Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 15 Question 15 of 401 Hard MCQ
Let [ ⋅ ] [\cdot ] [ ⋅ ] denote the greatest integer function, and let f ( x ) = m i n 2 x , x 2 f(x)=min{\sqrt{2}x,{x}^{2}} f ( x ) = min 2 x , x 2 .
Let S={x\in (-2,2) : the function g ( x ) = ∣ x ∣ [ x 2 ] g(x)=|x|[{x}^{2}] g ( x ) = ∣ x ∣ [ x 2 ] is discontinuous at x} .
Then ∑ x ∈ S f ( x ) \sum _{x\in S}f(x) ∑ x ∈ S f ( x ) equals
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g ( x ) = ∣ x ∣ [ x 2 ] g ( x ) ( ± 1 , ± 2 , ± 3 ) ∴ S = − 1 , 1 , − 2 , 2 ⋅ − 3 , 3 ∵ f ( x ) = m i n 2 x , x 2 ∴ ∑ x ∈ S f ( x ) = − 2 + 1 − 2 + 2 − 6 + 6 = 1 − 2 \begin{gathered} g(x)=|x|[{x}^{2}] \\ g(x)(\pm 1,\pm \sqrt{2},\pm \sqrt{3}) \\ \therefore S={-1,1,-\sqrt{2},\sqrt{2}\cdot -\sqrt{3},\sqrt{3}} \\ \because f(x)=min{\sqrt{2x},{x}^{2}} \\ \therefore \sum _{x\in S}f(x)=-\sqrt{2}+1-2+2-\sqrt{6}+\sqrt{6}=1-\sqrt{2} \end{gathered} g ( x ) = ∣ x ∣ [ x 2 ] g ( x ) ( ± 1 , ± 2 , ± 3 ) ∴ S = − 1 , 1 , − 2 , 2 ⋅ − 3 , 3 ∵ f ( x ) = min 2 x , x 2 ∴ x ∈ S ∑ f ( x ) = − 2 + 1 − 2 + 2 − 6 + 6 = 1 − 2 Answer the question to unlock