Let {\\mathbb{R}} be the set of all real numbers andf: {\\mathbb{R}} \\to{\\mathbb{R}} be a continuous function. Suppose |f(x)-f(y)| \\geq |x-y| for all real numbers x and {y}. Then

Correct Answer is : ƒ is one-one and onto

ƒ is one-one, but need not be onto

ƒ is onto, but need not be one-one

ƒ need not be either one-one or onto

KVPY 2020-21 SB SX Exam Previous Papers

Question : 14 of 120

Marks: +1, -0

Let

{\ℝ}

be the set of all real numbers and

f: {\ℝ} \→{\ℝ}

be a continuous function. Suppose

|f(x)-f(y)| \≥ |x-y|

for all real numbers

x

and

{y}

. Then

Solution:

Let

f({x})=f({y})

So,

|f(x)-f(y)| \≥|x-y|

\⇒0 \≥|x-y| \⇒x-y=0 \⇒x=y

\⇒f

is one-one Since,

f

is continuous So

f(0)

is finite Now,

|f(x)-f(0)| \≥|x-0|

\⇒\lim ↙{x \→\∞}|f(x)-f(0)| \≥ \lim ↙{x \→\∞}|x|

\⇒\lim ↙{x \→\∞}f(x)=\∞

\⇒f

is unbounded

\⇒f

is surjective

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