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Mathematics

Functions

Easy

Question

$text Let end text f left parenthesis x right parenthesis equals fraction numerator a x plus b over denominator c x plus d end fraction text then end text f left square bracket f left parenthesis x right parenthesis right square bracket equals x text provided that end text$

$d = - a$
$d = a$
$a = b = 1$
$a = b = c = d = 1$

The correct answer is: $d equals negative a$

$f o f left parenthesis x right parenthesis equals fraction numerator a open square brackets fraction numerator a x plus b over denominator c x plus d end fraction close square brackets plus b over denominator c open square brackets fraction numerator a x plus b over denominator c x plus d end fraction close square brackets plus d end fraction equals x therefore space of 1em left parenthesis a c plus d c right parenthesis x squared plus open parentheses b c plus d squared minus b c minus a squared close parentheses x minus a b minus b d equals 0 text It is true for all real end text x therefore space of 1em a c plus d c equals 0 comma a squared minus d squared equals 0 comma a b plus b d equals 0 s o space space a equals negative d$

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MathematicsFunctions

$text If end text f colon R not stretchy rightwards arrow A text defined by end text f left parenthesis x right parenthesis equals fraction numerator 1 over denominator x squared plus 2 x plus 2 end fraction straight for all x element of R text is surjective, then end text A equals$

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If x and {x} represents integral and fractional parts of x,then $left square bracket x right square bracket plus sum from r equals 1 to 2000 of fraction numerator left curly bracket x plus r right curly bracket over denominator 2000 end fraction equals$

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