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.You need to login to perform this action.
You will be redirected in
3 sec