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📐 Chapter 2: Introduction to Linear Polynomials • End-of-Chapter Exercises

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Exercise Set 2.1 5Exercise Set 2.2 7Exercise Set 2.3 5Exercise Set 2.4 4Exercise Set 2.5 3Exercise Set 2.6 1End-of-Chapter Exercises 14All Exercises 39
Q1

End-of-Chapter Exercises

1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is -7.

A possible example is: 5x³ - 7x² + 2x + 1. (Any polynomial where the highest power is x³ and the term containing x² is -7x² is correct).

Q2

End-of-Chapter Exercises

2. Find the values of the following polynomials at the indicated values of the variables.

(i) 5x² - 3x + 7 if x = 1

(ii) 4t³ - t² + 6 if t = a

  • (i) If x = 1: 5(1)² - 3(1) + 7 = 5 - 3 + 7 = 9
  • (ii) If t = a: Simply replace t with a to get 4a³ - a² + 6
Q3

End-of-Chapter Exercises

3. If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.

  1. Let the number be x.
  2. Set up the linear equation: (5/2)x + 2/3 = -7/12
  3. Subtract 2/3 from both sides:
    (5/2)x = -7/12 - 2/3
    (5/2)x = -7/12 - 8/12
    (5/2)x = -15/12
    (5/2)x = -5/4
  4. Multiply both sides by 2/5 to isolate x:
    x = (-5/4) × (2/5)
    x = -1/2
Q4

End-of-Chapter Exercises

4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

  1. Let the smaller number be x, so the larger number is 5x.
  2. After adding 21 to both, the new numbers are x + 21 and 5x + 21.
  3. Since 5x was already 5 times larger than x, 5x + 21 will be the larger of the two new numbers. So, it must be twice the smaller new number:
    5x + 21 = 2(x + 21)
  4. Solve for x:
    5x + 21 = 2x + 42
    3x = 21
    x = 7
  5. The numbers are:
    Smaller number: 7
    Larger number: 5 × 7 = 35
Q5

End-of-Chapter Exercises

5. If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

  • (i) After 6 months: 800 + (6 × 250) = 800 + 1500 = ₹2300
  • (ii) After 2 years (24 months): 800 + (24 × 250) = 800 + 6000 = ₹6800
  • Linear Pattern: The total amount A after m months is expressed as A(m) = 800 + 250m.
Q6

End-of-Chapter Exercises

6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

  1. Let the tens digit be t and the units digit be u. The original number is 10t + u.
  2. The interchanged number is 10u + t.
  3. Their sum is 143:
    (10t + u) + (10u + t) = 143
    11t + 11u = 143
    t + u = 13
  4. We are also given that the digits differ by 3, so either t - u = 3 or u - t = 3.
  5. Case 1: t - u = 3
    Adding the equations: (t + u) + (t - u) = 13 + 3 ⇒ 2t = 16 ⇒ t = 8
    If t = 8, then u = 5. The number is 85.
  6. Case 2: u - t = 3
    Adding the equations: (u + t) + (u - t) = 13 + 3 ⇒ 2u = 16 ⇒ u = 8
    If u = 8, then t = 5. The number is 58.
  7. The two possible numbers are 58 and 85.
Q7

End-of-Chapter Exercises

7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.

(i) y = -3x + 4      (ii) 2y = 4x + 7      (iii) 5y = 6x - 10      (iv) 3y = 6x - 11

Are any of the lines parallel?

Convert each to the form y = ax + b:

  • (i) y = -3x + 4:
    Slope = -3, y-intercept = 4, Coordinates = (0, 4).
  • (ii) 2y = 4x + 7 ⇒ y = 2x + 3.5:
    Slope = 2, y-intercept = 3.5, Coordinates = (0, 3.5).
  • (iii) 5y = 6x - 10 ⇒ y = (6/5)x - 2:
    Slope = 6/5 (or 1.2), y-intercept = -2, Coordinates = (0, -2).
  • (iv) 3y = 6x - 11 ⇒ y = 2x - 11/3:
    Slope = 2, y-intercept = -11/3, Coordinates = (0, -11/3).

Parallel lines: Lines (ii) and (iv) both have a slope of 2, so they are parallel to each other.

Q8

End-of-Chapter Exercises

8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9/5(x - 273) + 32.

(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.

(ii) If the temperature is 158 °F. then find the temperature in Kelvin.

  • (i) From Kelvin to Fahrenheit:
    Substitute x = 313 into the equation:
    y = 9/5(313 - 273) + 32
    y = 9/5(40) + 32
    y = 9(8) + 32 = 72 + 32 = 104 °F
  • (ii) From Fahrenheit to Kelvin:
    Substitute y = 158 into the equation:
    158 = 9/5(x - 273) + 32
    126 = 9/5(x - 273)
    126 × (5/9) = x - 273
    70 = x - 273
    x = 70 + 273 = 343 K
Q9

End-of-Chapter Exercises

9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.

  1. The relationship is Work = Force × distance.
  2. Given the constant force is 3 units, the linear equation is: w = 3d.
  3. When distance d = 2 units, the work done is: w = 3(2) = 6 units.
  4. (To verify on a graph: Plot the line passing through (0,0), (1,3), and (2,6). Reading the y-value at x=2 confirms it is 6).
Q10

End-of-Chapter Exercises

10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).

(i) Find the polynomial p(x).

(ii) Find the coordinates where the graph of p(x) cuts the axes.

(iii) Draw the graph of p(x) and verify your answers.

  1. (i) Find the polynomial p(x):
    Let p(x) = ax + b. We have:
    5 = a(1) + b ⇒ a + b = 5
    11 = a(3) + b ⇒ 3a + b = 11
    Subtracting the first from the second gives 2a = 6 ⇒ a = 3.
    Substitute a = 3 into the first: 3 + b = 5 ⇒ b = 2.
    The polynomial is p(x) = 3x + 2.
  2. (ii) Coordinates where it cuts the axes:
    y-axis intercept (set x = 0): p(0) = 3(0) + 2 = 2. Coordinates: (0, 2).
    x-axis intercept (set p(x) = 0): 0 = 3x + 2 ⇒ x = -2/3. Coordinates: (-2/3, 0).
Q11

End-of-Chapter Exercises

11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:

(i) p(0) = 5.

(ii) The polynomial p(x) - q(x) cuts the x-axis at (3, 0).

(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.

Find the polynomials p(x) and q(x).

  1. From (i), p(0) = 5 ⇒ a(0) + b = 5 ⇒ b = 5.
  2. From (iii), p(x) + q(x) = (a + c)x + (b + d) = 6x + 4.
    This means a + c = 6 and b + d = 4.
    Since b = 5, then 5 + d = 4 ⇒ d = -1.
  3. From (ii), p(3) - q(3) = 0 ⇒ p(3) = q(3).
    3a + b = 3c + d
    Substitute b = 5 and d = -1:
    3a + 5 = 3c - 1 ⇒ 3a - 3c = -6 ⇒ a - c = -2.
  4. We now have a system for a and c:
    a + c = 6
    a - c = -2
    Adding them: 2a = 4 ⇒ a = 2.
    Substituting back: 2 + c = 6 ⇒ c = 4.
  5. The polynomials are: p(x) = 2x + 5 and q(x) = 4x - 1.
Q12

End-of-Chapter Exercises

12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

(ii) Complete the following table.

(iii) Find a rule to determine the number of matchsticks required for the nth stage.

(iv) How many matchsticks will be required for the 15th stage of the pattern?

(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

  • (i) & (ii) Table Completion:
    Stage 1: 6 matchsticks.
    Stage 2: 6 + 5 = 11 matchsticks (shares 1 side).
    Stage 3: 11 + 5 = 16 matchsticks.
    Stage12345n
    Number of matchsticks6111621265n + 1
  • (iii) Rule: The pattern increases by 5 each time. The rule is 5n + 1.
  • (iv) 15th Stage: 5(15) + 1 = 75 + 1 = 76 matchsticks.
  • (v) Can 200 matchsticks form a stage?
    Set 5n + 1 = 200 ⇒ 5n = 199 ⇒ n = 199/5 = 39.8.
    No, because n must be a whole number integer (a specific stage), and 199 is not perfectly divisible by 5.
Stage 1 (6) Stage 2 (6+5 = 11) Stage 3 (11+5 = 16)
Growing Matchstick Hexagon Pattern: Each new hexagon adds 5 matchsticks (Formula: 5n + 1)
Q13

End-of-Chapter Exercises

13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:

(i) The graph of p(x) passes through the points (2, 3) and (6, 11).

(ii) The graph of q(x) passes through the point (4, -1).

(iii) The graph of q(x) is parallel to the graph of p(x).

Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

  1. Find p(x):
    3 = 2a + b
    11 = 6a + b
    Subtracting gives 4a = 8 ⇒ a = 2. Substituting back gives b = -1.
    So, p(x) = 2x - 1.
  2. Find q(x):
    Because q(x) is parallel to p(x), its slope c must equal a. So, c = 2.
    q(x) = 2x + d. It passes through (4, -1), so:
    -1 = 2(4) + d ⇒ -1 = 8 + d ⇒ d = -9.
    So, q(x) = 2x - 9.
  3. X-axis intercepts (where y = 0):
    For p(x): 2x - 1 = 0 ⇒ x = 1/2. Coordinates: (1/2, 0).
    For q(x): 2x - 9 = 0 ⇒ x = 9/2. Coordinates: (9/2, 0).
Q14

End-of-Chapter Exercises

14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Given f(x) = ax + a, where a > 0. We can factor this as f(x) = a(x + 1).

  1. Positive Slope: The slope is a. Since a > 0, all lines have a positive slope and rise from left to right (linear growth).
  2. Equal Slope and Y-intercept: When x = 0, f(0) = a. The y-intercept is (0, a), which is above the origin since a > 0.
  3. Common X-intercept: To find where the line cuts the x-axis, set f(x) = 0:
    a(x + 1) = 0
    Since a ≠ 0, we have x + 1 = 0 ⇒ x = -1.
    Therefore, all such lines pass through the identical fixed point (-1, 0) on the x-axis.
Family of Lines: f(x) = ax + a (a > 0) x y a = 0.5 a = 1 a = 2 Fixed Point (−1, 0)
Pencil of Lines: Every member of f(x) = a(x + 1) pivots through (−1, 0) with slope a