Let fx- 1 if x is rational 0 if x is irrational-Then-

The correct option is A f is discontinuous for every real xBetween every two rational nbsp;numbers nbsp;there will be irrational numbers and viceversa. T ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

Let fx= 1 ...

Question

Let $f (x) = {\begin{matrix} 1 & if x i s r a t i o n a l 0 & if x i s i r r a t i o n a l \end{matrix}$ .Then,

f is discontinuous for every real x

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f is continuous on R

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f is continuous' at the points where x is rational

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f is continuous at the points where x is irrational.

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Solution

The correct option is A f is discontinuous for every real x
Between every two rational numbers there will be irrational numbers and viceversa. There will be no point of continuity.

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Similar questions

Q. Let

f : R \to R

be defined as

f (x) = {\begin{matrix} 0, & x is irrational sin | x |, & x is rational \end{matrix}

Then which of the following is true?

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Q. Assertion :Let

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 1 + x & x < 0 1 + [x] + sin x & 0 \leq x \leq π / 2 3 & x \geq π / 2 \end{matrix}

f is continuous on R

- {1}

Reason: The greatest integer function is discontinuous at every integer.

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Q. Let

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(correct answer + 1, wrong answer - 0.25)

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Let f(x)={1 if x is rational0 if x is irrational.Then,

The correct option is A f is discontinuous for every real xBetween every two rational numbers there will be irrational numbers and viceversa. There will be no point of continuity.

Let $f (x) = {\begin{matrix} 1 & if x i s r a t i o n a l 0 & if x i s i r r a t i o n a l \end{matrix}$ .Then,

The correct option is A f is discontinuous for every real x
Between every two rational numbers there will be irrational numbers and viceversa. There will be no point of continuity.