dse 2010

[hs]

[Amit Goyal]

Note that y >= y* implies { x belongs to [0,1] | f(x) >= y} is a subset of { x belongs to [0,1] | f(x) >= y*}
Thus, min { x belongs to [0,1] | f(x) >= y} >= min { x belongs to [0,1] |f(x) >= y*}
Hence, g(y) >= g(y*).
So, g is non decreasing.

Consider f(x) = min{2x, 1}
For this f,
g(y) = y/2
Clearly, g is continuous but f is not strictly increasing.

Thu, the correct answer is (a)

[mrittik]

If f(x)>=y then x>=f^(-1) (y)....now g(y)=min(x)= f^(-1) (y).....it is clearly understandable that g(y) is non-decreasing....@Amit Sir, what do you think?

[X]

Thank you, so much Sir!

[Nupur]

Shouldn't the answer be (c) both are transitive??

[mrittik]

Nope....'>' is only Transitive bcoz it reveals only transitivity property...'~' is not transitive...bcoz it doesn't reveal the Transitivity....Only in Impossibility Theorem this Transitivity rule breaks