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Maths-

General

Easy

Question

The solution of the differential equation $left parenthesis 1 plus x to the power of 2 end exponent right parenthesis fraction numerator d y over denominator d x end fraction equals x$ is

$y = \tan^{- 1} x + c$
$y = - \tan^{- 1} x + c$
$y = \frac{1}{2} \log_{e} (1 + x^{2}) + c$
$y = - \frac{1}{2} \log_{e} (1 + x^{2}) + c$

The correct answer is: $y equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application left parenthesis 1 plus x to the power of 2 end exponent right parenthesis plus c$

$left parenthesis 1 plus x to the power of 2 end exponent right parenthesis fraction numerator d y over denominator d x end fraction equals x$ Þ $d y equals fraction numerator x over denominator 1 plus x to the power of 2 end exponent end fraction d x$
Þ $not stretchy integral d y equals not stretchy integral fraction numerator x over denominator 1 plus x to the power of 2 end exponent end fraction d x plus c$ Þ $y equals fraction numerator 1 over denominator 2 end fraction log subscript e end subscript invisible function application left parenthesis 1 plus x to the power of 2 end exponent right parenthesis plus c$ .

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Let f : R → R such that for all ‘x’ and y in R $vertical line f left parenthesis x right parenthesis minus f left parenthesis y right parenthesis vertical line less or equal than vertical line x minus y vertical line to the power of 3 end exponent text then end text f left parenthesis x right parenthesis$

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