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Maths-

General

Easy

Question

The function $f open parentheses x close parentheses equals cot to the power of negative 1 end exponent invisible function application x plus x$ increasing in the interval

$(1, \infty)$
$(- 1, \infty)$
$(- \infty, \infty)$
$(0, \infty)$

The correct answer is: $left parenthesis negative infinity comma blank infinity right parenthesis$

$f open parentheses x close parentheses equals cot to the power of negative 1 end exponent invisible function application x plus x$
$rightwards double arrow f to the power of ´ end exponent open parentheses x close parentheses equals negative fraction numerator 1 over denominator 1 plus x to the power of 2 end exponent end fraction plus 1$

$equals fraction numerator x to the power of 2 end exponent over denominator 1 plus x to the power of 2 end exponent end fraction comma blank c l e a r l y comma blank f ’ left parenthesis x right parenthesis greater than$ for all $x$ .

So, $f left parenthesis x right parenthesis$ increases in ( $negative infinity comma infinity$ ).

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