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Maths-

General

Easy

Question

$I f space a equals fraction numerator left parenthesis 25 minus x squared right parenthesis space over denominator 16 end fraction$ and log $fraction numerator open parentheses 24 minus 2 x minus x squared close parentheses over denominator 14 end fraction greater than 1$ then $x element of$

(0,1)
$(- 3, 1) \cup (3, 4)$
$(- \infty, - 17) \cup (1, \infty)$
(4,5)

The correct answer is: $left parenthesis negative 3 comma 1 right parenthesis space union space left parenthesis 3 comma 4 right parenthesis$

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 24 minus 2 x minus x squared greater than 0 comma a greater than 0 comma a not equal to 1 end cell row cell not stretchy rightwards double arrow x squared plus 2 x minus 24 less than 0 comma x squared minus 25 less than 0 comma x not equal to plus-or-minus 3 end cell row cell not stretchy rightwards double arrow x element of left parenthesis negative 5 comma negative 3 right parenthesis union left parenthesis negative 3 comma 3 right parenthesis union left parenthesis 3 comma 4 right parenthesis end cell row cell text case(i) : end text x element of left parenthesis negative 5 comma negative 3 right parenthesis union left parenthesis 3 comma 4 right parenthesis not stretchy rightwards double arrow fraction numerator 25 minus x squared over denominator 16 end fraction less than 1 not stretchy rightwards double arrow 0 end cell row cell space of 1em less than a less than 1 end cell row cell text so given equation becomes end text fraction numerator 24 minus 2 x minus x squared over denominator 14 end fraction less than a not stretchy rightwards double arrow end cell row cell table attributes columnspacing 1em end attributes row cell 24 minus 2 x minus x squared end cell row cell fraction numerator 14 over denominator not stretchy rightwards double arrow x squared plus 16 x minus 17 end fraction plus fraction numerator 25 minus x squared over denominator 16 end fraction end cell row cell not stretchy rightwards double arrow x element of x element of left parenthesis negative straight infinity comma negative 17 right parenthesis union left parenthesis 1 comma straight infinity right parenthesis end cell end table end cell end table$

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$text If end text f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign center center columnspacing 1em end attributes row cell open parentheses e to the power of x minus 1 close parentheses vertical line x sin invisible function application x vertical line end cell cell x greater than 0 end cell row 0 cell x to the power of n comma end cell end table space of 1em x equals 0 close$ is derivable at $x equals 0$ then interval of possible values of ‘n’ is

$text If end text f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign center center columnspacing 1em end attributes row cell open parentheses e to the power of x minus 1 close parentheses vertical line x sin invisible function application x vertical line end cell cell x greater than 0 end cell row 0 cell x to the power of n comma end cell end table space of 1em x equals 0 close$ is derivable at $x equals 0$ then interval of possible values of ‘n’ is

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