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Maths-

General

Easy

Question

Let $f open parentheses x close parentheses equals fraction numerator 1 over denominator 2 end fraction minus tan invisible function application open parentheses fraction numerator pi x over denominator 2 end fraction close parentheses$ , $negative 1 less than x less than 1$ and $g open parentheses x close parentheses equals square root of 3 plus 4 x minus 4 x to the power of 2 end exponent end root$ , then dom $left parenthesis f plus g right parenthesis$ is given by

$[\frac{1}{2}, 1]$
$(\frac{1}{2}, - 1)$
$(- \frac{1}{2}, 1)$
$[- \frac{1}{2}, - 1]$

The correct answer is: $open parentheses negative 1 half comma 1 close parentheses$

Given, $f open parentheses x close parentheses equals fraction numerator 1 over denominator 2 end fraction minus tan invisible function application open parentheses fraction numerator pi x over denominator 2 end fraction close parentheses comma blank minus 1 less than x less than 1$
Given, domain of $f left parenthesis x right parenthesis$ is $d subscript 1 end subscript equals left parenthesis negative 1 comma blank 1 right parenthesis$
For domain of $g open parentheses x close parentheses comma blank 3 plus 4 x minus 4 x to the power of 2 end exponent greater or equal than 0$
$rightwards double arrow blank open parentheses 2 x minus 3 close parentheses left parenthesis 2 x plus 1 right parenthesis less or equal than 0$
$therefore$ Domain of $g left parenthesis x right parenthesis$ is $d subscript 2 end subscript equals open square brackets negative fraction numerator 1 over denominator 2 end fraction comma blank fraction numerator 3 over denominator 2 end fraction close square brackets$
Hence, domain of $open parentheses f plus g close parentheses equals d subscript 1 end subscript intersection d subscript 2 end subscript equals open square brackets negative fraction numerator 1 over denominator 2 end fraction comma blank 1 close square brackets$

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