JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 24)
Let N be the set of natural numbers and two functions f and g be defined as f, g : N $$ \to $$ N such that
f(n) = $$\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$$;
and g(n) = n $$-$$($$-$$ 1)n.
Then fog is -
f(n) = $$\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$$;
and g(n) = n $$-$$($$-$$ 1)n.
Then fog is -
neither one-one nor onto
onto but not one-one
both one-one and onto
one-one but not onto
Explanation
f(x) = $$\left\{ {\matrix{
{{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr
{{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr
} \,\,} \right.$$;
g(x) = n $$-$$ ($$-$$ 1)n $$\left\{ {\matrix{ {n + 1;\,\,\,\,n\,\,is\,\,odd} \cr {n - 1;\,\,\,\,n\,\,is\,\,even} \cr } } \right.$$
f(g(n)) = $$\left\{ {\matrix{ {{n \over 2};\,\,\,\,n\,\,is\,\,even} \cr {{{n + 1} \over 2};\,\,\,\,n\,\,is\,\,odd} \cr } } \right.$$
$$ \therefore $$ many one but onto
g(x) = n $$-$$ ($$-$$ 1)n $$\left\{ {\matrix{ {n + 1;\,\,\,\,n\,\,is\,\,odd} \cr {n - 1;\,\,\,\,n\,\,is\,\,even} \cr } } \right.$$
f(g(n)) = $$\left\{ {\matrix{ {{n \over 2};\,\,\,\,n\,\,is\,\,even} \cr {{{n + 1} \over 2};\,\,\,\,n\,\,is\,\,odd} \cr } } \right.$$
$$ \therefore $$ many one but onto
Comments (0)
