Class 10 Maths Case Study Worksheet Chapter 2 Polynomials | Detailed Solutions & Marking Scheme
VEDANT WORKSHEET SERIES
CLASS X | MATHEMATICS (041)
CHAPTER 2: POLYNOMIALS (CASE STUDY SPECIAL)
Time Allowed: 1 Hour Maximum Marks: 30
GENERAL INSTRUCTIONS
This question paper contains 5 case study questions.
Each case study has three sub-questions of 1, 1 and 2 marks respectively.
All questions are compulsory.
Show necessary calculations clearly.
CASE STUDY 1 : THE ROLLER COASTER DESIGN
A roller coaster track is designed in the shape of a parabola. The equation of the track is:
p(x) = x² − 4x + 3
(i) Find the zeros of the polynomial. (1 mark)
(ii) What is the shape of the graph of p(x) = ax² + bx + c if a > 0? (1 mark)
(iii) If the new polynomial is
g(x) = x² − 4x + (3 + k),
find k so that the graph touches the x-axis at exactly one point. (2 marks)
CASE STUDY 2 : ARCHERY TARGET PRACTICE
The height of an arrow is given by:
h(t) = −5t² + 20t + 2
(i) Find the height at t = 0. (1 mark)
(ii) What is the maximum number of zeros of a quadratic polynomial? (1 mark)
(iii) If α and β are zeros of
f(t) = 5t² − 20t − 2,
find α² + β². (2 marks)
CASE STUDY 3 : HIGHWAY OVERPASS CONSTRUCTION
The equation of the overpass is:
p(x) = x² − 2x − 8
(i) Find the distance between the zeros of the polynomial. (1 mark)
(ii) If one zero of x² + (a + 1)x + b is 2 and the other is −3, find a. (1 mark)
(iii) Find a quadratic polynomial whose zeros are the reciprocals of the zeros of
p(x) = x² − 2x − 8. (2 marks)
CASE STUDY 4 : FUEL EFFICIENCY MODELING
The efficiency is given by:
E(v) = −(1/40)v² + (5/2)v
(i) Find the speeds (other than zero) where efficiency becomes zero. (1 mark)
(ii) If α and β are zeros of a quadratic polynomial, write its general form. (1 mark)
(iii) If the sum of zeros of
p(v) = kv² + 2v + 3k
is equal to their product, find k. (2 marks)
CASE STUDY 5 : BASKETBALL TRAJECTORY
The path of the ball is:
f(x) = −x² + 6x − 5
(i) Find the values of x where f(x) = 0. (1 mark)
(ii) Does the graph open upwards or downwards? (1 mark)
(iii) If new zeros are (α + 2) and (β + 2), where α and β are zeros of f(x), find the new polynomial. (2 marks)
VEDANT WORKSHEET SERIES – Designed for Academic Excellence
VEDANT WORKSHEET SERIES – ANSWER KEY & MARKING SCHEME
CLASS X | MATHEMATICS (041)
CHAPTER 2: POLYNOMIALS (CASE STUDY SPECIAL)
Maximum Marks: 30
CASE STUDY 1 : THE ROLLER COASTER DESIGN
(i)
p(x) = x² − 4x + 3
= (x − 1)(x − 3)
Zeros = 1, 3
Marking: Factorisation (½) + Correct zeros (½) = 1 mark
(ii)
For p(x) = ax² + bx + c, if a > 0 → graph opens upwards
Marking: Correct statement = 1 mark
(iii)
g(x) = x² − 4x + (3 + k)
Condition for one zero: D = 0
D = b² − 4ac
= (−4)² − 4(1)(3 + k)
= 16 − 4(3 + k)
= 16 − 12 − 4k
= 4 − 4k
For one solution:
4 − 4k = 0 ⇒ k = 1
Answer: k = 1
Marking: Discriminant (1) + Final answer (1) = 2 marks
CASE STUDY 2 : ARCHERY TARGET PRACTICE
(i)
h(0) = −5(0)² + 20(0) + 2 = 2
Marking: Substitution (½) + Answer (½) = 1 mark
(ii)
Maximum number of zeros of a quadratic polynomial = 2
Marking: Correct answer = 1 mark
(iii)
For f(t) = 5t² − 20t − 2
Sum of zeros:
α + β = −(−20)/5 = 20/5 = 4
Product of zeros:
αβ = (−2)/5
Now,
α² + β² = (α + β)² − 2αβ
= 4² − 2(−2/5)
= 16 + 4/5
= (80 + 4)/5
= 84/5
Answer: 84/5
Marking: Formula (1) + Calculation (1) = 2 marks
CASE STUDY 3 : HIGHWAY OVERPASS CONSTRUCTION
(i)
p(x) = x² − 2x − 8
= (x − 4)(x + 2)
Zeros = 4, −2
Distance = |4 − (−2)| = 6
Marking: Zeros (½) + Distance (½) = 1 mark
(ii)
Sum of zeros = −(a + 1)
Given zeros: 2 and −3
Sum = 2 + (−3) = −1
So,
−(a + 1) = −1
⇒ a + 1 = 1
⇒ a = 0
Answer: a = 0
Marking: Equation (½) + Answer (½) = 1 mark
(iii)
Zeros of p(x) = 4 and −2
Reciprocal zeros = 1/4 and −1/2
Sum = 1/4 − 1/2 = −1/4
Product = (1/4)(−1/2) = −1/8
Polynomial:
x² − (sum)x + product
= x² − (−1/4)x − 1/8
= x² + (1/4)x − 1/8
Multiply by 8:
8x² + 2x − 1
Answer: 8x² + 2x − 1
Marking: Reciprocal idea (1) + Final polynomial (1) = 2 marks
CASE STUDY 4 : FUEL EFFICIENCY MODELING
(i)
E(v) = −(1/40)v² + (5/2)v
Set E(v) = 0:
v[−(1/40)v + (5/2)] = 0
Non-zero solution:
−v/40 + 5/2 = 0
⇒ v/40 = 5/2
⇒ v = 100
Answer: 100 km/h
Marking: Factorisation (½) + Answer (½) = 1 mark
(ii)
Quadratic polynomial with zeros α and β:
= x² − (α + β)x + αβ
Marking: Correct form = 1 mark
(iii)
For p(v) = kv² + 2v + 3k
Sum = −2/k
Product = 3k/k = 3
Given: Sum = Product
⇒ −2/k = 3
⇒ −2 = 3k
⇒ k = −2/3
Answer: k = −2/3
Marking: Formula (1) + Answer (1) = 2 marks
CASE STUDY 5 : BASKETBALL TRAJECTORY
(i)
f(x) = −x² + 6x − 5 = 0
Multiply by −1:
x² − 6x + 5 = 0
= (x − 1)(x − 5)
x = 1, 5
Answer: 1, 5
Marking: Factorisation (½) + Answer (½) = 1 mark
(ii)
Coefficient of x² is negative → graph opens downwards
Marking: Correct statement = 1 mark
(iii)
Original zeros: 1 and 5
New zeros:
1 + 2 = 3
5 + 2 = 7
Polynomial:
x² − (3 + 7)x + (3×7)
= x² − 10x + 21
Answer: x² − 10x + 21
Marking: New zeros (1) + Polynomial (1) = 2 marks
FINAL MARK DISTRIBUTION
Case Study 1: 4 marks
Case Study 2: 4 marks
Case Study 3: 4 marks
Case Study 4: 4 marks
Case Study 5: 4 marks
Total = 20 marks (Case Study Section)
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