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Chapter 2: Polynomials

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This page provides comprehensive Chapter 2: Polynomials - Assertion Reason Worksheet - SJMaths. Assertion and Reason type questions for Class 10 Polynomials. Practice for CBSE Board Exams.

Assertion-Reason Worksheet

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Directions:

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  • (C) Assertion (A) is true but Reason (R) is false.
  • (D) Assertion (A) is false but Reason (R) is true.
  1. Question 1:
    Assertion (A): The polynomial $x^2 + 4x + 5$ has no real zeroes.
    Reason (R): A quadratic polynomial $ax^2 + bx + c$ has no real zeroes if its discriminant $D = b^2 - 4ac < 0$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Calculate $D$ for $x^2 + 4x + 5$. Here $a=1, b=4, c=5$.
    $D = (4)^2 - 4(1)(5) = 16 - 20 = -4$.
    Step 2: Since $D < 0$, the polynomial has no real zeroes. Assertion is true.
    Step 3: The Reason correctly states the condition for no real zeroes. Thus, R explains A.
  2. Question 2:
    Assertion (A): If the sum of zeroes of the quadratic polynomial $x^2 - 2kx + 8$ is 2, then the value of $k$ is 1.
    Reason (R): Sum of zeroes of a quadratic polynomial $ax^2 + bx + c$ is given by $-b/a$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Using Reason, Sum $= -(-2k)/1 = 2k$.
    Step 2: Given Sum $= 2$. So, $2k = 2 \Rightarrow k = 1$.
    Step 3: Both A and R are true, and R is the correct explanation.
  3. Question 3:
    Assertion (A): The degree of the polynomial $x^3 - 2x^2 + \sqrt{x} + 1$ is 3.
    Reason (R): The highest power of the variable in a polynomial is called its degree.
    (A)
    (B)
    (C)
    (D)
    Solution: (D)
    Step 1: Check if the expression is a polynomial. The term $\sqrt{x} = x^{1/2}$ has a fractional exponent.
    Step 2: For a polynomial, exponents must be non-negative integers. Thus, A is false (it is not a polynomial).
    Step 3: Reason is a true definition.
  4. Question 4:
    Assertion (A): The zeroes of the polynomial $x^2 - 2$ are $\sqrt{2}$ and $-\sqrt{2}$.
    Reason (R): $x^2 - 2$ can be factored as $(x - \sqrt{2})(x + \sqrt{2})$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Factorize $x^2 - 2$ using $a^2 - b^2 = (a-b)(a+b)$.
    $x^2 - (\sqrt{2})^2 = (x-\sqrt{2})(x+\sqrt{2})$.
    Step 2: Zeroes are found by setting factors to 0: $x = \sqrt{2}, -\sqrt{2}$.
    Step 3: Both are true and R explains A.
  5. Question 5:
    Assertion (A): A quadratic polynomial can have at most 2 real zeroes.
    Reason (R): A polynomial of degree $n$ has at most $n$ real zeroes.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: A quadratic polynomial has degree 2.
    Step 2: According to the fundamental theorem (Reason), it can have at most 2 real zeroes.
    Step 3: R is the correct explanation for A.
  6. Question 6:
    Assertion (A): The graph of $y = x^2 + 1$ does not intersect the x-axis.
    Reason (R): The graph of a quadratic polynomial is a parabola.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: For $x^2 + 1 = 0$, $x^2 = -1$, which has no real solutions. So it does not intersect the x-axis. A is true.
    Step 2: R is true (shape is parabola).
    Step 3: However, being a parabola doesn't guarantee it won't intersect (some parabolas do). The reason is $D < 0$. So, B is correct.
  7. Question 7:
    Assertion (A): If the product of zeroes of the polynomial $kx^2 - 6x - 6$ is 4, then $k = -1.5$.
    Reason (R): Product of zeroes of $ax^2 + bx + c$ is given by $c/a$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Using Reason, Product $= c/a = -6/k$.
    Step 2: Given Product $= 4$. So, $-6/k = 4 \Rightarrow k = -6/4 = -1.5$.
    Step 3: Both are true and R explains A.
  8. Question 8:
    Assertion (A): $x = -1$ is a zero of the polynomial $p(x) = x^2 - 3x - 4$.
    Reason (R): A real number $k$ is a zero of polynomial $p(x)$ if $p(k) = 0$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Calculate $p(-1) = (-1)^2 - 3(-1) - 4 = 1 + 3 - 4 = 0$.
    Step 2: Since $p(-1) = 0$, $x=-1$ is a zero.
    Step 3: R provides the definition used in Step 2.
  9. Question 9:
    Assertion (A): The polynomial $x^2 + x + 1$ has two distinct real roots.
    Reason (R): A quadratic polynomial has distinct real roots if its discriminant $D > 0$.
    (A)
    (B)
    (C)
    (D)
    Solution: (D)
    Step 1: Calculate $D$ for $x^2 + x + 1$. $D = 1^2 - 4(1)(1) = 1 - 4 = -3$.
    Step 2: Since $D < 0$, it has no real roots. Assertion is false.
    Step 3: Reason is a standard true statement.
  10. Question 10:
    Assertion (A): $2x + 3$ is a linear polynomial.
    Reason (R): A polynomial of degree 1 is called a linear polynomial.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: The degree of $2x + 3$ is 1 (power of $x$).
    Step 2: By definition (Reason), degree 1 implies linear polynomial.
    Step 3: Both are true and R explains A.
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