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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 22

  • question_answer
    Let \[\operatorname{f}: \left\{ x, y, z \right\}\,\to  \{1,2,3 \}\] be a one-one mapping such that only one of the following three statements is true and remaining two are false: \[f(x)\ne 2,\,\,f(y)=2,\,\,f(z)\ne 1\], then

    A) \[f\left( x \right)>f\left( y \right)>f\left( z \right)\]          

    B) \[f\left( x \right)<f\left( y \right)<f\left( z \right)\]

    C) \[f\left( y \right)<f\left( x \right)<f\left( z \right)\]          

    D) \[f\left( y \right)<f\left( z \right)<f\left( x \right)\]

    Correct Answer: C

    Solution :

    Let \[\operatorname{f}(x)\ne 2\] be true and\[\operatorname{f}(y) = 2\], \[\operatorname{f}(z)\ne 1\] are false \[\Rightarrow \,\,\,\,\,\operatorname{f}(x)\ne 2,\,\,f(y)\ne 2,\,\,f(z)=1\] \[\Rightarrow \,\,\,\,\,\operatorname{f}(x)=3,\,\,f(y)=3,\,\,f(z)=1\] but then function is many one. Now, let \[\operatorname{f}(z)\ne 1\] is true & \[\operatorname{f}(x)\ne 2\] & \[\operatorname{f}\left( y \right) = 2\] are false \[\Rightarrow \,\,\,\,\operatorname{f}(x)=\,\,2,\,\,f(y)\ne 2\,\,and\,\,f(z)\ne 1.\] \[\Rightarrow \,\,\,\operatorname{f}(x)=2,\,\,f(z)=3\,\,and\,\,f(y)=1.\] Hence one-one.


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