If f x - begin cases x- - if - when x - is - rational 0- - text if - when x - is irrationalend cases - g x - begin cases 0- - text if - when x - is rational x - - if - when x - is irrational end cases - thenf-g isA- One-one ontoB- One-one not ontoC- Not one-one but ontoD- Not one-one not onto

The correct option is A One one onto (f g)(x)= x, amp; x∈ Q x, amp; x∉ Q; ...

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Byju's Answer

Standard XII

Mathematics

Properties of Set Operation

If f x = begi...

Question

If $f (x) = {\begin{matrix} x, & if w h e n x i s r a t i o n a l 0, & if w h e n x i s i r r a t i o n a l \end{matrix}$ ;

$g (x) = {\begin{matrix} 0, & if w h e n x i s r a t i o n a l x, & if w h e n x i s i r r a t i o n a l \end{matrix}$ ;

then $(f - g$ ) is

One-one onto

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One-one not onto

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Not one-one but onto

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Not one-one not onto

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Solution

The correct option is A
One-one onto

$(f - g) (x) = {\begin{matrix} x, & x \in Q - x, & x \notin Q \end{matrix}$ ;

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If $f (x) =$ ${\begin{matrix} x, & when x is rational 0, & when x is irrational \end{matrix}$ ; $g (x) =$ ${\begin{matrix} 0, & when x is rational x, & when x is irrational \end{matrix}$ then $(f - g)$ is

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,

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and

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Q. Let

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f_{3} (x) = {\begin{matrix} sin x if x < 0, x if x \geq 0; \end{matrix}

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f_{4} (x) = {\begin{matrix} f_{2} (f_{1} (x)) if x < 0, f_{2} (f_{1} (x)) - 1 if x \geq 0. \end{matrix}

\begin{matrix} L i s t I & L i s t I I P. f_{4} is & 1. onto but not one-one Q. f_{3} is & 2. neither continuous nor one-one R. f_{2} \circ f_{1} is & 3. differentiable but not one-one S. f_{2} is & 4. continuous and not one-one \end{matrix}

which of the following option is correct ?

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If f(x)={x, if whenx is rational0, if whenx is irrational; g(x)={0, if whenx is rationalx, if whenx is irrational; then(f−g) is

The correct option is A One-one onto (f−g)(x)={x,x∈Q−x,x∉Q;

If $f (x) = {\begin{matrix} x, & if w h e n x i s r a t i o n a l 0, & if w h e n x i s i r r a t i o n a l \end{matrix}$ ;

$g (x) = {\begin{matrix} 0, & if w h e n x i s r a t i o n a l x, & if w h e n x i s i r r a t i o n a l \end{matrix}$ ;

then $(f - g$ ) is

The correct option is A
One-one onto

$(f - g) (x) = {\begin{matrix} x, & x \in Q - x, & x \notin Q \end{matrix}$ ;