Assertion (A): f : R → R, f(x) = sin x, is not one-one. Reason (R): sin 0 = sin π.

Relations & Functions · Question 95 of 125

Question 95

HardOriginal question

Assertion (A): $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = \sin x$ , is not one-one.
Reason (R): $\sin 0 = \sin \pi$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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CorrectNot quiteAnswer A

Correct answer: A

A is true: $\sin x$ repeats its values.
R is true and is a pair of different inputs with the same output, which is exactly what makes A true.

Concept: A counter-example to one-one
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsOne-one functions

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Assertion (A): f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R, f(x)=sin⁡xf(x) = \sin xf(x)=sinx, is not one-one.Reason (R): sin⁡0=sin⁡π\sin 0 = \sin \pisin0=sinπ.

Explanation

Assertion (A): $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = \sin x$ , is not one-one.
Reason (R): $\sin 0 = \sin \pi$ .