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

f ( x ) = [ x ] 2 + [ x ] + 1 − 3 = {[ x ] + 2 } {[ x ] − 1 } So, x = 1 , 1.1 , 1.2 , … ⇒ f ( x ) = 0 Only integral values will be attained.

Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityContinuity- MiscellaneousEasy2 minQB

MathematicsEasymultiple choice

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

Answer:

A, B

Solution:

$f (x) = [x]^{2} + [x] + 1 - 3 = {[x] + 2} {[x] - 1} So, x = 1, 1.1, 1.2, \dots \Rightarrow f (x) = 0$

Only integral values will be attained.

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Continuity- Miscellaneous

⏱ 2mℹ️ Source: QB

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