Q1. Page 36: Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is -7.

Answer:

A polynomial of degree 3 has the form ax³ + bx² + cx + d, where a ≠ 0.

We need the coefficient of x² to be -7, so b = -7.

One example is: x³ - 7x² + 2x + 1.

Q2. Page 36: Find the values of the following polynomials at the indicated values of the variables. 5x² - 3x + 7 if x = 1 4t³ - t² + 6 if t = a

Answer:

For 5x² - 3x + 7 at x = 1:

5(1)² - 3(1) + 7 = 5 - 3 + 7 = 9

For 4t³ - t² + 6 at t = a:

4(a)³ - (a)² + 6 = 4a³ - a² + 6

Q3. Page 37: If we multiply a number by 5/2 and add 2/3 to the product, we get -7/12. Find the number.

Answer:

Let the number be x.

According to the problem:

(5/2)x + 2/3 = -7/12

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

Multiply both sides by 2/5:

x = (-5/4) × (2/5) = -10/20 = -1/2

The number is -1/2.

Q4. Page 37: 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?

Answer:

Let the smaller number be x.

Then the larger number is 5x.

After adding 21, the new numbers are x + 21 and 5x + 21.

According to the problem, one of the new numbers becomes twice the other.

Since 5x is already larger, we have:

5x + 21 = 2(x + 21)

5x + 21 = 2x + 42

3x = 21

x = 7

The numbers are 7 and 35.

Q5. Page 37: 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.

Answer:

Linear Pattern: Let n be the number of months. The total amount A is:

A(n) = 800 + 250n

(i) After 6 months:

A(6) = 800 + 250(6) = 800 + 1500 = ₹2300

(ii) After 2 years (24 months):

A(24) = 800 + 250(24) = 800 + 6000 = ₹6800

Q6. Page 37: 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.

Answer:

Let the two-digit number be 10x + y, where x is the tens digit and y is the units digit.

The digits differ by 3, so |x - y| = 3.

When the digits are interchanged, the new number is 10y + x.

Sum of the original and new number is 143:

(10x + y) + (10y + x) = 143

11x + 11y = 143

11(x + y) = 143

x + y = 13

Now we have two equations:

x + y = 13

x - y = 3 (assuming x > y) or y - x = 3

Case 1: x - y = 3

Adding the two equations: 2x = 16 => x = 8, y = 5. Number = 85.

Case 2: y - x = 3

Adding x + y = 13 and y - x = 3: 2y = 16 => y = 8, x = 5. Number = 58.

The two numbers are 58 and 85.

Q7. Page 37: 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?

Answer:

(i) y = -3x + 4

• Slope (a) = -3
• y-intercept (b) = 4
• Point where it cuts the y-axis: (0, 4)

(ii) 2y = 4x + 7

Dividing by 2: y = 2x + 7/2

• Slope (a) = 2
• y-intercept (b) = 7/2
• Point where it cuts the y-axis: (0, 7/2)

(iii) 5y = 6x - 10

Dividing by 5: y = (6/5)x - 2

• Slope (a) = 6/5
• y-intercept (b) = -2
• Point where it cuts the y-axis: (0, -2)

(iv) 3y = 6x - 11

Dividing by 3: y = 2x - 11/3

• Slope (a) = 2
• y-intercept (b) = -11/3
• Point where it cuts the y-axis: (0, -11/3)

Parallel Lines: Lines (ii) and (iv) have the same slope (a = 2). Therefore, they are parallel.

Q8. Page 37: 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.

Answer:

(i) To find °F when K = 313:

y = (9/5)(313 - 273) + 32

y = (9/5)(40) + 32

y = (9 × 8) + 32

y = 72 + 32 = 104 °F

(ii) To find K when °F = 158:

158 = (9/5)(x - 273) + 32

158 - 32 = (9/5)(x - 273)

126 = (9/5)(x - 273)

126 × (5/9) = x - 273

14 × 5 = x - 273

70 = x - 273

x = 343 K

Q9. Page 37: 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.

Answer:

Linear Equation: Work (w) = Force × Distance (d). If the constant force is 3 units, then:

w = 3d

Graph: To draw the graph, plot two points. For d=0, w=0 (0,0). For d=1, w=3 (1,3). Connect these points to form a straight line passing through the origin.

Work done for distance = 2 units:

w = 3(2) = 6 units

On the graph, the point (2, 6) lies on the line w=3d, verifying the answer.

Q10. Page 38: 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.

Answer:

(i) Find the polynomial p(x):

Let p(x) = ax + b.

Using the points (1, 5) and (3, 11):

a(1) + b = 5 => a + b = 5 ...(1)

a(3) + b = 11 => 3a + b = 11 ...(2)

Subtracting (1) from (2): 2a = 6 => a = 3

Substitute a = 3 in (1): 3 + b = 5 => b = 2

Therefore, p(x) = 3x + 2.

(ii) Coordinates where the graph cuts the axes:

• y-axis (x=0): p(0) = 3(0) + 2 = 2. Point: (0, 2)
• x-axis (y=0): 3x + 2 = 0 => x = -2/3. Point: (-2/3, 0)

(iii) Graph verification: Plot points (1,5), (3,11), (0,2), and (-2/3, 0). They all lie on the line y=3x+2.

Q11. Page 38: 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).

Answer:

From (i): p(0) = a(0) + b = b = 5. So, b = 5.

From (iii): p(x) + q(x) = (a + c)x + (b + d) = 6x + 4.

Therefore, a + c = 6 and b + d = 4.

Since b = 5, 5 + d = 4 => d = -1.

From (ii): p(x) - q(x) cuts the x-axis at (3,0), so p(3) - q(3) = 0.

p(3) - q(3) = (3a + b) - (3c + d) = 0

3a - 3c + b - d = 0

3(a - c) + 5 - (-1) = 0

3(a - c) + 6 = 0

3(a - c) = -6

a - c = -2

Now we have two equations:

a + c = 6 ...(1)

a - c = -2 ...(2)

Adding (1) and (2): 2a = 4 => a = 2

Substitute a = 2 in (1): 2 + c = 6 => c = 4

Therefore:

p(x) = 2x + 5

q(x) = 4x - 1

Q12. Page 38: 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. [IMAGE NEEDED: The first three stages of a growing pattern of hexagons made using matchsticks. Stage 1: 1 hexagon. Stage 2: 2 hexagons sharing a side. Stage 3: 3 hexagons in a row, each sharing a side.] (i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages? (ii) Complete the following table.

Stage Number	1	2	3	4	5	...	n
Number of matchsticks

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

Answer:

This is a classic matchstick pattern problem. A hexagon is made of 6 matchsticks. When two hexagons share a side, they share one matchstick.

Pattern Analysis:

• Stage 1: 1 hexagon. Matchsticks = 6
• Stage 2: 2 hexagons sharing one side. Matchsticks = 6 + (6 - 1) = 11
• Stage 3: 3 hexagons in a row. Matchsticks = 6 + 2(6 - 1) = 6 + 10 = 16

We see that each additional hexagon adds 5 matchsticks (since one side is shared).

(i) Next two stages:

• Stage 4: Matchsticks = 16 + 5 = 21
• Stage 5: Matchsticks = 21 + 5 = 26

(ii) Completed Table:

(iii) Rule for the n^th stage:

Number of matchsticks = 5n + 1

Q13. Page 39: 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.

Answer:

Find p(x):

p(x) = ax + b passes through (2, 3) and (6, 11):

2a + b = 3 ...(1)

6a + b = 11 ...(2)

Subtracting (1) from (2): 4a = 8 => a = 2

Substitute a = 2 in (1): 4 + b = 3 => b = -1

So, p(x) = 2x - 1.

Find q(x):

q(x) = cx + d is parallel to p(x), so their slopes are equal: c = a = 2.

q(x) = 2x + d passes through (4, -1):

-1 = 2(4) + d => -1 = 8 + d => d = -9

So, q(x) = 2x - 9.

Points where the lines meet the x-axis (y=0):

• p(x): 2x - 1 = 0 => x = 1/2. Point: (1/2, 0)
• q(x): 2x - 9 = 0 => x = 9/2. Point: (9/2, 0)

Q14. Page 39: What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Answer:

All linear functions of the form f(x) = ax + a can be rewritten as f(x) = a(x + 1).

They all pass through the point (-1, 0).

This is because when x = -1, f(-1) = a(-1) + a = 0, regardless of the value of a.

Therefore, all these lines intersect the x-axis at the same point, (-1, 0).