Class 10 CBSE Chapter 2 Practice Test

Format: MCQs (5m) + Short (10m) + Long (10m) + Case Study (5m) = 30 Marks

CBSE CLASS X • MATHEMATICAL ASSESSMENT

Chapter 2: Polynomials

Standard / Basic Practice Test Paper

Time Allowed: 1 Hour Maximum Marks: 30 Marks

General Instructions:

1. This question paper contains 4 Sections: A, B, C, and D.
2. Section A consists of 5 Multiple Choice Questions (MCQs) carrying 1 mark each.
3. Section B consists of 5 Short Answer Questions carrying 2 marks each.
4. Section C consists of 2 Long Answer Questions carrying 5 marks each.
5. Section D consists of 1 Case Study based question carrying 5 marks.
6. All questions are compulsory. Use of calculators is not permitted.

Section A (1 Mark Each)

5 Marks

1.

If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = x^2 - 5x + 6$, then the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is:

(a) $\frac{5}{6}$

(b) $-\frac{5}{6}$

(c) $\frac{6}{5}$

(d) $-\frac{6}{5}$

2.

If one zero of the quadratic polynomial $x^2 + 3x + k$ is $2$, then the value of $k$ is:

(a) $10$

(b) $-10$

(c) $-7$

(d) $-2$

3.

A quadratic polynomial whose sum and product of zeroes are $-3$ and $2$ respectively is:

(a) $x^2 + 3x + 2$

(b) $x^2 - 3x + 2$

(c) $x^2 + 3x - 2$

(d) $x^2 - 3x - 2$

4.

If one zero of the polynomial $5x^2 + 13x + k$ is the reciprocal of the other, then the value of $k$ is:

(a) $0$

(b) $5$

(c) $\frac{1}{5}$

(d) $6$

5.

If the zeroes of a quadratic polynomial $ax^2 + bx + c$ (where $c \neq 0$) are equal, then:

(a) $c$ and $a$ have opposite signs

(b) $c$ and $b$ have opposite signs

(a) $c$ and $a$ have the same sign

(d) $c$ and $b$ have the same sign

Section B (2 Marks Each)

5 × 2 = 10 Marks

6. Find the zeroes of the quadratic polynomial $6x^2 - 3 - 7x$ and verify the relationship between the zeroes and the coefficients.

[2 Marks]

7. Find a quadratic polynomial whose zeroes are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

[2 Marks]

8. If $\alpha$ and $\beta$ are the zeroes of $p(x) = 2x^2 - 4x + 5$, evaluate the expression $\alpha^2 + \beta^2$.

[2 Marks]

9. If one zero of the polynomial $(a^2 + 9)x^2 + 13x + 6a$ is reciprocal of the other, find the value of $a$.

[2 Marks]

10. If the sum of the zeroes of the quadratic polynomial $p(x) = (k^2 - 14)x^2 - 2x - 12$ is equal to $1$, find the value(s) of $k$.

[2 Marks]

Section C (5 Marks Each)

2 × 5 = 10 Marks

11. If $\alpha$ and $\beta$ are zeroes of the quadratic polynomial $f(x) = x^2 - p(x + 1) - c$:

[5 Marks]

1. Show that $(\alpha + 1)(\beta + 1) = 1 - c$.
2. If $(\alpha + 1)(\beta + 1) = 0$, determine the value of $c$.

12. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 3x^2 - 6x + 4$:

[5 Marks]

1. Form a new quadratic polynomial whose zeroes are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
2. Find the exact numerical value of $\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right) + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta$.

Section D (Case Study)

1 × 5 = 5 Marks

Case Study: Architectural Bridge Design

An engineering firm is designing a parabolic arch bridge over a highway bypass. The vertical profile of the arch bridge is modeled mathematically by the quadratic function $p(x) = -x^2 + 2x + 8$, where $x$ represents the horizontal displacement from a central reference marker (in meters), and $p(x)$ represents the height of the bridge structure above ground level (in meters).

Based on the above scenario, answer the following questions (1 Mark each):

1. What is the geometrical shape formed by the graph of the quadratic polynomial $p(x)$?
2. Calculate the zeroes of the polynomial $p(x) = -x^2 + 2x + 8$.
3. Find the total horizontal width (span) between the two base foundation points of the arch.
4. At what horizontal distance $x$ does the arch bridge reach its maximum height?
5. Calculate the maximum vertical height achieved by the bridge above ground level.

Answer Key & Detailed Solutions

Section A: MCQs

• 1. (a) $\frac{5}{6}$ — Explanation: $\alpha + \beta = 5$, $\alpha\beta = 6$. Therefore, $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{5}{6}$.
• 2. (b) $-10$ — Explanation: $p(2) = 0 \implies (2)^2 + 3(2) + k = 0 \implies 4 + 6 + k = 0 \implies k = -10$.
• 3. (a) $x^2 + 3x + 2$ — Explanation: $p(x) = x^2 - (\text{Sum})x + (\text{Product}) = x^2 - (-3)x + 2 = x^2 + 3x + 2$.
• 4. (b) $5$ — Explanation: Let zeroes be $\alpha$ and $\frac{1}{\alpha}$. Product of zeroes $= 1 = \frac{c}{a} = \frac{k}{5} \implies k = 5$.
• 5. (c) $c$ and $a$ have the same sign — Explanation: Equal zeroes $\implies D = b^2 - 4ac = 0 \implies 4ac = b^2 > 0 \implies ac > 0$.

Section B: Short Answers

6. Solution: Rearrange polynomial: $6x^2 - 7x - 3 = 0$.
By splitting middle term: $6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)$.
Zeroes: $\alpha = \frac{3}{2}, \beta = -\frac{1}{3}$.
$\text{Sum } \alpha + \beta = \frac{3}{2} - \frac{1}{3} = \frac{7}{6} = -\frac{b}{a}$.
$\text{Product } \alpha\beta = \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{1}{2} = \frac{-3}{6} = \frac{c}{a}$. (Verified)

7. Solution: Zeroes $\alpha = 2 + \sqrt{3}$, $\beta = 2 - \sqrt{3}$.
Sum $S = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.
Product $P = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
Polynomial $p(x) = x^2 - Sx + P = \mathbf{x^2 - 4x + 1}$.

8. Solution: For $2x^2 - 4x + 5$: $a=2, b=-4, c=5$.
$\alpha + \beta = -\frac{-4}{2} = 2$, $\alpha\beta = \frac{5}{2}$.
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (2)^2 - 2\left(\frac{5}{2}\right) = 4 - 5 = \mathbf{-1}$.

9. Solution: Zeroes are $\alpha$ and $\frac{1}{\alpha}$. Product $= 1 = \frac{c}{a}$.
$\implies \frac{6a}{a^2 + 9} = 1 \implies a^2 + 9 = 6a \implies a^2 - 6a + 9 = 0 \implies (a - 3)^2 = 0 \implies \mathbf{a = 3}$.

10. Solution: Sum of zeroes $= -\frac{b}{a} = -\frac{-2}{k^2 - 14} = \frac{2}{k^2 - 14}$.
Given sum $= 1 \implies \frac{2}{k^2 - 14} = 1 \implies k^2 - 14 = 2 \implies k^2 = 16 \implies \mathbf{k = \pm 4}$.

Section C: Long Answers

11. Solution:
Expand $f(x) = x^2 - px - p - c = x^2 - px - (p + c)$.
Here $a = 1, b = -p, c' = -(p + c)$.
Sum $\alpha + \beta = p$, Product $\alpha\beta = -(p + c)$.
(a) $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = -(p + c) + p + 1 = -p - c + p + 1 = \mathbf{1 - c}$. (Proved)
(b) If $(\alpha + 1)(\beta + 1) = 0 \implies 1 - c = 0 \implies \mathbf{c = 1}$.

12. Solution:
For $3x^2 - 6x + 4$: $\alpha + \beta = 2$, $\alpha\beta = \frac{4}{3}$.
(a) New zeroes $S' = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{4 - 8/3}{4/3} = \frac{4/3}{4/3} = 1$.
New product $P' = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$.
Polynomial: $x^2 - S'x + P' = \mathbf{x^2 - x + 1}$.
(b) Expression $= 1 + 2\left(\frac{\alpha+\beta}{\alpha\beta}\right) + 3\alpha\beta = 1 + 2\left(\frac{2}{4/3}\right) + 3\left(\frac{4}{3}\right) = 1 + 2\left(\frac{3}{2}\right) + 4 = 1 + 3 + 4 = \mathbf{8}$.

Section D: Case Study

1. Parabola (opening downwards since coefficient of $x^2$ is negative).
2. $-x^2 + 2x + 8 = 0 \implies x^2 - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0 \implies \mathbf{x = 4 \text{ and } x = -2}$.
3. Total width (span) $= 4 - (-2) = \mathbf{6 \text{ meters}}$.
4. Maximum height occurs at midpoint of zeroes: $x = \frac{4 + (-2)}{2} = \mathbf{1 \text{ meter}}$.
5. Maximum height $p(1) = -(1)^2 + 2(1) + 8 = -1 + 2 + 8 = \mathbf{9 \text{ meters}}$.

Class 10 Mathematics Practice Assessment • Polynomials (30 Marks) • NCERT/CBSE Aligned