NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2 covers different cases of the relationship between Zeroes and Coefficients of a Polynomial. The exercise has 2 questions with 6 cases each.

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Class 10 Chapter 2 Polynomials Topics:

CBSE Class 10 Maths Study Guides:

CBSE X Related Questions

  • 1.
    The dimensions of a window are 156 cm \(\times\) 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.


      • 2.
        Assertion (A) : \((\sqrt{3} + \sqrt{5})\) is an irrational number.
        Reason (R) : Sum of the any two irrational numbers is always irrational.

          • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
          • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
          • Assertion (A) is true, but Reason (R) is false.
          • Assertion (A) is false, but Reason (R) is true.

        • 3.
          Solve the linear equations \(3x + y = 14\) and \(y = 2\) graphically.


            • 4.
              A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

                • \(314 \sqrt{2}\) \(\text{cm}^{2}\)
                • \(314\) \(\text{cm}^{2}\)
                • \(\frac{3140}{3}\) \(\text{cm}^{2}\)
                • \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

              • 5.
                Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)


                  • 6.
                    An ice-cream cone of radius \(r\) and height \(h\) is completely filled by two spherical scoops of ice-cream. If radius of each spherical scoop is \(\frac{r}{2}\), then \(h : 2r\) equals

                      • \(1 : 8\)
                      • \(1 : 2\)
                      • \(1 : 1\)
                      • \(2 : 1\)

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