ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 12 - 5. continuity and differentiation

ધારોકે \([x]\) એ મહત્તમ પૂર્ણાંક \(\leq x\) છે. તો વિધેય \(f(x)=|[x]|+\sqrt{x-[x]}\) અંતરાલ \((-2,1)\) માં જ્યાં અસતત હોય તેવા બિંદુુઓની સંખ્યા \(......\) છે.

  1. A \(4\)
  2. B \(6\)
  3. C \(8\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2\)

Step-by-step Solution

Detailed explanation

Need to check at doubtful points discont at \(x \in I\) only at \(x=-1 \Rightarrow f\left(-1^{+}\right)=1+0=1\) \(\Rightarrow f\left(-1^{-}\right)=2+1=3\) at \(x=0 \Rightarrow f\left(0^{+}\right)=0+0=0\) \(\Rightarrow f \left(0^{-}\right)=1+1=2\) at…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app