10th Class Mathematics Polynomials Question Bank Case Based (MCQs) - Polynomials

  • question_answer
    Case Study : Q. 26 to 30
    Ramesh was asked by one of his friends to find the polynomial whose zeroes are \[\frac{-2}{\sqrt{3}}\] and \[\frac{\sqrt{3}}{4}\]. He obtained the polynomial as shown below:
    Let \[\alpha =\frac{-2}{\sqrt{3}}\] and \[\alpha =\frac{\sqrt{3}}{4}\]
    \[\Rightarrow \,\,\,\,\,\,\,\alpha +\beta =\frac{-2}{\sqrt{3}}+\frac{\sqrt{3}}{4}=\frac{-8+1}{4\sqrt{3}}=\frac{-7}{4\sqrt{3}}\]
    \[\Rightarrow \,\,\,\,\,\,\,\alpha \beta =\frac{-2}{\sqrt{3}}\times \frac{\sqrt{3}}{4}=\frac{-1}{2}\]
    Required polynomial \[={{x}^{2}}-(\alpha +\beta )x+\alpha \beta \]
    \[={{x}^{2}}-\left( \frac{-7}{4\sqrt{3}} \right)x+\left( \frac{-1}{2} \right)\]
    His another friend Kavita pointed out that the polynomial obtained is not correct.
    Based on the above situation, answer the following questions:
    Is the claim of Kavita correct?

    A) Yes

    B) No

    C) Can't say

    D) None of these

    Correct Answer: A

    Solution :

    Given, \[\alpha =\frac{-2}{\sqrt{3}}\]and \[\beta =\frac{\sqrt{3}}{4}\]
    \[\alpha +\beta =\frac{-2}{\sqrt{3}}+\frac{\sqrt{3}}{4}=\frac{-8+3}{4\sqrt{3}}=\frac{-5}{4\sqrt{3}}\]
    \[\alpha \beta =\frac{-2}{\sqrt{3}}\times \frac{\sqrt{3}}{4}=-\frac{1}{2}\]
    Yes, because value of \[(\alpha +\beta )\] calculated by Anirudh is incorrect.
    So, option [a] is correct.

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