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JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    Let \[g:[-2,2]\to R\] where\[g(x)={{x}^{2015}}+\sgn (x)+\left[ \frac{{{x}^{2}}+1}{p} \right]\] be an odd function for all \[x\in [-2,2]\] then the smallest integral value of p is equal to [Note: [k] denote the greatest integer less than or equal to k.]

    A)  6                                

    B)  5

    C)  3                                

    D)  2

    Correct Answer: A

    Solution :

    We must have \[\left[ \frac{{{x}^{2}}+1}{P} \right]=0\,\,\forall \,\,x\in \,\,\{-2,\,\,3]\] \[\Rightarrow \,0\le \,\frac{{{x}^{2}}+1}{P}<1\] (Note that p > 0) \[\Rightarrow \,\frac{5}{p}<1\,\Rightarrow \,p>5\]


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