NCERT p.39]
1. Find $p(x)$
$$a = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$$
$$3 = 2(2) + b \implies b = -1 \implies p(x) = 2x - 1$$
2. Find $q(x)$ Parallel $\implies$ same slope $c = 2$.
$$-1 = 2(4) + d \implies -1 = 8 + d \implies d = -9 \implies q(x) = 2x - 9$$
3. $x$-Intercepts
• For $p(x)$: $2x - 1 = 0 \implies x = 0.5 \implies (0.5, 0)$.
• For $q(x)$: $2x - 9 = 0 \implies x = 4.5 \implies (4.5, 0)$.
🎯 Final Answer: $\mathbf{p(x) = 2x - 1}$, $\mathbf{q(x) = 2x - 9}$; $x$-intercepts are $\mathbf{(0.5, 0)}$ and $\mathbf{(4.5, 0)}$.
1. $p(x)$ निकालना
$$a = \frac{11 - 3}{6 - 2} = 2 \implies 3 = 2(2) + b \implies b = -1 \implies p(x) = 2x - 1$$
2. $q(x)$ निकालना Parallel होने के कारण slope $c = 2$ होगा।
$$-1 = 2(4) + d \implies d = -9 \implies q(x) = 2x - 9$$
3. $x$-Intercepts
• $p(x)$ के लिए: $2x - 1 = 0 \implies x = 0.5 \implies (0.5, 0)$।
• $q(x)$ के लिए: $2x - 9 = 0 \implies x = 4.5 \implies (4.5, 0)$।
🎯 Final Answer: $\mathbf{p(x) = 2x - 1}$, $\mathbf{q(x) = 2x - 9}$; $x$-intercepts $\mathbf{(0.5, 0)}$ और $\mathbf{(4.5, 0)}$ हैं।
End Ex · Q14 What geometric property do all linear functions of the form $f(x) = ax + a$ ($a > 0$) have in common? [NCERT p.39]
End Ex · Q14 $f(x) = ax + a$ ($a > 0$) के रूप वाले सभी linear functions में कौन सी property common होती है? [NCERT p.39]
1. Find $x$-Intercept (Zero of Function)
$$f(x) = 0 \implies ax + a = 0 \implies a(x + 1) = 0$$
Since $a > 0$, we have $x + 1 = 0 \implies x = -1$.
2. Common Geometric Invariant For every value of $a > 0$, the line passes through the fixed common point $(-1, 0)$ on the $x$-axis.
🎯 Final Answer: Every such linear function passes through the fixed point $(-1, 0)$ on the $x$-axis.
1. $x$-Intercept (Zero) निकालना
$$ax + a = 0 \implies a(x + 1) = 0 \implies x = -1$$
2. Common Invariant $a > 0$ की हर value के लिए यह line $x$-axis पर स्थित fixed point $(-1, 0)$ से होकर गुजरती है।
🎯 Final Answer: ये सभी linear functions निश्चित point $(-1, 0)$ से होकर गुजरते हैं।