CSIR NET MATHEMATICS | Real Analysis | Riemann Integration DEC 2011 (हिंदी में)

CSIR NET MATHEMATICS | Real Analysis | Riemann Integration DEC 2011 (हिंदी में)
Real Analysis, Riemann Integration,
Example (marks 3)

Define f :[0,1]  [0,1] by f(x)=(2^k-1)/2^k for x∈[(2^(k-1)-1)/2^(k-1) ,(2^k-1)/2^k ],k≥1.

Then f is a Riemann-integral function such that

CSIR NET Answer Key December 2011,

CSIR NET Answer Key June 2019, Mathematical Science,

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