Chapter 2 : Class 10 : Polynomials : Case Study Question Bank

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Chapter 2: Polynomials

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Roller Coaster Design 4 Marks

Roller Coaster Track

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An amusement park structural engineer is designing a new high-speed roller coaster. A specific dip in the track is modeled as a parabolic curve to ensure safe gravitational forces on the passengers. In a 2D coordinate system, where the ground level represents the x-axis (horizontal distance in meters) and the y-axis represents the height relative to the ground (in meters), the profile of this section of the track is modeled by the quadratic polynomial:

$$p(x) = x^2 - 4x - 5$$

Q1. Conceptual Understanding 1 Mark

Find the horizontal distances from the origin (points A and B on the x-axis) where the roller coaster track is exactly at ground level.

(A) x = −5 m and x = 1 m

(B) x = 5 m and x = −1 m

(C) x = 4 m and x = −1 m

(D) x = 2 m and x = −3 m

Q2. Application 1 Mark

Calculate the value of p(2) and interpret what this value represents physically in terms of the roller coaster's design.

(A) −5 meters (surface contact)

(B) −7 meters (entry depth)

(C) −9 meters (lowest dip below ground)

(D) −12 meters (tunnel base)

Q3. Analytical Reasoning 2 Marks

If another section of the track is represented by g(x) = x² − (k+3)x + (3k−1), find the value of k such that the sum of its zeroes is equal to half of their product.

(A) k = 7

(B) k = 5

(C) k = 9

(D) k = 3

OR (Alternative Q3)

For the original track polynomial p(x) = x² − 4x − 5 with zeroes α and β, form a new quadratic polynomial in standard form whose zeroes are 2α and 2β.

(A) x² − 4x − 10

(B) x² − 16x − 20

(C) x² − 8x − 10

(D) x² − 8x − 20

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Set p(x) = 0: x² − 4x − 5 = 0. (x − 5)(x + 1) = 0 ⇒ x = 5 or x = −1. Horizontal distances are 5 m and −1 m (Points B and A).	[0.5 Mark] for factorizing. [0.5 Mark] for values x = 5, −1.
Q2	p(2) = (2)² − 4(2) − 5 = 4 − 8 − 5 = −9 m. Physically, this represents the lowest dip (vertex of the parabola), situated 9 meters below ground level.	[0.5 Mark] for p(2) = −9. [0.5 Mark] for physical interpretation.
Q3	For g(x) = x² − (k+3)x + (3k−1): Sum = k + 3, Product = 3k − 1. Given: k + 3 = ½(3k − 1) ⇒ 2k + 6 = 3k − 1 ⇒ k = 7.	[1 Mark] for sum and product relations. [1 Mark] for solving k = 7.
Q3 (OR)	Original: α + β = 4, αβ = −5. New zeroes: 2α, 2β. New Sum S = 2(α + β) = 8. New Product P = 4αβ = 4(−5) = −20. Polynomial: x² − Sx + P = x² − 8x − 20.	[1 Mark] for new sum and product. [1 Mark] for x² − 8x − 20.

Case Study 2 Space Exploration 4 Marks

Sounding Rocket Launch

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ISRO launches a sounding rocket to study upper atmospheric winds. The height of the rocket h(t) in meters above sea level, t seconds after launch, is modeled by the quadratic polynomial:

$$h(t) = -5t^2 + 40t + 45 \quad (t \ge 0)$$

Q1. Conceptual Understanding 1 Mark

Find the zeroes of the quadratic polynomial h(t) by factorization.

(A) t = 5 and t = −9

(B) t = 9 and t = −1

(C) t = 8 and t = −5

(D) t = 6 and t = −3

Q2. Application 1 Mark

What is the physical significance of the positive zero t = 9 in this launch scenario?

(A) Rocket splashes down/lands after 9 s

(B) Rocket engine shuts down at 9 s

(C) Max speed achieved at 9 s

(D) Rocket reaches peak height at 9 s

Q3. Analytical Reasoning 2 Marks

Calculate the maximum height reached by the sounding rocket and the time taken to reach this peak height.

(A) 115 m at t = 3 s

(B) 120 m at t = 5 s

(C) 125 m at t = 4 s

(D) 130 m at t = 4.5 s

OR (Alternative Q3)

If a test rocket's path is q(t) = t² − 6t + (m−4), find the value of m if one of its zeroes is double the other.

(A) m = 10

(B) m = 12

(C) m = 8

(D) m = 16

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Set h(t) = 0: −5(t² − 8t − 9) = 0. (t − 9)(t + 1) = 0 ⇒ t = 9 or t = −1.	[0.5 Mark] for simplification. [0.5 Mark] for roots t = 9, −1.
Q2	The positive zero t = 9 represents the time in seconds when the rocket returns to the ground (height = 0). The negative zero is discarded as time cannot be negative.	[0.5 Mark] for landing time. [0.5 Mark] for discarding negative root.
Q3	Vertex occurs at t = −b / (2a) = −40 / (2 × −5) = 4 seconds. h(4) = −5(16) + 40(4) + 45 = −80 + 160 + 45 = 125 meters.	[1 Mark] for finding t = 4 s. [1 Mark] for max height 125 m.
Q3 (OR)	Let zeroes be α, 2α. Sum = α + 2α = 6 ⇒ 3α = 6 ⇒ α = 2. Product = (2)(4) = m − 4 ⇒ 8 = m − 4 ⇒ m = 12.	[1 Mark] for finding α = 2. [1 Mark] for evaluating m = 12.

Case Study 3 Sports Analytics 4 Marks

Basketball Shot

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A sports analytics team tracks basketball trajectories. The height f(x) (in feet) at a horizontal distance x (in feet) from release is modeled by the quadratic polynomial:

$$f(x) = -x^2 + 10x - 16$$

Q1. Conceptual Understanding 1 Mark

Find the horizontal distances where the ball is mathematically at a height of 0 feet (f(x) = 0).

(A) 2 ft and 8 ft

(B) 1 ft and 16 ft

(C) 4 ft and 4 ft

(D) 3 ft and 7 ft

Q2. Application 1 Mark

If the zeroes of the trajectory polynomial are α and β, calculate the value of α² + β².

(A) 64

(B) 100

(C) 72

(D) 68

Q3. Analytical Reasoning 2 Marks

A flatter trajectory g(x) = x² − px + q has zeroes equal to the reciprocals of the zeroes of f(x). Find p and q.

(A) p = 1/2, q = 1/8

(B) p = 5/8, q = 1/16

(C) p = 3/4, q = 1/12

(D) p = 5/16, q = 1/8

OR (Alternative Q3)

For the original trajectory polynomial f(x) with zeroes α and β, evaluate (α/β) + (β/α).

(A) 17/4 (or 4.25)

(B) 15/4

(C) 19/4

(D) 21/4

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Set −x² + 10x − 16 = 0 ⇒ x² − 10x + 16 = 0. (x − 8)(x − 2) = 0 ⇒ x = 2 ft and x = 8 ft.	[0.5 Mark] for splitting middle term. [0.5 Mark] for roots x = 2, 8.
Q2	α + β = 10, αβ = 16. α² + β² = (α + β)² − 2αβ = 10² − 2(16) = 100 − 32 = 68.	[1 Mark] for evaluation to 68.
Q3	Reciprocal zeroes: 1/2 and 1/8. Sum p = 1/2 + 1/8 = 5/8. Product q = (1/2)(1/8) = 1/16.	[1 Mark] for finding p = 5/8. [1 Mark] for finding q = 1/16.
Q3 (OR)	(α/β) + (β/α) = (α² + β²) / (αβ). Using α² + β² = 68 and αβ = 16: 68 / 16 = 17/4 = 4.25.	[1 Mark] for algebraic expression. [1 Mark] for evaluating 17/4.

Case Study 4 Civil Engineering 4 Marks

Suspension Bridge Cable

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A structural engineer designs a suspension bridge. The cable profile between two towers is modeled by the quadratic polynomial:

$$p(x) = \frac{1}{20}x^2 - 2x + 25 \quad (0 \le x \le 40)$$

Q1. Conceptual Understanding 1 Mark

Find the discriminant of p(x) and determine if the cable ever touches the bridge deck (y = 0).

(A) D = 0; touches at midpoint

(B) D = −1; never touches the deck

(C) D = 4; crosses at two points

(D) D = −5; touches once

Q2. Application 1 Mark

Find the height of the supporting cable at the left tower (x = 0) and right tower (x = 40 m).

(A) 25 m at both towers

(B) 20 m and 25 m

(C) 30 m at both towers

(D) 15 m and 20 m

Q3. Analytical Reasoning 2 Marks

For an arch q(x) = x² − 6x + (m−2), find the range of m for which it touches or crosses the deck (real zeroes).

(A) m ≥ 11

(B) m ≤ 9

(C) m ≤ 11

(D) m ≥ 7

OR (Alternative Q3)

If an arch's zeroes satisfy α + β = 8 and α² + β² = 40, find the polynomial (leading coefficient 1).

(A) x² − 8x + 16

(B) x² − 8x + 20

(C) x² − 6x + 12

(D) x² − 8x + 12

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	D = b² − 4ac = (−2)² − 4(1/20)(25) = 4 − 5 = −1. Since D < 0, no real zeroes exist; the cable never touches the bridge deck.	[0.5 Mark] for D = −1. [0.5 Mark] for interpretation.
Q2	p(0) = 25 m. p(40) = (1/20)(1600) − 2(40) + 25 = 80 − 80 + 25 = 25 m.	[0.5 Mark] for p(0) = 25 m. [0.5 Mark] for p(40) = 25 m.
Q3	Real roots require D ≥ 0: (−6)² − 4(1)(m − 2) ≥ 0. 36 − 4m + 8 ≥ 0 ⇒ 44 ≥ 4m ⇒ m ≤ 11.	[1 Mark] for setting D ≥ 0. [1 Mark] for m ≤ 11.
Q3 (OR)	(α + β)² = α² + β² + 2αβ ⇒ 8² = 40 + 2αβ ⇒ 2αβ = 24 ⇒ αβ = 12. Polynomial: x² − 8x + 12.	[1 Mark] for αβ = 12. [1 Mark] for x² − 8x + 12.

Case Study 5 Fintech Profit Model 4 Marks

Fintech Profit Model

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A fintech start-up models its daily profit P(x) (in ₹'000) as a function of premium users x (in hundreds):

$$P(x) = -x^2 + 12x - 20$$

Q1. Prime Break-Even 1 Mark

Find the active premium user counts (in hundreds) where the start-up breaks even (P(x) = 0).

(A) 4 and 8 hundred users

(B) 2 and 10 hundred users

(C) 3 and 9 hundred users

(D) 1 and 12 hundred users

Q2. Reciprocal Roots 1 Mark

Form a polynomial Q(x) whose zeroes are reciprocals of the zeroes of P(x).

(A) 10x² − 6x + 1

(B) 20x² − 10x + 1

(C) 20x² − 12x + 1

(D) 12x² − 20x + 1

Q3. Optimization 2 Marks

Find the maximum daily profit the start-up can achieve and the user count required.

(A) ₹16,000 at 6 hundred users

(B) ₹18,000 at 6 hundred users

(C) ₹20,000 at 8 hundred users

(D) ₹15,000 at 5 hundred users

OR (Alternative Q3)

If R(x) = P(x) + k has exactly one break-even point (equal roots), find the value of k.

(A) k = 16

(B) k = −8

(C) k = 12

(D) k = −16

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	−x² + 12x − 20 = 0 ⇒ x² − 12x + 20 = 0. (x − 2)(x − 10) = 0 ⇒ x = 2 and x = 10 hundred users.	[0.5 Mark] for quadratic setup. [0.5 Mark] for roots 2, 10.
Q2	Reciprocal zeroes: 1/2 and 1/10. S = 1/2 + 1/10 = 3/5, P = 1/20. Q(x) = 20(x² − (3/5)x + 1/20) = 20x² − 12x + 1.	[0.5 Mark] for S and P. [0.5 Mark] for 20x² − 12x + 1.
Q3	Vertex x = −12 / (2 × −1) = 6 hundred users. Max profit P(6) = −(6)² + 12(6) − 20 = −36 + 72 − 20 = ₹16 thousand (₹16,000).	[1 Mark] for x = 6. [1 Mark] for ₹16,000 profit.
Q3 (OR)	R(x) = −x² + 12x + (k − 20). D = 0 ⇒ 12² − 4(−1)(k − 20) = 0. 144 + 4k − 80 = 0 ⇒ 64 + 4k = 0 ⇒ k = −16.	[1 Mark] for setting D = 0. [1 Mark] for k = −16.

Case Study 6 Green Energy 4 Marks

Parabolic Solar Reflector

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A solar concentrator mirror is modeled by the quadratic polynomial:

$$f(x) = x^2 - 6x + k$$

Zeroes α and β represent the horizontal mounting offsets on a support frame.

Q1. Coefficients Relation 1 Mark

Write the sum of zeroes (α + β) and product of zeroes (αβ) in terms of k.

(A) α + β = 6, αβ = k

(B) α + β = −6, αβ = k

(C) α + β = 6, αβ = −k

(D) α + β = 3, αβ = k/2

Q2. Difference Evaluation 1 Mark

If the distance between mounting support points is |α − β| = 4 dm, calculate k.

(A) k = 4

(B) k = 5

(C) k = 6

(D) k = 8

Q3. Algebraic Operations 2 Marks

Using the value of k = 5, find the value of α³ + β³.

(A) 116

(B) 136

(C) 126

(D) 144

OR (Alternative Q3)

Find a new quadratic polynomial whose zeroes are (α + 2) and (β + 2).

(A) x² − 10x + 21

(B) x² − 8x + 15

(C) x² − 12x + 25

(D) x² − 10x + 25

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	α + β = −(−6)/1 = 6, αβ = k/1 = k.	[0.5 Mark] for sum. [0.5 Mark] for product.
Q2	(α − β)² = (α + β)² − 4αβ ⇒ 4² = 6² − 4k. 16 = 36 − 4k ⇒ 4k = 20 ⇒ k = 5.	[0.5 Mark] for identity. [0.5 Mark] for k = 5.
Q3	α³ + β³ = (α + β)[(α + β)² − 3αβ]. = 6[6² − 3(5)] = 6[36 − 15] = 6(21) = 126.	[1 Mark] for cubic identity. [1 Mark] for evaluating 126.
Q3 (OR)	New sum = (α + 2) + (β + 2) = (α + β) + 4 = 10. New product = (α + 2)(β + 2) = αβ + 2(α + β) + 4 = 5 + 12 + 4 = 21. Polynomial: x² − 10x + 21.	[1 Mark] for new S and P. [1 Mark] for x² − 10x + 21.

Case Study 7 Game Development 4 Marks

Game Jump Dynamics

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A video game programming team models a hero's jump trajectory by:

$$y = g(x) = -a(x - 1)(x - 5)$$

where y is height (m) and x is horizontal distance (m).

Q1. Scale Constant 1 Mark

If maximum height is 4 m at x = 3 m, determine a and write g(x) in standard form.

(A) a = 1, g(x) = −x² + 6x − 5

(B) a = 2, g(x) = −2x² + 12x − 10

(C) a = 0.5, g(x) = −0.5x² + 3x − 2.5

(D) a = 1, g(x) = −x² + 5x − 6

Q2. Height Evaluation 1 Mark

Calculate the height of the hero when horizontal distance covered is 2 m (g(2)).

(A) 2 meters

(B) 3 meters

(C) 3.5 meters

(D) 4 meters

Q3. Roots Ratio 2 Marks

If α and β are zeroes of g(x), calculate the numerical value of (α/β) + (β/α).

(A) 24/5

(B) 5.4

(C) 28/5

(D) 26/5 (or 5.2)

OR (Alternative Q3)

Find a quadratic polynomial whose zeroes are double the zeroes of g(x).

(A) x² − 6x + 10

(B) x² − 12x + 20

(C) x² − 10x + 24

(D) x² − 12x + 40

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	At x = 3, y = 4: 4 = −a(3 − 1)(3 − 5) = −a(2)(−2) = 4a ⇒ a = 1. g(x) = −(x − 1)(x − 5) = −x² + 6x − 5.	[0.5 Mark] for a = 1. [0.5 Mark] for standard form.
Q2	g(2) = −(2)² + 6(2) − 5 = −4 + 12 − 5 = 3 meters.	[1 Mark] for evaluation to 3 m.
Q3	Zeroes are α = 1, β = 5. (α/β) + (β/α) = (α² + β²) / (αβ) = (6² − 2(5)) / 5 = 26/5 = 5.2.	[1 Mark] for formula expansion. [1 Mark] for 26/5.
Q3 (OR)	New zeroes: 2(1) = 2 and 2(5) = 10. Sum = 12, Product = 20. Polynomial: x² − 12x + 20.	[1 Mark] for new zeroes 2, 10. [1 Mark] for x² − 12x + 20.

Case Study 8 Precision Agriculture 4 Marks

Crop Yield vs Nitrogen

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Wheat yield Y(x) (t/ha) based on nitrogen application x (in 10 kg/ha) is modeled by:

$$Y(x) = -\frac{1}{2}x^2 + 8x - 14$$

Q1. Crop Failure Limits 1 Mark

Find nitrogen rates x (in 10 kg/ha) where crop yield is zero (Y(x) = 0).

(A) x = 4 and x = 12

(B) x = 2 and x = 14

(C) x = 3 and x = 10

(D) x = 1 and x = 15

Q2. Optimal Nitrogen Rate 1 Mark

Find the optimized application rate of nitrogen x that yields the maximum possible harvest.

(A) x = 8 (80 kg/ha)

(B) x = 10 (100 kg/ha)

(C) x = 6 (60 kg/ha)

(D) x = 7.5 (75 kg/ha)

Q3. Product & Sum Roots 2 Marks

If α and β are roots of Y(x), calculate the numerical value of α²β + αβ².

(A) 224

(B) 392

(C) 448

(D) 512

OR (Alternative Q3)

Find Z(x) with roots (α + 2) and (β + 2), keeping leading coefficient as −1/2.

(A) −½x² + 10x − 32

(B) −½x² + 12x − 36

(C) −½x² + 8x − 28

(D) −½x² + 10x − 24

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	−½x² + 8x − 14 = 0 ⇒ x² − 16x + 28 = 0. (x − 14)(x − 2) = 0 ⇒ x = 2 and x = 14.	[0.5 Mark] for simplification. [0.5 Mark] for roots 2, 14.
Q2	x = −b / (2a) = −8 / (2 × −½) = 8 (80 kg/ha).	[1 Mark] for evaluating x = 8.
Q3	α + β = 16, αβ = 28. α²β + αβ² = αβ(α + β) = 28 × 16 = 448.	[1 Mark] for factorization. [1 Mark] for 448.
Q3 (OR)	New roots sum = 16 + 4 = 20. New roots product = 28 + 2(16) + 4 = 64. Z(x) = −½(x² − 20x + 64) = −½x² + 10x − 32.	[1 Mark] for new sum and product. [1 Mark] for Z(x).

Live Practice: Chapter 2 Polynomials

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Case Study 1

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