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Maths-

General

Easy

Question

The function $x square root of 1 minus x to the power of 2 end exponent end root comma left parenthesis x greater than 0 right parenthesis$ has

A local maxima
A local minima
Neither a local maxima nor a local minima
None of the above

The correct answer is: A local maxima

Let $f open parentheses x close parentheses equals x square root of 1 minus x to the power of 2 end exponent end root$
$rightwards double arrow blank f to the power of ´ end exponent open parentheses x close parentheses equals fraction numerator 1 minus 2 x to the power of 2 end exponent over denominator square root of 1 minus x to the power of 2 end exponent end root end fraction$

$rightwards double arrow blank$ For maxima or minima, put $fraction numerator d y over denominator d x end fraction equals 0$

$x equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction$

But $x greater than 0$ , therefore we have $x equals fraction numerator 1 over denominator square root of 2 blank end root end fraction$

Now, $f to the power of ´ ´ end exponent open parentheses x close parentheses equals fraction numerator square root of 1 minus x to the power of 2 end exponent end root open parentheses negative 4 x close parentheses minus open parentheses 1 minus 2 x to the power of 2 end exponent close parentheses fraction numerator negative x over denominator square root of 1 minus x to the power of 2 end exponent end root end fraction over denominator left parenthesis 1 minus x to the power of 2 end exponent right parenthesis end fraction$

$equals fraction numerator 2 x to the power of 3 end exponent minus 3 x over denominator open parentheses 1 minus x to the power of 2 end exponent close parentheses to the power of 3 divided by 2 end exponent end fraction$

$rightwards double arrow blank f ´ ´ open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses$ = $minus$ ve, maximum.

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