Chapter 10 Exercise 10.2 Solutions

Question 1 Tangent & Gradient

Plot the graph of y = 2x² - 4x + 3 from -1 to 3. Draw tangent at (2,3) and find its gradient.

Solution Step-by-Step

Substitute values for x to find corresponding y:

x	-1	0	1	2	3
y	9	3	1	3	9

Plot these coordinates and connect with a smooth parabolic curve opening upward.

To find the gradient of the curve at (2,3):
1. Draw the tangent line touching the curve exactly at point A(2,3).
2. Select a second point on the tangent line, for example B(3,7).
3. Compute the slope m of the tangent line:

m = y₂ - y₁x₂ - x₁

= 7 - 33 - 2 = 41 = 4

Thus, the gradient of the tangent line at (2,3) is 4. (Algebraic check: derivative y' = 4x - 4. At x=2, y' = 8 - 4 = 4. Verified!)

Question 2 Tangent & Gradient

Plot the graph of y = 3x² + x + 1 and draw tangent at (1, 5). Also find gradient of the tangent line at this point.

Solution Step-by-Step

Substitute values:

x	-2	-1	0	1	2
y	11	3	1	5	15

Plot coordinates and connect with a smooth parabola.

Find the gradient at (1,5):
1. Draw the tangent line at point A(1,5).
2. Select a second point on the tangent, for example B(2,12).
3. Compute gradient:

m = 12 - 52 - 1 = 71 = 7

Thus, the gradient of the tangent line is 7. (Algebraic check: derivative y' = 6x + 1. At x=1, y' = 6 + 1 = 7. Verified!)

Question 3 School Decay

The strength of students in a school was 1000 in 2016. If the strength decays according to the equation S = 1000e^-t, where S is the number of students at time t (in years since 2016).
(a) Graph the equation from t = 0 (2016) to t = 9 (2025).
(b) From the graph, estimate the student strength in 2019 and in 2023.

Solution Step-by-Step

(a) Calculate coordinates:

Year	2016 (t=0)	2017 (t=1)	2018 (t=2)	2019 (t=3)	2020 (t=4)	2021 (t=5)	2022 (t=6)	2023 (t=7)	2024 (t=8)	2025 (t=9)
Students S	1000	368	135	50	18	7	3	1	0	0

Plot points and join them with an exponential decay curve. The scale is 1 division on x-axis = 1 year, and 1 division on y-axis = 200 students.

(b) Estimations:
• In 2019 (t = 3): From the graph, student strength is approximately 50 students (calculated exactly: 1000 × e^-3 ≈ 49.78).
• In 2023 (t = 7): From the graph, student strength is approximately 1 student (calculated exactly: 1000 × e^-7 ≈ 0.91).

Question 4 Demand & Supply

The demand and supply functions for a product are given by the equations P_d = 400 - 5Q and P_s = 3Q + 24. Plot the graph of each function over the interval Q = 0 to Q = 100. Find the market equilibrium quantity and price.

Solution Step-by-Step

Calculate coordinates for demand (P_d) and supply (P_s) functions:

Quantity Q	0	25	47	50	100
Demand P_d	400	275	165	150	-100
Supply P_s	24	99	165	174	324

Plot the two lines. Demand is a downward sloping line and supply is an upward sloping line.

The equilibrium point occurs where Demand equals Supply:

P_d = P_s

400 - 5Q = 3Q + 24

8Q = 376 ⇒ Q = 47 units

Equilibrium Price P = 3(47) + 24 = Rs. 165

Thus, the equilibrium quantity is 47 units and equilibrium price is Rs. 165.

Question 5 Salary Function

Shahid's salary S(x) in rupees is based on the following formula: S(x) = 45000 + 4500x, where x is the number of years he has been with the company. Sketch and interpret the graph of the salary function for 0 ≤ x ≤ 5.

Solution Step-by-Step

Substitute values:

Years x	0	1	2	3	4	5
Salary S(x) (Rs.)	45,000	49,500	54,000	58,500	63,000	67,500

Plot these coordinates. The graph is a straight line starting at (0, 45000) with a positive slope.

Interpretation:
• The y-intercept represents Shahid's starting salary of Rs. 45,000.
• The slope (gradient) represents Shahid's annual salary increment of Rs. 4,500 per year.

Question 6 Cost & Revenue

A company manufactures school bags. The cost function of producing x bags is C(x) = 1200 + 20x and the revenue from selling x bags is R(x) = 50x.
(a) Find the break-even point.
(b) Determine the profit or loss when 250 bags are sold.
(c) Plot graphs of both functions and identify the break-even point.

Solution Step-by-Step

(a) Find break-even point

Break-even occurs when total revenue equals total cost:

R(x) = C(x)

50x = 1200 + 20x

30x = 1200 ⇒ x = 40 bags

Break-even Revenue R(40) = 50 × 40 = Rs. 2,000

(b) Profit/loss at x = 250 bags

Profit function is Revenue - Cost:

Profit P(x) = R(x) - C(x) = 50x - (1200 + 20x) = 30x - 1200

P(250) = 30(250) - 1200

= 7500 - 1200 = Rs. 6,300 profit

(c) Plotting and Intersection

Plot lines: Cost starting at (0, 1200) and Revenue starting at (0, 0). The intersection point (break-even point) is at (40, 2000).

Question 7 Newspaper Profit

A newspaper agency has a fixed cost of Rs. 70 per edition and marginal printing and distribution costs of Rs. 40 per copy. If selling price is Rs. 50, the profit function is P(x) = 10x - 70, where x is the number of newspapers. Plot the graph and find profit for 500 newspapers.

Solution Step-by-Step

Plot coordinates for P(x) = 10x - 70:

x (Newspapers)	0	100	200	300	400	500
Profit (Rs.)	-70	930	1930	2930	3930	4930

To find profit for x = 500 copies:

P(500) = 10(500) - 70

= 5000 - 70 = Rs. 4,930

∴ The profit for selling 500 newspapers is Rs. 4,930.

Question 8 Expensive Shirts

Ali manufactures expensive shirts. The cost (in rupees) for producing x shirts is C(x) = 1500 + 10x + 0.2x², for 0 ≤ x ≤ 150. Plot the graph and find the cost of producing 200 shirts.

Solution Step-by-Step

Calculate coordinates for the cost function:

Shirts x	0	25	50	75	100	125	150
Cost C(x) (Rs.)	1500	1875	2500	3375	4500	5875	7500

To find the cost of producing x = 200 shirts:

C(200) = 1500 + 10(200) + 0.2(200)²

= 1500 + 2000 + 0.2(40000)

= 1500 + 2000 + 8000 = Rs. 11,500

∴ The total cost for 200 shirts is Rs. 11,500.

Interactive Sandbox Tangent & Gradient Explorer

Select Parabola

Tangent Point (x): 2.0

9th Mathematics

Exercise 10.2 Solutions