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NTSE Sample Paper NTSE SAT Practice Test-18

  • question_answer
    If\[x=t+\frac{1}{t}\]and\[y={{t}^{2}}+\frac{1}{{{t}^{2}}}\], then which one of the following is correct relation of x and

    A) \[{{x}^{2}}-y=1\]                   

    B) \[{{x}^{2}}+y=1\]

    C) \[{{x}^{2}}-y+1=0\]  

    D) \[{{x}^{2}}-y=2\]

    Correct Answer: D

    Solution :

    [d] \[{{t}^{2}}+\frac{1}{{{t}^{2}}}={{\left( t+\frac{1}{t} \right)}^{2}}-2\] \[\Rightarrow \,\,y={{x}^{2}}-2\,\,or\,{{x}^{2}}-y=2.\]


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