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Question

The function $f open parentheses x close parentheses equals a sin invisible function application x plus fraction numerator 1 over denominator 3 end fraction sin invisible function application 3 x$ has maximum value at $x equals fraction numerator pi over denominator 3 end fraction$ . The value of $a$ is

3
$\frac{1}{3}$
2
$\frac{1}{2}$

The correct answer is: 2

Given, $f open parentheses x close parentheses equals a sin invisible function application x plus fraction numerator 1 over denominator 3 end fraction sin invisible function application 3 x$ has minimum value at $x equals fraction numerator pi over denominator 3 end fraction$ , then $f to the power of ´ end exponent open parentheses x close parentheses equals 0$ at $x equals fraction numerator pi over denominator 3 end fraction$
$rightwards double arrow blank f to the power of ´ end exponent open parentheses x close parentheses equals a cos invisible function application x plus cos invisible function application 3 x$

$rightwards double arrow f to the power of ´ end exponent open parentheses fraction numerator pi over denominator 3 end fraction close parentheses equals a cos invisible function application fraction numerator pi over denominator 3 end fraction plus cos invisible function application open parentheses 3 cross times fraction numerator pi over denominator 3 end fraction close parentheses equals 0$

$rightwards double arrow blank a cross times fraction numerator 1 over denominator 2 end fraction plus cos invisible function application left parenthesis pi right parenthesis equals 0$

$rightwards double arrow blank fraction numerator a over denominator 2 end fraction minus 1 equals 0 blank rightwards double arrow blank a equals 2$

Related Questions to study

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The number of values of $x$ where $f open parentheses x close parentheses equals cos invisible function application x plus cos invisible function application square root of 2 x$ attains its maximum is

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parallel

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