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Maths-

General

Easy

Question

Let $f to the power of ´ end exponent$ be an injective function with domain $left curly bracket x comma y comma z right curly bracket$ and range $left curly bracket 12 comma 3 right curly bracket$ such that exactly one of the following statements is correct and the remaining are flase $f left parenthesis x right parenthesis equals 1$ , $f left parenthesis y right parenthesis not equal to$ $1 comma f left parenthesis z right parenthesis not equal to 2$ the value of $f to the power of negative 1 end exponent$ (1) is

x
y
z
$N o n e$

The correct answer is: y

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The domain of the function $f left parenthesis x right parenthesis equals root index 3 of fraction numerator 2 x minus 1 over denominator x to the power of 2 end exponent minus 10 x minus 11 end fraction end root$

The domain of the function $f left parenthesis x right parenthesis equals root index 3 of fraction numerator 2 x minus 1 over denominator x to the power of 2 end exponent minus 10 x minus 11 end fraction end root$

If $f left parenthesis x right parenthesis equals l o g invisible function application open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses$ then $f open parentheses x subscript 1 end subscript close parentheses plus f open parentheses x subscript 2 end subscript close parentheses equals$

If $f left parenthesis x right parenthesis equals l o g invisible function application open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses$ then $f open parentheses x subscript 1 end subscript close parentheses plus f open parentheses x subscript 2 end subscript close parentheses equals$

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$A r e a blank o f blank t h e blank r e g i o n blank b o u n d e d blank b y blank y equals open vertical bar x close vertical bar blank a n d blank y equals 2 blank i s$

$A r e a blank o f blank t h e blank r e g i o n blank b o u n d e d blank b y blank y equals open vertical bar x close vertical bar blank a n d blank y equals 2 blank i s$

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$T h e blank a r e a blank b o u n d e d blank b y blank y to the power of 2 end exponent equals 4 a x blank a n d blank y equals m x blank i s fraction numerator a to the power of 2 end exponent over denominator 3 end fraction s q. blank u n i t s blank t h e n blank m equals blank$

$T h e blank a r e a blank b o u n d e d blank b y blank y to the power of 2 end exponent equals 4 a x blank a n d blank y equals m x blank i s fraction numerator a to the power of 2 end exponent over denominator 3 end fraction s q. blank u n i t s blank t h e n blank m equals blank$

$I f blank z to the power of 2 end exponent plus z plus 1 equals 0 comma$ where z is a complex number, then the value of $open parentheses z plus fraction numerator 1 over denominator z end fraction close parentheses to the power of 2 end exponent plus open parentheses z to the power of 2 end exponent plus fraction numerator 1 over denominator z to the power of 2 end exponent end fraction close parentheses to the power of 2 end exponent plus open parentheses z to the power of 1 end exponent plus fraction numerator 1 over denominator z to the power of 3 end exponent end fraction close parentheses to the power of 2 end exponent plus midline horizontal ellipsis plus open parentheses z to the power of 6 end exponent plus fraction numerator 1 over denominator z to the power of 6 end exponent end fraction close parentheses to the power of 2 end exponent text is end text$

$I f blank z to the power of 2 end exponent plus z plus 1 equals 0 comma$ where z is a complex number, then the value of $open parentheses z plus fraction numerator 1 over denominator z end fraction close parentheses to the power of 2 end exponent plus open parentheses z to the power of 2 end exponent plus fraction numerator 1 over denominator z to the power of 2 end exponent end fraction close parentheses to the power of 2 end exponent plus open parentheses z to the power of 1 end exponent plus fraction numerator 1 over denominator z to the power of 3 end exponent end fraction close parentheses to the power of 2 end exponent plus midline horizontal ellipsis plus open parentheses z to the power of 6 end exponent plus fraction numerator 1 over denominator z to the power of 6 end exponent end fraction close parentheses to the power of 2 end exponent text is end text$

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$T h e blank a r e a blank o f blank t h e blank g r e a t e r blank r e g i o n blank b o u n d e d blank b y blank y equals C o s blank x semicolon blank y equals x plus 1 blank a n d blank y equals 0 blank i s$

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