Let f ( x ) = [ x ] 2 + [ x + 1 ] − 3 , where [ x ] = the greatest integer ≤ x . Then

f ( x ) = [ x ] 2 + [ x + 1 ] − 3 = {[ x ] + 2 } {[ x ] − 1 } So, x = 1 , 1.1 , 1.2 , … f ( x ) = 0 ∴ f ( x ) is many one Only integral values will be attained ∴ f ( x ) is into

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Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsTypes of Function (Mapping)Medium2 minQB

MathematicsMediummultiple choice

Let $f (x) = [x]^{2} + [x + 1] - 3$ , where $[x] =$ the greatest integer $\leq x$ . Then

Options:(select one or more)

Answer:

A, B

Solution:

$f (x) = [x]^{2} + [x + 1] - 3 = {[x] + 2} {[x] - 1}$

So, $x = 1, 1.1, 1.2, \dots f (x) = 0 ∴ f (x)$ is many one Only integral values will be attained $∴ f (x) is into$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Types of Function (Mapping)

⏱ 2mℹ️ Source: QB

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