General Instructions
- The question paper has 21 questions divided into six sections, carrying a total of 40 marks.
- Section A: Questions 1–8, one mark each (8 marks). Section B: Questions 9–10, one mark each (2 marks).
- Section C: Questions 11–14, two marks each (8 marks). Section D: Questions 15–18, three marks each (12 marks).
- Section E: Questions 19–20, four marks each (8 marks). Section F: Question 21, two marks (2 marks).
- In Question 20, attempt either part (a) or part (b). All other questions are compulsory.
- Write the full solution, not just the final answer, wherever marks are awarded for working.
Section A – Multiple Choice Questions
8 questions × 1 mark = 8 marks
Each question below has four options (A, B, C, D). Choose the one correct answer.
The graph of a quadratic polynomial \(p(x)\) touches (but does not cross) the x-axis at exactly one point. How many distinct real zeroes does \(p(x)\) have?
Which of the following is a quadratic polynomial?
The area of a rectangle is \((x^2 + 9x + 20)\) square units. If its length is \((x + 5)\) units, its breadth is
The zeroes of the quadratic polynomial \(x^2 + bx + c\) are 5 and \(-3\). Then
Can the numbers \(-3\) and \(-4\) be the zeroes of the polynomial \(x^2 - 7x + 12\)?
A quadratic polynomial whose zeroes are \(-2\) and 5 is
If one zero of the polynomial \(p(x) = 6x^2 + 37x - (k - 2)\) is the reciprocal of the other, then the value of \(k\) is CBSE PYQ · 2023
The graph of the quadratic polynomial \(p(x)\) intersects the x-axis at \((-2, 0)\) and \((4, 0)\). Then \(p(x)\) is
Section B – Assertion and Reason
2 questions × 1 mark = 2 marks
In each of the following questions, an Assertion (A) and a Reason (R) are given. Choose the correct option from (a) to (d).
- (a) Both A and R are true, and R is the correct explanation of A.
- (b) Both A and R are true, but R is not the correct explanation of A.
- (c) A is true, but R is false.
- (d) A is false, but R is true.
Assertion (A): If the sum of the zeroes of the quadratic polynomial \(kx^2 + 2x + 3k\) is equal to their product, then \(k = 1\).
Reason (R): For the polynomial \(ax^2 + bx + c\), the sum of the zeroes is \(-\frac{b}{a}\) and the product of the zeroes is \(\frac{c}{a}\).
Assertion (A): The graph of \(p(x) = x^2 - 5x + 6\) intersects the x-axis at two distinct points.
Reason (R): The zeroes of \(p(x)\) are 2 and 3, which are two distinct real numbers.
Section C – Short Answer Questions
4 questions × 2 marks = 8 marks
Find the zeroes of the quadratic polynomial \(6x^2 - 3 - 7x\) and verify the relationship between the zeroes and the coefficients.
The area of a rectangular park is \((x^2 + 11x + 30)\) square metres. If its length is \((x + 6)\) metres, find its breadth.
If \(-3\) is a zero of the polynomial \(x^2 + 2x + k\), find the value of \(k\) and the other zero of the polynomial.
The graph of a quadratic polynomial \(p(x)\) cuts the x-axis at the points \((1, 0)\) and \((5, 0)\). Write the polynomial \(p(x)\) in the standard form \(ax^2 + bx + c\).
Section D – Application-Based Questions
4 questions × 3 marks = 12 marks
These questions test application of concepts. Show your working clearly.
A rectangular garden has length \((x + 4)\) metres and breadth \((x + 2)\) metres.
(a) Write the area of the garden as a quadratic polynomial in \(x\). [1]
(b) If the area of the garden is 48 square metres, find the value of \(x\) and hence the actual dimensions of the garden. [1½]
(c) State which value of \(x\) must be rejected when solving, and why. [½]
A rectangular display board has a perimeter of 34 units and an area of 60 square units. Let the length of the board be \(L\) units and its breadth be greater than zero.
(a) Write the breadth in terms of \(L\), and hence form the quadratic polynomial in \(L\) for the area. [1½]
(b) Find the length and the breadth of the board, and interpret the two zeroes of the quadratic in \(L\) in the context of the board. [1½]
The product of two consecutive positive integers is 240. Take the smaller number as \(x\).
(a) Form the quadratic polynomial in \(x\). [1]
(b) Find the two integers, giving a reason for rejecting the other value of \(x\). [2]
A stall owner's profit, in rupees, from selling \(x\) dozen cups of juice is modelled by the polynomial \(P(x) = 90x - x^2\).
(a) Find the zeroes of \(P(x)\) and interpret what each zero means for the business. [2]
(b) For what range of values of \(x\) is the profit positive? Explain. [1]
Section E – Long Answer Questions
2 questions × 4 marks = 8 marks
If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(x^2 - 7x + k\) such that \(\alpha - \beta = 3\):
(a) Find the value of \(k\). [2½]
(b) Hence, find the quadratic polynomial whose zeroes are \(\alpha - 1\) and \(\beta - 1\). [1½]
Attempt either part (a) or part (b).
(a) A rectangular garden has length \((x + 8)\) metres and breadth \((x + 3)\) metres. If its area is 84 square metres, form the quadratic in \(x\), find \(x\) and the actual dimensions, and verify your answer using the sum of the zeroes of the polynomial. [4]
(b) Find the value of \(k\) for which the zeroes of the polynomial \(3x^2 - 10x + k\) are reciprocal of each other. For this value of \(k\), obtain the zeroes and verify the relationship between the zeroes and the coefficients. CBSE PYQ · 2019 [4]
Section F – Case-Based Question
1 question × 2 marks = 2 marks
The sports ground near a school is rectangular. Its length is \((2x + 1)\) metres and its breadth is \((x + 2)\) metres. The area of the ground is given by the polynomial \(2x^2 + 5x + 2\) square metres. A contractor uses this polynomial to plan the ground's fencing.
(a) Find the zeroes of the polynomial \(2x^2 + 5x + 2\) and explain why neither zero is meaningful as a dimension of the ground. [1]
(b) If the actual area of the ground is 77 square metres, form the required quadratic equation in \(x\), find the value of \(x\), and obtain the length and breadth of the ground. [1]
Check your work: Section A (8) + Section B (2) + Section C (8) + Section D (12) + Section E (8) + Section F (2) = 40 marks. Attempt the paper fully before revealing the answers below.
Answer Key
Section A – Multiple Choice Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 1 | C | 1 | Original |
| 2 | B | 1 | Original |
| 3 | A | 1 | Original |
| 4 | B | 1 | Original |
| 5 | B | 1 | Original |
| 6 | B | 1 | Original |
| 7 | D | 1 | CBSE PYQ (2023) |
| 8 | B | 1 | Original |
Section B – Assertion and Reason
| Question | Answer | Marks | Source |
|---|---|---|---|
| 9 | D | 1 | Original |
| 10 | A | 1 | Original |
Section C – Short Answer Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 11 | \(-\frac{1}{3}\) and \(\frac{3}{2}\); sum = \(\frac{7}{6}\), product = \(-\frac{1}{2}\) | 2 | Original (NCERT-based) |
| 12 | \((x + 5)\) metres | 2 | Original |
| 13 | \(k = -3\); other zero is 1 | 2 | Original |
| 14 | \(p(x) = x^2 - 6x + 5\) | 2 | Original |
Section D – Application-Based Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 15 | (a) \(x^2 + 6x + 8\) (b) \(x = 4\), garden is 8 m × 6 m (c) \(x = -10\) rejected | 3 | Original |
| 16 | (a) breadth = \(17 - L\); \(L^2 - 17L + 60 = 0\) (b) length 12, breadth 5 | 3 | Original |
| 17 | (a) \(x^2 + x - 240 = 0\) (b) 15 and 16 | 3 | Original |
| 18 | (a) \(x = 0\), \(x = 90\) (b) \(0 < x < 90\) | 3 | Original |
Section E – Long Answer Questions
| Question | Answer | Marks | Source |
|---|---|---|---|
| 19 | (a) \(k = 10\) (b) \(x^2 - 5x + 4\) | 4 | Original |
| 20 | (a) \(x = 4\), 12 m × 7 m OR (b) \(k = 3\); zeroes \(\frac{1}{3}\), 3 | 4 | (a) Original (b) CBSE PYQ (2019) |
Section F – Case-Based Question
| Question | Answer | Marks | Source |
|---|---|---|---|
| 21 | (a) \(-\frac{1}{2}\), \(-2\) (rejected) (b) \(x = 5\); 11 m × 7 m | 2 | Original |
Detailed Solutions
Source: Original question
Concept Used: Geometrical meaning of the zeroes — where the graph meets the x-axis.
A quadratic polynomial has at most two zeroes. When its graph touches the x-axis at exactly one point, the two zeroes are real and equal (both equal to the x-coordinate of the point of contact). It does not cross the axis, so the two zeroes are equal, not distinct.
Source: Original question
Concept Used: Definition of a polynomial and its degree.
A quadratic polynomial has degree 2, so it must contain a term in \(x^2\) with a non-zero coefficient. \(4x + 7\) has degree 1, 8 has degree 0, and \(x^2 - \frac{3}{x}\) is not a polynomial because it contains division by the variable.
Source: Original question
Concept Used: Area of a rectangle = length × breadth, and factorisation.
Breadth = \(\frac{\text{Area}}{\text{Length}} = \frac{x^2 + 9x + 20}{x + 5}\).
\(x^2 + 9x + 20 = (x + 5)(x + 4)\), so breadth = \(x + 4\) units.
Source: Original question
Concept Used: Sum of zeroes \(= -\frac{b}{a}\) and product of zeroes \(= \frac{c}{a}\).
Sum = \(5 + (-3) = 2\), so \(-b = 2 \Rightarrow b = -2\).
Product = \(5 \times (-3) = -15\), so \(c = -15\).
Source: Original question
Concept Used: Testing candidate zeroes against the sum and product of zeroes.
For \(x^2 - 7x + 12\), sum of zeroes \(= -\frac{b}{a} = 7\) and product \(= \frac{c}{a} = 12\).
If the zeroes were \(-3\) and \(-4\), their sum would be \(-7\), but the required sum is 7. The product condition would be satisfied, but the sum condition is not, so they cannot be the zeroes.
Source: Original question
Concept Used: Forming a quadratic from its zeroes: \((x - \alpha)(x - \beta)\).
\(p(x) = (x - (-2))(x - 5) = (x + 2)(x - 5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10\).
Source: CBSE PYQ (2023)
Concept Used: Product of zeroes of \(ax^2 + bx + c\) is \(\frac{c}{a}\); reciprocal zeroes have product 1.
Let the zeroes be \(\alpha\) and \(\frac{1}{\alpha}\). Then their product is 1.
\(\frac{c}{a} = \frac{-(k - 2)}{6} = 1\)
\(-(k - 2) = 6 \Rightarrow k - 2 = -6 \Rightarrow k = -4\).
Source: Original question
Concept Used: If a graph cuts the x-axis at \(\alpha\) and \(\beta\), the zeroes are \(\alpha\) and \(\beta\).
Zeroes are \(-2\) and 4, so \(p(x) = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8\).
Source: Original question
Concept Used: Sum \(= -\frac{b}{a}\), product \(= \frac{c}{a}\).
Sum \(= -\frac{2}{k}\), product \(= \frac{3k}{k} = 3\).
Given sum = product: \(-\frac{2}{k} = 3 \Rightarrow k = -\frac{2}{3}\).
So the Assertion (which claims \(k = 1\)) is false. The Reason correctly states the standard relationships, so it is true.
Source: Original question
Concept Used: Zeroes of a quadratic and the points where its graph meets the x-axis.
\(x^2 - 5x + 6 = (x - 2)(x - 3)\), so the zeroes are 2 and 3, two distinct real numbers. The graph of a quadratic therefore meets the x-axis at these two distinct points. Both A and R are true and R correctly explains A.
Source: Original question (based on NCERT Exercise 2.2, Q1)
Concept Used: Factorising a quadratic and verifying sum/product of zeroes.
\(6x^2 - 3 - 7x = 6x^2 - 7x - 3\).
Splitting the middle term: \(6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3)\).
Zeroes: \(3x + 1 = 0 \Rightarrow x = -\frac{1}{3}\) and \(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\).
Verification:
Sum \(= -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6} = -\frac{b}{a} = \frac{7}{6}\). ✓
Product \(= -\frac{1}{3} \times \frac{3}{2} = -\frac{1}{2} = \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}\). ✓
Source: Original question
Concept Used: Area = length × breadth, factorisation by splitting the middle term.
\(x^2 + 11x + 30 = (x + 6)(x + 5)\).
Breadth = \(\frac{(x + 6)(x + 5)}{x + 6} = x + 5\) metres.
Source: Original question
Concept Used: A number \(\alpha\) is a zero of \(p(x)\) if \(p(\alpha) = 0\).
\(p(-3) = (-3)^2 + 2(-3) + k = 9 - 6 + k = 3 + k = 0 \Rightarrow k = -3\).
So \(p(x) = x^2 + 2x - 3 = (x + 3)(x - 1)\).
The other zero is given by \(x - 1 = 0 \Rightarrow x = 1\).
Check: Sum of zeroes \(= -3 + 1 = -2 = -\frac{b}{a}\). ✓
Source: Original question
Concept Used: The graph meets the x-axis at the zeroes of the polynomial.
The zeroes are 1 and 5, so \(p(x) = (x - 1)(x - 5) = x^2 - 5x - x + 5 = x^2 - 6x + 5\).
Source: Original question
Concept Used: Area of a rectangle, forming and solving a quadratic, rejecting non-meaningful roots.
(a) Area = \((x + 4)(x + 2) = x^2 + 2x + 4x + 8 = x^2 + 6x + 8\).
(b) Given area is 48: \(x^2 + 6x + 8 = 48 \Rightarrow x^2 + 6x - 40 = 0\).
\(x^2 + 6x - 40 = (x + 10)(x - 4) = 0 \Rightarrow x = -10\) or \(x = 4\).
Hence \(x = 4\), so length = \(4 + 4 = 8\) m and breadth = \(4 + 2 = 6\) m.
(c) \(x = -10\) must be rejected because it would make the length \((-10 + 4) = -6\) m and breadth \((-10 + 2) = -8\) m, which are negative and meaningless as dimensions.
Check: \(8 \times 6 = 48\) m². ✓
Source: Original question
Concept Used: Perimeter and area of a rectangle; forming a quadratic from a real-world constraint.
(a) Perimeter = 34, so half the perimeter (length + breadth) = 17. With length \(L\), breadth = \(17 - L\).
Area = \(L(17 - L) = 60 \Rightarrow 17L - L^2 = 60 \Rightarrow L^2 - 17L + 60 = 0\).
(b) \(L^2 - 17L + 60 = (L - 12)(L - 5) = 0 \Rightarrow L = 12\) or \(L = 5\).
Taking the longer side as length, \(L = 12\) and breadth = \(17 - 12 = 5\). The two zeroes 12 and 5 are simply the two side lengths — whether we call either one "length" or "breadth", the rectangle is the same (12 × 5).
Check: Perimeter \(= 2(12 + 5) = 34\) and area \(= 12 \times 5 = 60\). ✓
Source: Original question
Concept Used: Forming and solving a quadratic from a word problem.
(a) Smaller number \(= x\), next consecutive integer \(= x + 1\). Their product: \(x(x + 1) = 240 \Rightarrow x^2 + x - 240 = 0\).
(b) \(x^2 + x - 240 = (x + 16)(x - 15) = 0 \Rightarrow x = -16\) or \(x = 15\).
Since the integers are positive, \(x = 15\), and the numbers are 15 and 16.
Check: \(15 \times 16 = 240\). ✓
Source: Original question
Concept Used: Zeroes of a polynomial and the sign of a quadratic between its zeroes.
(a) \(P(x) = 90x - x^2 = x(90 - x)\). Zeroes are \(x = 0\) and \(x = 90\).
Interpretation: the profit is ₹0 when no cups are sold (\(x = 0\)) and again when 90 dozen cups are sold — the first zero is the starting point, and the second is the point where the rising-then-falling profit returns to zero.
(b) The graph of \(P(x) = -x^2 + 90x\) is a downward-opening parabola, which lies above the x-axis between its two zeroes. Since \(x\) (dozens) cannot be negative, the profit is positive for \(0 < x < 90\). Beyond 90 dozen, profit becomes negative (a loss).
Source: Original question
Concept Used: Sum of zeroes \(= -\frac{b}{a}\), product \(= \frac{c}{a}\); forming a quadratic from new zeroes.
(a) For \(x^2 - 7x + k\): sum of zeroes \(= \alpha + \beta = 7\) and product \(= \alpha\beta = k\).
Given \(\alpha - \beta = 3\). Adding the two equations:
\((\alpha + \beta) + (\alpha - \beta) = 7 + 3 \Rightarrow 2\alpha = 10 \Rightarrow \alpha = 5\).
Then \(\beta = 7 - 5 = 2\). Hence \(k = \alpha\beta = 10\).
(b) New zeroes are \(\alpha - 1 = 4\) and \(\beta - 1 = 1\).
Sum of new zeroes = 5, product = 4.
Polynomial = \(x^2 - (\text{sum})x + \text{product} = x^2 - 5x + 4\).
Source: Original question
Concept Used: Area of a rectangle, solving a quadratic, verifying with the sum of zeroes.
Area = \((x + 8)(x + 3) = x^2 + 3x + 8x + 24 = x^2 + 11x + 24\).
Given area 84: \(x^2 + 11x + 24 = 84 \Rightarrow x^2 + 11x - 60 = 0\).
\(x^2 + 11x - 60 = (x + 15)(x - 4) = 0 \Rightarrow x = -15\) or \(x = 4\).
Since dimensions are positive, \(x = 4\). Length = \(4 + 8 = 12\) m, breadth = \(4 + 3 = 7\) m.
Verification using zeroes: the two zeroes of \(x^2 + 11x - 60\) are \(-15\) and 4, and their sum is \(-11\). The sum of zeroes of \(ax^2 + bx + c\) is \(-\frac{b}{a} = -\frac{11}{1} = -11\). ✓ (Also \(12 \times 7 = 84\) m².)
Source: CBSE PYQ (Delhi 2019)
Concept Used: Reciprocal zeroes have product 1 = \(\frac{c}{a}\); verifying sum/product relationships.
For \(3x^2 - 10x + k\): product of zeroes \(= \frac{k}{3}\).
For reciprocal zeroes, product \(= 1\), so \(\frac{k}{3} = 1 \Rightarrow k = 3\).
With \(k = 3\), \(3x^2 - 10x + 3 = 3x^2 - 9x - x + 3 = 3x(x - 3) - 1(x - 3) = (3x - 1)(x - 3)\).
Zeroes are \(\frac{1}{3}\) and 3 (which are reciprocals).
Verification:
Sum \(= \frac{1}{3} + 3 = \frac{10}{3} = -\frac{b}{a} = \frac{10}{3}\). ✓
Product \(= \frac{1}{3} \times 3 = 1 = \frac{c}{a} = \frac{3}{3} = 1\). ✓
Source: Original question
Concept Used: Zeroes of a quadratic, rejecting non-meaningful roots, forming a quadratic from a given area.
(a) \(2x^2 + 5x + 2 = 2x^2 + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)\).
Zeroes: \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\) and \(x + 2 = 0 \Rightarrow x = -2\).
Both zeroes are negative. A dimension (length or breadth) must be positive, so neither zero can represent a side of the ground — in fact the polynomial above represents the area in terms of the positive quantity \(x\), whose valid value is found in part (b).
(b) If area is 77: \(2x^2 + 5x + 2 = 77 \Rightarrow 2x^2 + 5x - 75 = 0\).
\(2x^2 + 5x - 75 = 2x^2 + 15x - 10x - 75 = x(2x + 15) - 5(2x + 15) = (2x + 15)(x - 5)\).
So \(x = 5\) (the value \(x = -\frac{15}{2}\) is rejected as negative).
Length = \(2(5) + 1 = 11\) m and breadth = \(5 + 2 = 7\) m.
Check: \(11 \times 7 = 77\) m². ✓