Class 10 Polynomials Mock Test (CBSE)

Topics in Polynomials

• Geometrical meaning of the zeroes of a polynomial
• Relationship between zeroes and coefficients of a polynomial

Polynomials practice questions with answers

1. Q1Multiple choice·1 mark
   If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 - 5x + k$ such that $\alpha - \beta = 1$, then the value of $k$ is:

   1. (a)$4$
   2. (b)$5$
   3. (c)$6$
   4. (d)$7$
   Show answer and step marks

   1. Correct option: (c) $6$. We have $\alpha + \beta = 5$ and $\alpha - \beta = 1$. Solving these gives $\alpha = 3$ and $\beta = 2$. Thus, $k = \alpha\beta = 3 \times 2 = 6$.+1

   Answer: (c)
2. Q2Multiple choice·1 mark
   If one zero of the quadratic polynomial $x^2 + 3x + k$ is $2$, then the value of $k$ is:

   1. (a)$10$
   2. (b)$-10$
   3. (c)$5$
   4. (d)$-5$
   Show answer and step marks

   1. Correct option: (b) $-10$. Substitute $x = 2$ into the polynomial and equate to zero: $(2)^2 + 3(2) + k = 0$, which yields $4 + 6 + k = 0$, giving $k = -10$.+1

   Answer: (b)
3. Q3Short answer·2 marks
   Find the zeroes of the quadratic polynomial $f(x) = x^2 - 2x - 8$ and verify the relationship between the zeroes and its coefficients.
   Show answer and step marks

   1. $x^2 - 2x - 8 = (x - 4)(x + 2)$, so the zeroes are $4$ and $-2$.+1
   2. Sum of zeroes: $4 + (-2) = 2 = -\frac{-2}{1} = -\frac{b}{a}$.+0.5
   3. Product of zeroes: $4 \times (-2) = -8 = \frac{-8}{1} = \frac{c}{a}$.+0.5

   Answer: Zeroes: $4, -2$; sum $= 2 = -\frac{b}{a}$, product $= -8 = \frac{c}{a}$
4. Q4Short answer·3 marks
   Two quadratic polynomials are given: $p(x) = 6x^2 - 5x - 6$ and $q(x) = 6x^2 + 5x - 6$. (a) Find the sum of the zeroes of $p(x)$. [1] (b) Find the product of the zeroes of $q(x)$. [1] (c) Compare the sum of the zeroes of $q(x)$ with the sum of the zeroes of $p(x)$. [1]
   Show answer and step marks

   1. (a) For $ax^2+bx+c$, sum of zeroes $= -\dfrac{b}{a} = -\dfrac{(-5)}{6}$+0.5
   2. (a) Sum of zeroes of $p(x)$ $= \dfrac{5}{6}$+0.5
   3. (b) Product of zeroes $= \dfrac{c}{a} = \dfrac{-6}{6}$ for $q(x)$+0.5
   4. (b) Product of zeroes of $q(x)$ $= -1$+0.5
   5. (c) Sum of zeroes of $q(x) = -\dfrac{5}{6}$+0.5
   6. (c) $-\dfrac{5}{6}$ is equal in magnitude to $\dfrac{5}{6}$ but opposite in sign, so the sum for $q(x)$ is the negative of the sum for $p(x)$+0.5

   Answer: (a) $\dfrac{5}{6}$; (b) $-1$; (c) The sum of the zeroes of $q(x)$ is $-\dfrac{5}{6}$, which is equal in magnitude but opposite in sign to the sum of the zeroes of $p(x)$, which is $\dfrac{5}{6}$.
5. Q5Long answer·5 marks
   Find all the zeroes of the polynomial $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$, given that two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$. Verify the relationship between the zeroes and the coefficients of the polynomial.
   Show answer and step marks

   1. Since $\sqrt{2}$ and $-\sqrt{2}$ are zeroes, $(x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2$ is a factor of $p(x)$.+0.5
   2. Divide $2x^4 - 3x^3 - 3x^2 + 6x - 2$ by $x^2 - 2$: quotient $2x^2 - 3x + 1$, remainder $0$.+1.5
   3. Factorise the quotient: $2x^2 - 3x + 1 = (2x - 1)(x - 1)$.+0.5
   4. The other zeroes are $1$ and $\frac{1}{2}$, so the zeroes are $\sqrt{2}, -\sqrt{2}, 1, \frac{1}{2}$.+0.5
   5. Sum of zeroes: $\sqrt{2} - \sqrt{2} + 1 + \frac{1}{2} = \frac{3}{2} = -\frac{b}{a} = -\frac{-3}{2}$.+0.5
   6. Sum of products two at a time: $(\sqrt{2})(-\sqrt{2}) + (\sqrt{2} - \sqrt{2})(1 + \frac{1}{2}) + (1)(\frac{1}{2}) = -2 + 0 + \frac{1}{2} = -\frac{3}{2} = \frac{c}{a}$.+0.5
   7. Sum of products three at a time: $(\sqrt{2})(-\sqrt{2})(1 + \frac{1}{2}) + (1)(\frac{1}{2})(\sqrt{2} - \sqrt{2}) = -3 = -\frac{d}{a} = -\frac{6}{2}$.+0.5
   8. Product of zeroes: $(\sqrt{2})(-\sqrt{2})(1)(\frac{1}{2}) = -1 = \frac{e}{a} = \frac{-2}{2}$.+0.5

   Answer: Zeroes are $\sqrt{2}, -\sqrt{2}, 1, \frac{1}{2}$. Verified: sum $= \frac{3}{2} = -\frac{b}{a}$; sum of products two at a time $= -\frac{3}{2} = \frac{c}{a}$; three at a time $= -3 = -\frac{d}{a}$; product $= -1 = \frac{e}{a}$.
6. Q6Long answer·5 marks
   If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 2x^2 - 5x + 2$, find: (i) The values of $\alpha + \beta$ and $\alpha\beta$ using the relationship between zeroes and coefficients. (ii) The value of $\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta$. (iii) A quadratic polynomial whose zeroes are $2\alpha$ and $2\beta$.
   Show answer and step marks

   1. (i) $\alpha + \beta = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$.+1
   2. (i) $\alpha\beta = \frac{c}{a} = \frac{2}{2} = 1$.+1
   3. (ii) $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{1} = \frac{5}{2}$.+1
   4. (ii) $\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta = \frac{5}{2} - 2(1) = \frac{1}{2}$.+0.5
   5. (iii) For zeroes $2\alpha$ and $2\beta$: sum $= 2(\alpha + \beta) = 5$ and product $= 4\alpha\beta = 4$.+0.5
   6. (iii) The required polynomial is $k(x^2 - 5x + 4)$, where $k$ is a non-zero real constant (e.g. $x^2 - 5x + 4$).+1

   Answer: (i) $\alpha + \beta = \frac{5}{2}, \alpha\beta = 1$ (ii) $\frac{1}{2}$ (iii) $k(x^2 - 5x + 4)$

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