JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer 84) DIRECTION (Qs. 84): Each of these questions contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Choose the correct answer (ONLY ONE option is correct) from the following-
    Statement-1 : If \[{{x}^{2}}+x+1=0\] then the value of\[{{\left( x+\frac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}+...+\left( {{x}^{27}}+\frac{1}{{{x}^{27}}} \right)\,\,\text{is}\,\,54.\]
    Statement-2: \[\omega ,\,\,\,{{\omega }^{2}}\] are the roots of given equation and\[x+\frac{1}{x}=-1,\,\,{{x}^{2}}+\frac{1}{{{x}^{2}}}=-1,\,\,{{x}^{3}}+\frac{1}{{{x}^{3}}}=2\]

    A)  Statement-1 is false, Statement-2 is true.

    B)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

    C)  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

    D)  Statement-1 is true, Statement-2 is false.

    Correct Answer: B

    Solution :

    \[x+\frac{1}{x}=-1,\,\,{{x}^{2}}+\frac{1}{{{x}^{2}}}=-1\],                 \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=2,\,\,{{x}^{4}}+\frac{1}{{{x}^{4}}}=x+\frac{1}{x},\]                 \[{{x}^{5}}+\frac{1}{{{x}^{5}}}=-1,\,\,{{x}^{6}}+\frac{1}{{{x}^{6}}}=2,\,\,\,etc.\] \[\Rightarrow \]\[{{\left( x+\frac{1}{x} \right)}^{2}}+{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{3}}+{{\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)}^{2}}\]\[+{{\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)}^{2}}+{{\left( {{x}^{5}}+\frac{1}{{{x}^{5}}} \right)}^{2}}\] \[+{{\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)}^{2}}+{{\left( {{x}^{7}}+\frac{1}{{{x}^{7}}} \right)}^{2}}+....+{{\left( {{({{x}^{3}})}^{9}}+\frac{1}{{{({{x}^{3}})}^{9}}} \right)}^{2}}\]\[=(1+1+4)+(1+1+4)(1+1+4)+...9\]times

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