JEE Main & Advanced Sample Paper JEE Main Sample Paper-14

  • question_answer
    Let, S be a relation on \[{{R}^{+}}\] defined as \[xSy\Leftrightarrow {{x}^{2}}-{{y}^{2}}=2(y-x)\]. Then, S is

    A)  reflexive on \[{{R}^{+}}\]                          

    B)  symmetric on \[{{R}^{+}}\]

    C)  antisymmetric on \[{{R}^{+}}\]                     

    D)  equivalence relation on \[{{R}^{+}}\]

    Correct Answer: D

    Solution :

     Given \[x\,Sy\,\Leftrightarrow \,{{x}^{2}}-{{y}^{2}}=2(y-x),\]  \[\forall \,x,\,\,y\in \,{{R}^{+}}\] \[\because \] \[{{x}^{2}}-{{y}^{2}}=2(x-x),\,\]     \[\forall \,x,\,\,y\,\in {{R}^{+}}\] 0 = 0 \[\therefore \] R is reflexive. For symmetric of \[S,\,\,xSy\] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}=2(y-x)\] \[\Rightarrow \] \[-({{y}^{2}}-{{x}^{2}})=-2\,(x-y)\] \[\Rightarrow \]  \[{{y}^{2}}-{{x}^{2}}=2\,(x-y)\] \[\Rightarrow \] \[y\,Sx\] \[\therefore \] S is symmetric. Now, by definition of S, \[{{x}^{2}}-{{y}^{2}}=2(y-x)\] and        \[{{y}^{2}}-{{z}^{2}}=2(z-y)\] On adding, we get \[({{x}^{2}}-{{z}^{2}})=2(z-x)\] \[\Rightarrow \]               \[xSz\] \[\therefore \] S is transitive. Clearly, S is an equivalence relation on\[{{R}^{+}}\].

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