Navigating through the newly revised CBSE Class 9 Mathematics curriculum (Ganita Manjari Part 1) requires a thorough algebraic and graphical understanding of polynomials, terms, coefficients, constant polynomials, degree of polynomials, classification into monomials, binomials, and trinomials, linear polynomials (P(x) = ax + b with a \ne 0), evaluating polynomials at specific values, computing the zero of a linear polynomial (x = -b / a), linear growth and decay patterns, formulating real-world linear algebraic equations (y = ax + b), finding slopes and intercepts, and constructing straight-line graphs on the Cartesian plane. Chapter 2 of Class 9 Mathematics, “Introduction to Linear Polynomials”, bridges arithmetic number patterns with formal linear algebra. It explores why the degree of a non-zero constant polynomial is 0 while the degree of the zero polynomial is undefined; demonstrates how real-world variable costs (like taxi fares, gym memberships, and digital subscriptions) follow linear equations; proves that every linear polynomial has exactly one real zero corresponding to its x-intercept on a graph; and details the step-by-step algebraic methods for solving linear equations. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and step-by-step responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from in-text “Think and Reflect” prompts and Exercise Sets 2.1, 2.2, 2.3, and 2.4 to the complete End-of-Chapter Exercises (Questions 1 to 14 on Pages 32–36)—has been solved with exhaustive step-by-step derivations. All algebraic evaluations, zero-finding calculations, and linear equation systems follow a structured box format with clear ASCII coordinate graphs using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Classification of Polynomials by Degree & Number of Terms
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STRUCTURE OF A POLYNOMIAL EXPRESSION
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P(x) = 5x³ - 4x² + 7x - 9
│ │ │ │
Leading Coefficient ─────┘ │ │ └─── Constant Term (Degree 0)
Quadratic Term (Degree 2) ─────┘ │
Linear Term (Degree 1) ─────────────┘
Highest Power of Variable = 3 ===> Degree of Polynomial = 3 (Cubic)
================================================================================
| Classification Basis | Category Name | Defining Mathematical Condition | Standard General Form | Concrete Algebraic Example | Number of Real Zeroes |
|---|---|---|---|---|---|
| By Degree | Zero Polynomial | Constant term is strictly zero (0). | P(x) = 0 | 0 | Infinite (Every real number is a zero). |
| By Degree | Constant Polynomial | Degree is Zero (0) (a \ne 0). | P(x) = c = c·x⁰ | P(x) = -9 or P(x) = 7 | No Zeroes (Never equals zero). |
| By Degree | Linear Polynomial | Highest exponent power is One (1). | P(x) = ax + b (a \ne 0) | P(x) = 3x - 5 | Exactly 1 Zero (x = -b / a). |
| By Degree | Quadratic Polynomial | Highest exponent power is Two (2). | P(x) = ax² + bx + c (a \ne 0) | P(x) = 2x² - 4x + 6 | At most 2 Zeroes. |
| By Degree | Cubic Polynomial | Highest exponent power is Three (3). | P(x) = ax³ + bx² + cx + d | P(x) = x³ - 3x² + 2x - 1 | At most 3 Zeroes. |
| By Number of Terms | Monomial | Consists of exactly 1 term. | P(x) = a·xⁿ | 5x³, -7x, 12 | Depends on degree. |
| By Number of Terms | Binomial | Consists of exactly 2 terms. | P(x) = a·xⁿ + b·xᵐ | 3x - 8, 4x² + 5 | Depends on degree. |
| By Number of Terms | Trinomial | Consists of exactly 3 terms. | P(x) = a·x² + bx + c | x² - 5x + 6 | Depends on degree. |
2. Graphical Properties of a Linear Polynomial: y = ax + b
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GRAPH OF A LINEAR POLYNOMIAL: y = ax + b
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y-axis
▲
│ / (Straight Line Graph)
│ /
│ /
y-intercept ──►│(0, b)
│ /
│ / Slope m = a = (Change in y) / (Change in x)
│ /
◄───────────────────────────────┼─/─────────────────────────────► x-axis
(-b/a, 0) │(0,0)
▲ │
│ │
Zero of Polynomial ─┘
(x-intercept)
================================================================================
| Key Feature of Graph | Mathematical Representation | Geometric Interpretation on Cartesian Plane |
|---|---|---|
| Geometric Shape | y = ax + b | Always a Straight Line across the coordinate plane. |
| Slope / Rate of Change (m) | Slope m = a | Measures the steepness and direction of the line (a > 0 slopes up; a < 0 slopes down). |
| Y-Intercept (Constant Term) | (0, b) | The exact point where the straight line crosses the vertical y-axis (when x = 0). |
| X-Intercept (Zero of Polynomial) | (-b/a, 0) | The exact point where the straight line crosses the horizontal x-axis (when y = 0). |
NCERT In-Text Questions: “Think and Reflect”
Page No. 18: Think and Reflect (Questions 1 to 3)
Question 1 Why is the exponent of the variable in a polynomial restricted strictly to whole numbers (0, 1, 2, 3, …)? Can expressions like 1/x or square root of x be polynomials? [Exam Favorite]
Answer:
- Rule: An algebraic expression is defined as a polynomial if and only if all exponents of the variable are non-negative integers (Whole Numbers).
- Why Expressions with Negative or Fractional Powers are NOT Polynomials:
- Expression
1 / x = x⁻¹: The exponent of x is-1(a negative integer). Hence, it is not a polynomial. - Expression
√x = x^(1/2): The exponent of x is1/2(a fractional power, not a whole number). Hence, it is not a polynomial. - Expression
x² + 3 / (x - 1): Contains a variable in the denominator, resulting in non-polynomial behavior and undefined points.
- Expression
Question 2 What is the difference between a polynomial and an algebraic equation? [Exam Favorite]
Answer:
- Polynomial (P(x)): An algebraic expression formed by variables, coefficients, and non-negative integer exponents without an equality sign (e.g.,
P(x) = 2x + 5or3x² - 4x + 1). It represents a mathematical value that changes as x varies. - Algebraic Equation (P(x) = 0 or P(x) = Q(x)): A mathematical statement of equality relating two expressions using an equals sign (
=) (e.g.,2x + 5 = 0or3x - 4 = 11). An equation can be solved to find specific values of the variable that satisfy the equality.
Question 3 Why is the degree of a non-zero constant polynomial zero, while the degree of the zero polynomial is undefined? [Exam Favorite]
Answer:
- Non-Zero Constant Polynomial (e.g., P(x) = 7): Any non-zero constant c can be written as
c = c · x⁰(since x^0 = 1). The highest power of the variable is clearly0. Therefore, its degree is strictly 0. - The Zero Polynomial (P(x) = 0): Zero can be written with any power of x without changing its value:
0 = 0·x⁰ = 0·x¹ = 0·x² = 0·x¹⁰⁰ ...Because no unique highest power can be assigned, the degree of the zero polynomial is not defined (undefined).
NCERT Exercise Set 2.1: Terms, Coefficients & Degrees (Page No. 22)
Question 1 State the degree of each of the following polynomials: (i) P(x) = 2x² - 5x + 3 (ii) Q(y) = y³ + 2y - 1 (iii) R(x) = -9 (iv) S(z) = 4z - 3 [Exam Favorite]
Answer:
- (i)
2x² - 5x + 3: The highest power of the variable x is 2. Degree = 2 (Quadratic Polynomial). - (ii)
y³ + 2y - 1: The highest power of the variable y is 3. Degree = 3 (Cubic Polynomial). - (iii)
-9: Can be written as-9·x⁰. The highest power of x is 0. Degree = 0 (Constant Polynomial). - (iv)
4z - 3: The highest power of the variable z is 1 (4z¹). Degree = 1 (Linear Polynomial).
Question 2 Write one example of each: (a) A linear binomial in variable x (b) A quadratic monomial in variable y (c) A cubic trinomial in variable z [Exam Favorite]
Answer:
- (a) Linear Binomial in x (Degree 1, 2 terms):
P(x) = 5x - 7 - (b) Quadratic Monomial in y (Degree 2, 1 term):
Q(y) = -3y² - (c) Cubic Trinomial in z (Degree 3, 3 terms):
R(z) = 2z³ + 4z - 9
Question 3 For the polynomial P(x) = 7x³ - 4x² + 5x - 8: (i) Write the coefficient of x³. (ii) Write the coefficient of x². (iii) Write the coefficient of x. (iv) State the constant term. [Exam Favorite]
Answer:
- (i) Coefficient of x³ = +7
- (ii) Coefficient of x² = -4 (always include the negative sign)
- (iii) Coefficient of x = +5
- (iv) Constant term = -8
Question 4 Which of the following expressions are polynomials in one variable? State reasons for your answers: (a) 4x² - 3x + 7 (b) y + 2 / y (c) 3√t + t√2 (d) x¹⁰ + y³ + t⁵⁰ [Exam Favorite]
Answer:
- (a)
4x² - 3x + 7: YES, it is a polynomial in one variable. All powers of x (2 and 1) are non-negative integers (whole numbers), and only one variable x is present. - (b)
y + 2 / y = y¹ + 2·y⁻¹: NO, it is NOT a polynomial. The second term contains the exponent-1, which is not a whole number. - (c)
3√t + t√2 = 3·t^(1/2) + t¹·√2: NO, it is NOT a polynomial. The first term contains the exponent1/2, which is a fraction and not a whole number. - (d)
x¹⁰ + y³ + t⁵⁰: It is a polynomial, but NOT in one variable. All powers are whole numbers (10, 3, 50), but it involves three distinct variables (x, y, t).
NCERT Exercise Set 2.2: Evaluating Polynomials & Zeroes (Page No. 27)
Question 1 Evaluate the linear polynomial P(x) = 5x - 3 at: (i) x = 0 (ii) x = 2 (iii) x = -1 [Exam Favorite]
Answer:
================================================================================
STEP-BY-STEP EVALUATION OF LINEAR POLYNOMIAL P(x) = 5x - 3:
--------------------------------------------------------------------------------
(i) AT x = 0:
P(0) = 5(0) - 3 = 0 - 3 = -3
(ii) AT x = 2:
P(2) = 5(2) - 3 = 10 - 3 = 7
(iii) AT x = -1:
P(-1) = 5(-1) - 3 = -5 - 3 = -8
FINAL ANSWER SUMMARY:
• P(0) = -3
• P(2) = 7
• P(-1) = -8
================================================================================
Question 2 Find the value of the quadratic polynomial Q(s) = 7s² - 4s + 6 at s = 4. [Exam Favorite]
Answer:
================================================================================ CALCULATION: -------------------------------------------------------------------------------- Given: Q(s) = 7s² - 4s + 6 Substitute s = 4: Q(4) = 7·(4)² - 4·(4) + 6 Q(4) = 7·(16) - 16 + 6 Q(4) = 112 - 16 + 6 = 96 + 6 = 102 FINAL ANSWER: The value of Q(4) is 102. ================================================================================
Question 3 Find the zero of each of the following linear polynomials: (i) P(x) = 2x + 5 (ii) Q(x) = 3x - 4 (iii) R(x) = ax + b (where a ≠ 0) (iv) S(x) = cx (where c ≠ 0) [Exam Favorite]
Answer:
================================================================================
COMPUTING ZEROES OF LINEAR POLYNOMIALS (Set P(x) = 0):
--------------------------------------------------------------------------------
(i) P(x) = 2x + 5:
2x + 5 = 0
2x = -5 ===> x = -5 / 2 = -2.5
(ii) Q(x) = 3x - 4:
3x - 4 = 0
3x = 4 ===> x = 4 / 3 ≈ 1.33
(iii) R(x) = ax + b (a ≠ 0):
ax + b = 0
ax = -b ===> x = -b / a
(iv) S(x) = cx (c ≠ 0):
cx = 0 ===> x = 0 / c = 0
FINAL ANSWER SUMMARY:
• Zero of P(x) = -5/2
• Zero of Q(x) = 4/3
• Zero of R(x) = -b/a
• Zero of S(x) = 0
================================================================================
NCERT Exercise Set 2.3 & 2.4: Linear Patterns & Growth (Pages 30–31)
Question 1 Ananya starts a monthly savings plan with an initial deposit of Rs 500. She decides to add Rs 150 every subsequent month. (i) Calculate the total amount saved month by month for the first 4 months. (ii) Formulate a linear expression representing the total amount saved in the nth month. (iii) Find the total amount saved at the end of 12 months (1 year). [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP LINEAR GROWTH MODELLING: -------------------------------------------------------------------------------- (i) MONTH-BY-MONTH SAVINGS TABLE: • Month 1 (n = 1): Initial amount = Rs 500 • Month 2 (n = 2): 500 + 150 = Rs 650 • Month 3 (n = 3): 500 + 150 + 150 = Rs 800 • Month 4 (n = 4): 500 + 150 + 150 + 150 = Rs 950 (ii) LINEAR EXPRESSION FOR NTH MONTH: • Pattern: Amount A(n) = Initial Deposit + (n - 1) × Monthly Addition A(n) = 500 + 150·(n - 1) A(n) = 500 + 150n - 150 A(n) = 150n + 350 (where n is the month number) (iii) SAVINGS AFTER 12 MONTHS (n = 12): A(12) = 150·(12) + 350 A(12) = 1800 + 350 = Rs 2150 FINAL ANSWER: (i) Savings for first 4 months: Rs 500, Rs 650, Rs 800, Rs 950 (ii) Linear Expression: A(n) = 150n + 350 (iii) Amount after 12 months = Rs 2,150 ================================================================================
NCERT End-of-Chapter Exercises (Pages 32–36, Questions 1 to 14)
Question 1 An online digital learning platform charges a fixed monthly fee and an additional fee per digital learning module accessed. A student observed that when she accessed 10 modules, her monthly bill was Rs 400. When she accessed 14 modules, her monthly bill was Rs 500. If the monthly bill y depends on the number of modules accessed x according to the linear relation y = ax + b, find the values of a and b. Write the complete linear equation. [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP SOLUTION (SIMULTANEOUS LINEAR EQUATIONS): -------------------------------------------------------------------------------- GIVEN LINEAR MODEL: y = ax + b • For x = 10 modules, Bill y = Rs 400 ===> 10a + b = 400 ...(Equation 1) • For x = 14 modules, Bill y = Rs 500 ===> 14a + b = 500 ...(Equation 2) STEP 1: Eliminate b by subtracting Equation 1 from Equation 2 (14a + b) - (10a + b) = 500 - 400 14a - 10a = 100 4a = 100 a = 100 / 4 = 25 STEP 2: Substitute a = 25 into Equation 1 to find b 10·(25) + b = 400 250 + b = 400 b = 400 - 250 = 150 STEP 3: Physical Interpretation of Parameters • a = Rs 25 ===> Variable cost per module (Slope). • b = Rs 150 ===> Fixed monthly subscription fee (Y-intercept). FINAL ANSWER: • Value of a = 25 • Value of b = 150 • Complete Linear Equation: y = 25x + 150 ================================================================================
Question 2 A sports gym charges a fixed monthly maintenance fee and an additional cost per hour for using the badminton court. A student observed that when she used the court for 10 hours, her bill was Rs 800. When she used it for 15 hours, her bill was Rs 1100. If the monthly bill y depends on court usage hours x according to y = ax + b, find the values of a and b. How much will a student pay if she uses the court for 20 hours in a month? [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP SOLUTION (GYM BILL LINEAR MODEL): -------------------------------------------------------------------------------- GIVEN LINEAR MODEL: y = ax + b • For x = 10 hours, Bill y = Rs 800 ===> 10a + b = 800 ...(Equation 1) • For x = 15 hours, Bill y = Rs 1100 ===> 15a + b = 1100 ...(Equation 2) STEP 1: Subtract Equation 1 from Equation 2 (15a + b) - (10a + b) = 1100 - 800 5a = 300 a = 300 / 5 = 60 (Hourly court fee = Rs 60/hr) STEP 2: Substitute a = 60 into Equation 1 10·(60) + b = 800 600 + b = 800 b = 800 - 600 = 200 (Fixed monthly fee = Rs 200) STEP 3: Calculate Bill for 20 Hours (x = 20) Linear Equation: y = 60x + 200 y = 60·(20) + 200 y = 1200 + 200 = Rs 1400 FINAL ANSWER: • a = 60, b = 200 • Linear Equation: y = 60x + 200 • Bill for 20 hours = Rs 1,400 ================================================================================
Question 3 Express the linear equation 3x + 2y = 12 in the standard slope-intercept form y = ax + b. Identify the slope and y-intercept of the line. [Exam Favorite]
Answer:
================================================================================ ALGEBRAIC REARRANGEMENT: -------------------------------------------------------------------------------- Given Equation: 3x + 2y = 12 STEP 1: Isolate the term containing y 2y = -3x + 12 STEP 2: Divide both sides by 2 y = (-3 / 2)x + (12 / 2) y = (-1.5)x + 6 COMPARING WITH y = ax + b: • Slope (a) = -3/2 (or -1.5) • Y-intercept (b) = 6 (Crosses y-axis at (0, 6)) FINAL ANSWER: • Standard form : y = (-3/2)x + 6 • Slope = -1.5 (or -3/2) • Y-intercept = 6 (Point (0, 6)) ================================================================================
Question 4 Draw a table of solutions and find the coordinates of intersection with both coordinate axes for the linear polynomial y = -3x + 4. [Exam Favorite]
Answer:
================================================================================ FINDING AXES INTERSECTIONS & SOLUTION TABLE: -------------------------------------------------------------------------------- GIVEN EQUATION: y = -3x + 4 STEP 1: Y-Intercept (Set x = 0) y = -3(0) + 4 = 4 Intersection with Y-axis = (0, 4) STEP 2: X-Intercept (Set y = 0) 0 = -3x + 4 3x = 4 ===> x = 4 / 3 ≈ 1.33 Intersection with X-axis = (4/3, 0) or (1.33, 0) SOLUTION TABLE FOR PLOTTING: ┌───────────┬────────┬────────┬────────┬────────┐ │ x │ -1 │ 0 │ 1 │ 4/3 │ ├───────────┼────────┼────────┼────────┼────────┤ │ y = -3x+4 │ 7 │ 4 │ 1 │ 0 │ └───────────┴────────┴────────┴────────┴────────┘ FINAL ANSWER: • Point of intersection with Y-axis = (0, 4) • Point of intersection with X-axis = (4/3, 0) ================================================================================
Question 5 The work done by a body on application of a constant force of 3 units is directly proportional to the distance travelled by the body. (i) Express this situation as a linear equation in two variables (let Work done = y and Distance = x). (ii) Find the work done when the distance travelled by the body is: (a) 2 units, (b) 0 units. [Exam Favorite]
Answer:
================================================================================
WORK-DISTANCE LINEAR MODEL:
--------------------------------------------------------------------------------
(i) LINEAR EQUATION:
Formula: Work Done = Force × Distance
Given: Constant Force = 3 units
Let Work Done = y, Distance Travelled = x
Linear Equation: y = 3x
(ii) COMPUTING WORK DONE:
(a) When Distance x = 2 units:
y = 3 × 2 = 6 units
(b) When Distance x = 0 units:
y = 3 × 0 = 0 units
FINAL ANSWER:
(i) Linear Equation: y = 3x
(ii) Work Done: (a) 6 units for 2 units distance; (b) 0 units for 0 units distance.
================================================================================
Question 6 The taxi fare in a city is structured as follows: For the first kilometre, the fare is Rs 8, and for the subsequent distance, it is Rs 5 per kilometre. Taking the total distance covered as x km and the total fare as Rs y, write a linear equation for this information and draw its graph. [Exam Favorite]
Answer:
================================================================================ TAXI FARE LINEAR EQUATION DERIVATION: -------------------------------------------------------------------------------- GIVEN DATA: • Total Distance = x km • Total Fare = Rs y • Fare for 1st km = Rs 8 • Remaining Distance = (x - 1) km • Fare for remaining distance = Rs 5 per km STEP 1: Formulate Total Fare Equation Total Fare = (Fare for 1st km) + (Fare for remaining distance) y = 8 + 5·(x - 1) y = 8 + 5x - 5 y = 5x + 3 STEP 2: Coordinates Table for Graph • If x = 1 km ===> y = 5(1) + 3 = Rs 8 ===> Point (1, 8) • If x = 2 km ===> y = 5(2) + 3 = Rs 13 ===> Point (2, 13) • If x = 3 km ===> y = 5(3) + 3 = Rs 18 ===> Point (3, 18) • If x = 0 km ===> y = 3 (Fixed Base Fare intercept) ===> Point (0, 3) FINAL ANSWER: Linear Equation: y = 5x + 3 (or 5x - y + 3 = 0). ================================================================================
Question 7 A water storage tank holds 10,000 litres of water. Water is drained out through an outlet pipe at a constant rate of 800 litres per hour. (i) Write a linear decay expression for the volume of water remaining V(t) after t hours. (ii) Find the volume of water left after 5 hours. (iii) How long will it take for the tank to be completely emptied? [Exam Favorite]
Answer:
================================================================================ LINEAR DECAY WATER TANK MODEL: -------------------------------------------------------------------------------- (i) LINEAR DECAY EXPRESSION: • Initial Volume at t = 0 : V₀ = 10,000 litres • Rate of Drainage : 800 litres/hour • Volume remaining V(t) = 10,000 - 800t (ii) VOLUME AFTER t = 5 HOURS: V(5) = 10,000 - 800·(5) V(5) = 10,000 - 4000 = 6,000 litres (iii) TIME TO EMPTY COMPLETELY (Set V(t) = 0): 10,000 - 800t = 0 800t = 10,000 t = 10,000 / 800 = 100 / 8 = 12.5 hours (12 hours 30 minutes) FINAL ANSWER: (i) Linear Expression : V(t) = 10,000 - 800t (ii) Volume after 5 h : 6,000 litres (iii) Time to empty : 12.5 hours ================================================================================
Question 8 Find the value of k if x = 2, y = 1 is a solution of the linear equation 2x + 3y = k. [Exam Favorite]
Answer:
================================================================================ SUBSTITUTION METHOD: -------------------------------------------------------------------------------- Given Equation: 2x + 3y = k Substitute x = 2 and y = 1: 2·(2) + 3·(1) = k 4 + 3 = k k = 7 FINAL ANSWER: The value of k is 7. ================================================================================
Question 9 If P(x) = x³ - 3x² + 4x - 12, check whether x = 3 is a zero of the polynomial. [Exam Favorite]
Answer:
================================================================================ ZERO VERIFICATION: -------------------------------------------------------------------------------- Given: P(x) = x³ - 3x² + 4x - 12 Substitute x = 3: P(3) = (3)³ - 3·(3)² + 4·(3) - 12 P(3) = 27 - 3·(9) + 12 - 12 P(3) = 27 - 27 + 0 = 0 CONCLUSION: Since P(3) = 0, x = 3 is strictly a ZERO of the polynomial P(x). ================================================================================
Question 10 Write two solutions for each of the following linear equations: (a) 4x + 3y = 12 (b) 2x + y = 7 [Exam Favorite]
Answer:
- (a)
4x + 3y = 12:- If x = 0:
4(0) + 3y = 12 ===> 3y = 12 ===> y = 4. Solution 1 = (0, 4). - If y = 0:
4x + 3(0) = 12 ===> 4x = 12 ===> x = 3. Solution 2 = (3, 0).
- If x = 0:
- (b)
2x + y = 7:- If x = 0:
2(0) + y = 7 ===> y = 7. Solution 1 = (0, 7). - If x = 2:
2(2) + y = 7 ===> 4 + y = 7 ===> y = 3. Solution 2 = (2, 3).
- If x = 0:
Question 11 How many linear equations in x and y can be satisfied by x = 1 and y = 2? Give two examples. [Exam Favorite]
Answer:
- Number of Equations: Infinitely Many linear equations can pass through a single point (1, 2) on the Cartesian plane.
- Two Examples:
x + y = 3(since 1 + 2 = 3)2x + y = 4(since 2(1) + 2 = 4)y - 2x = 0(since 2 – 2(1) = 0)
Question 12 The temperature of a city in degrees Fahrenheit (F) is related to degrees Celsius (C) by the linear equation F = (9/5)C + 32. (i) If the temperature is 30°C, what is the temperature in Fahrenheit? (ii) If the temperature is 95°F, what is the temperature in Celsius? (iii) Is there a numerical temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it. [Exam Favorite]
Answer:
================================================================================ FAHRENHEIT-CELSIUS LINEAR CONVERSION: -------------------------------------------------------------------------------- (i) CONVERTING 30°C TO FAHRENHEIT: F = (9 / 5)·(30) + 32 F = 9·(6) + 32 = 54 + 32 = 86°F (ii) CONVERTING 95°F TO CELSIUS: 95 = (9 / 5)C + 32 95 - 32 = (9 / 5)C 63 = (9 / 5)C C = 63 × (5 / 9) = 7 × 5 = 35°C (iii) TEMPERATURE NUMERICALLY IDENTICAL IN BOTH SCALES (Set F = C = x): x = (9 / 5)x + 32 x - (9 / 5)x = 32 (5x - 9x) / 5 = 32 -4x / 5 = 32 -4x = 32 × 5 = 160 x = 160 / (-4) = -40 FINAL ANSWER: (i) 30°C = 86°F (ii) 95°F = 35°C (iii) At -40° (-40°C = -40°F), both scales show the exact same numerical value. ================================================================================
Question 13 Find the coordinates where the straight line 2x - 5y = 10 intersects the x-axis and y-axis. [Exam Favorite]
Answer:
================================================================================ INTERSECTION OF 2x - 5y = 10 WITH AXES: -------------------------------------------------------------------------------- • Intersection with X-axis (Set y = 0): 2x - 5(0) = 10 2x = 10 ===> x = 5 Point = (5, 0) • Intersection with Y-axis (Set x = 0): 2(0) - 5y = 10 -5y = 10 ===> y = -2 Point = (0, -2) FINAL ANSWER: • X-axis intersection: (5, 0) • Y-axis intersection: (0, -2) ================================================================================
Question 14 Explain why a linear equation in two variables ax + by + c = 0 (where a, b are not both zero) has infinitely many solutions. [Exam Favorite]
Answer:
- Explanation:
- Geometrically, the equation
ax + by + c = 0represents a continuous straight line extending infinitely in both directions on the 2D Cartesian plane. - Every single point (x, y) that lies on this straight line is a valid mathematical solution to the equation.
- For every chosen value of x, there exists a corresponding unique value of y = (-ax – c) / b.
- Since the line contains an infinite number of points, a linear equation in two variables has infinitely many solutions.
- Geometrically, the equation
Frequently Asked Questions (FAQs) – Class 9 Maths Chapter 2
Question 1: What is a Linear Polynomial? [Exam Favorite] Answer: A linear polynomial is a polynomial of degree 1 having the standard form P(x) = ax + b, where a and b are real numbers and a \ne 0.
Question 2: How many zeroes can a linear polynomial have? [Exam Favorite] Answer: A linear polynomial has exactly one real zero given by x = -b / a.
Question 3: What is the geometric meaning of the zero of a linear polynomial? [Exam Favorite] Answer: The zero of a linear polynomial is the x-coordinate of the point where its straight-line graph intersects the x-axis (the point where y = 0).
Question 4: What is the degree of the zero polynomial? [Exam Favorite] Answer: The degree of the zero polynomial P(x) = 0 is undefined (not defined).
Question 5: What is the degree of a non-zero constant polynomial? [Exam Favorite] Answer: The degree of any non-zero constant polynomial (like P(x) = 8 or -15) is strictly 0.
Question 6: Is the expression x² + 2√x + 5 a polynomial? Why? [Exam Favorite] Answer: No. The middle term contains √x = x^(1/2), where the exponent is a fraction (1/2) and not a whole number.
Question 7: Find the zero of the polynomial P(x) = 4x - 7. [Exam Favorite] Answer: 4x - 7 = 0 ===> 4x = 7 ===> x = 7 / 4 (or 1.75).
Question 8: Find the value of P(x) = x² - 2x + 1 at x = 1. [Exam Favorite] Answer: P(1) = (1)² - 2(1) + 1 = 1 - 2 + 1 = 0.
Question 9: What is the coefficient of x² in P(x) = π/2 · x² + x - 5? [Exam Favorite] Answer: The coefficient of x² is π / 2.
Question 10: Express the linear relation y = -4x + 9 in standard form ax + by + c = 0. [Exam Favorite] Answer: 4x + y - 9 = 0 (where a = 4, b = 1, c = -9).
Question 11: Write a linear polynomial whose only zero is 5. [Exam Favorite] Answer: P(x) = x - 5 (or 2x - 10).
Question 12: Does a constant non-zero polynomial have any zeroes? [Exam Favorite] Answer: No. A constant non-zero polynomial (like P(x) = 5) never equals 0 for any value of x; hence, it has no zeroes.
Question 13: What does the graph of x = 4 represent on a Cartesian plane? [Exam Favorite] Answer: A vertical straight line parallel to the y-axis at a constant perpendicular distance of 4 units to the right of the origin.
Question 14: What does the graph of y = -3 represent on a Cartesian plane? [Exam Favorite] Answer: A horizontal straight line parallel to the x-axis at a constant distance of 3 units below the origin.
Question 15: If P(x) = 3x + 1, verify if x = -1/3 is a zero of P(x). [Exam Favorite] Answer: P(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0. Yes, x = -1/3 is a zero of P(x).
Mastering the NCERT Solutions for Class 9 Mathematics Chapter 2 (Ganita Manjari Part 1), “Introduction to Linear Polynomials”, equips students with the polynomial definitions, degree rules, zero evaluation methods, linear modeling equations, and graphing techniques required for top performance in CBSE examinations. Review the step-by-step algebraic box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.