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Maths-

General

Easy

Question

The set of all values of $a$ for which the function
$f open parentheses x close parentheses equals open parentheses fraction numerator square root of a plus 4 end root over denominator 1 minus a end fraction minus 1 close parentheses x to the power of 5 end exponent minus 3 blank x plus log invisible function application 5$

Decreases for all real $x$ is

$(- \infty, \infty)$
$[- 4, \frac{3 - \sqrt{21}}{2}] \cup [1, \infty)$
$(- 3, 5 - \frac{\sqrt{27}}{2}) \cup (2, \infty)$
$[1, \infty)$

The correct answer is: $open square brackets negative 4 comma fraction numerator 3 minus square root of 21 over denominator 2 end fraction close square brackets union left square bracket 1 comma infinity right parenthesis$

For $f left parenthesis x right parenthesis$ to be decreasing for all $x comma$ we must have
$f to the power of ´ end exponent open parentheses x close parentheses less than 0$ for all $x$

$rightwards double arrow 5 open parentheses fraction numerator square root of a plus 4 end root over denominator 1 minus a end fraction minus 1 close parentheses x to the power of 4 end exponent minus 3 less than 0 blank f o r blank a l l blank x$

$rightwards double arrow open parentheses fraction numerator square root of a plus 4 end root over denominator 1 minus a end fraction minus 1 close parentheses x to the power of 4 end exponent less than fraction numerator 3 over denominator 5 end fraction f o r blank a l l blank x$

$rightwards double arrow open parentheses fraction numerator square root of a plus 4 end root over denominator 1 minus a end fraction minus 1 close parentheses less or equal than 0 rightwards double arrow fraction numerator square root of a plus 4 end root over denominator 1 minus a end fraction less or equal than 1$

This inequality is trivially true for all a satisfying $a greater than 1$ i.e. $a element of left parenthesis 1 comma infinity right parenthesis$

So let us take $a less than 1$

Since $square root of a plus 4 end root$ is real, therefore $a greater than negative 4$

Thus, we have $negative 4 less or equal than a less than 1$

For these values of $a comma$ we must have

$fraction numerator square root of a plus 4 end root over denominator 1 minus a end fraction less or equal than 1$

$rightwards double arrow square root of a plus 4 end root less or equal than 1 minus a$

$rightwards double arrow open parentheses a plus 4 close parentheses less or equal than 1 plus a to the power of 2 end exponent minus 2 blank a$

$rightwards double arrow 0 less or equal than a to the power of 2 end exponent minus 3 blank a minus 3$

$rightwards double arrow a greater or equal than fraction numerator 3 plus square root of 21 over denominator 2 end fraction o r comma blank a less or equal than fraction numerator 3 minus square root of 21 over denominator 2 end fraction$

$rightwards double arrow negative 4 less or equal than a less or equal than fraction numerator 3 minus square root of 21 over denominator 2 end fraction blank left square bracket because a greater or equal than negative 4 right square bracket$

$H e n c e comma blank a element of open square brackets negative 4 comma fraction numerator 3 minus square root of 21 over denominator 2 end fraction close square brackets union left parenthesis 1 comma infinity right parenthesis$

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