Assertion (A): f : R → R, f(x) = x/(1 + | x |), is not onto. Reason (R): | f(x) | < 1 for every real x.

Question 124

HardOriginal question

Assertion (A): $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = \frac{x}{1 + \lvert x \rvert}$ , is not onto.
Reason (R): $\lvert f(x) \rvert < 1$ for every real $x$ .

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

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\lvert f(x) \rvert = \frac{\lvert x \rvert}{1 + \lvert x \rvert} < 1$ , since the denominator is larger than the numerator. R is true.
So 1, for one, is never reached, and $f$ is not onto $\mathbb{R}$ : A is true, and R is exactly why.

Worth noting

$f$ is one-one, but that does not make it onto: its values stay inside $(-1, 1)$ .

Concept: Values a function never reaches
Typical timeTypical: 120 s
Your timeYours: —

The rule this testsOnto and bijective functions

Practise this chapter →

Ten questions from Relations & Functions, at your own pace.

Assertion (A): f:R→Rf : \mathbb{R} \to \mathbb{R}f:R→R, f(x)=x1+∣x∣f(x) = \frac{x}{1 + \lvert x \rvert}f(x)=1+∣x∣x​, is not onto.Reason (R): ∣f(x)∣<1\lvert f(x) \rvert < 1∣f(x)∣<1 for every real xxx.

Explanation

Assertion (A): $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = \frac{x}{1 + \lvert x \rvert}$ , is not onto.
Reason (R): $\lvert f(x) \rvert < 1$ for every real $x$ .