Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

The domain of the function $f left parenthesis x right parenthesis equals exp invisible function application open parentheses square root of 5 x minus 3 minus 2 x squared end root close parentheses$ is

$[3 / 2, \infty)$
$[1, 3 / 2]$
$(- \infty, A)$
$(1, 3 / 2)$

The correct answer is: $left square bracket 1 comma 3 divided by 2 right square bracket$

Given, $f left parenthesis x right parenthesis equals e to the power of square root of 5 x minus 3 minus 2 x squared end root end exponent$
For domain of $f left parenthesis x right parenthesis$
$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 2 x squared minus 5 x plus 3 less or equal than 0 end cell row cell not stretchy rightwards double arrow left parenthesis 2 x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis less or equal than 0 end cell row cell not stretchy rightwards double arrow 1 less or equal than x less or equal than 3 over 2 end cell end table$
$therefore text Domain of end text f left parenthesis x right parenthesis equals open square brackets 1 comma 3 over 2 close square brackets.$

Related Questions to study

The value of $tan invisible function application open curly brackets cos to the power of negative 1 end exponent invisible function application open parentheses negative 2 over 7 close parentheses minus pi over 2 close curly brackets$ is

The value of $tan invisible function application open curly brackets cos to the power of negative 1 end exponent invisible function application open parentheses negative 2 over 7 close parentheses minus pi over 2 close curly brackets$ is

$sin invisible function application open parentheses 1 half cos to the power of negative 1 end exponent invisible function application 4 over 5 close parentheses$ is equal to

$sin invisible function application open parentheses 1 half cos to the power of negative 1 end exponent invisible function application 4 over 5 close parentheses$ is equal to

If $a greater than b greater than 0$ , then the value of $tan to the power of negative 1 end exponent invisible function application open parentheses a over b close parentheses plus tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a plus b over denominator a minus b end fraction close parentheses$ depends on

If $a greater than b greater than 0$ , then the value of $tan to the power of negative 1 end exponent invisible function application open parentheses a over b close parentheses plus tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator a plus b over denominator a minus b end fraction close parentheses$ depends on

parallel

$cot to the power of negative 1 end exponent invisible function application left parenthesis square root of cos invisible function application alpha end root right parenthesis minus tan to the power of negative 1 end exponent invisible function application left parenthesis square root of cos invisible function application alpha end root right parenthesis equals x$ then is equal to

$cot to the power of negative 1 end exponent invisible function application left parenthesis square root of cos invisible function application alpha end root right parenthesis minus tan to the power of negative 1 end exponent invisible function application left parenthesis square root of cos invisible function application alpha end root right parenthesis equals x$ then is equal to

Number of solutions of the equation $tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator 2 x plus 1 end fraction close parentheses plus tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator 4 x plus 1 end fraction close parentheses equals tan to the power of negative 1 end exponent invisible function application open parentheses 2 over x squared close parentheses$ is

Number of solutions of the equation $tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator 2 x plus 1 end fraction close parentheses plus tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator 4 x plus 1 end fraction close parentheses equals tan to the power of negative 1 end exponent invisible function application open parentheses 2 over x squared close parentheses$ is

$text If end text sec to the power of negative 1 end exponent invisible function application square root of 1 plus x squared end root plus cosec to the power of negative 1 end exponent invisible function application fraction numerator square root of 1 plus y squared end root over denominator y end fraction plus cot to the power of negative 1 end exponent invisible function application 1 over z equals pi$ $text then end text x plus y plus z$ is equal to

$text If end text sec to the power of negative 1 end exponent invisible function application square root of 1 plus x squared end root plus cosec to the power of negative 1 end exponent invisible function application fraction numerator square root of 1 plus y squared end root over denominator y end fraction plus cot to the power of negative 1 end exponent invisible function application 1 over z equals pi$ $text then end text x plus y plus z$ is equal to

parallel

$cos invisible function application open square brackets cos to the power of negative 1 end exponent invisible function application open parentheses negative 1 over 7 close parentheses plus sin to the power of negative 1 end exponent invisible function application open parentheses negative 1 over 7 close parentheses close square brackets text is equal to end text$

$cos invisible function application open square brackets cos to the power of negative 1 end exponent invisible function application open parentheses negative 1 over 7 close parentheses plus sin to the power of negative 1 end exponent invisible function application open parentheses negative 1 over 7 close parentheses close square brackets text is equal to end text$

$tan to the power of negative 1 end exponent invisible function application fraction numerator x over denominator square root of a squared minus x squared end root end fraction$ is equal to

$tan to the power of negative 1 end exponent invisible function application fraction numerator x over denominator square root of a squared minus x squared end root end fraction$ is equal to

If $theta equals sin to the power of negative 1 end exponent invisible function application x plus cos to the power of negative 1 end exponent invisible function application x minus tan to the power of negative 1 end exponent invisible function application x comma 1 less or equal than x less than straight infinity$ , then the smallest interval in which θ lies is

If $theta equals sin to the power of negative 1 end exponent invisible function application x plus cos to the power of negative 1 end exponent invisible function application x minus tan to the power of negative 1 end exponent invisible function application x comma 1 less or equal than x less than straight infinity$ , then the smallest interval in which θ lies is

parallel

If $A equals open square brackets table attributes columnalign center center center center columnspacing 1em end attributes row cell cos squared end cell alpha cell cos invisible function application horizontal ellipsis horizontal ellipsis end cell cell alpha sin invisible function application horizontal ellipsis horizontal ellipsis end cell row cell cos invisible function application horizontal ellipsis end cell cell alpha sin end cell alpha cell sin squared invisible function application alpha end cell end table close square brackets$ and $A equals open square brackets table attributes columnalign center center center center columnspacing 1em end attributes row cell cos squared end cell beta cell cos invisible function application horizontal ellipsis sin end cell alpha row cos cell beta sin end cell cell sin squared invisible function application beta end cell beta end table close square brackets$ are two matrices such that the product AB is null matrix, then $alpha minus beta$ is

If $A equals open square brackets table attributes columnalign center center center center columnspacing 1em end attributes row cell cos squared end cell alpha cell cos invisible function application horizontal ellipsis horizontal ellipsis end cell cell alpha sin invisible function application horizontal ellipsis horizontal ellipsis end cell row cell cos invisible function application horizontal ellipsis end cell cell alpha sin end cell alpha cell sin squared invisible function application alpha end cell end table close square brackets$ and $A equals open square brackets table attributes columnalign center center center center columnspacing 1em end attributes row cell cos squared end cell beta cell cos invisible function application horizontal ellipsis sin end cell alpha row cos cell beta sin end cell cell sin squared invisible function application beta end cell beta end table close square brackets$ are two matrices such that the product AB is null matrix, then $alpha minus beta$ is

If $A equals left square bracket x y z right square bracket comma B equals open square brackets table attributes columnspacing 1em end attributes row cell a straight h invisible function application g end cell row cell h b f end cell row cell g f c end cell end table close square brackets$ and $C equals open square brackets table attributes columnalign left columnspacing 1em end attributes row x row y row z end table close square brackets$ Then, $B to the power of straight prime left parenthesis A B right parenthesis equals 0$ if

If $A equals left square bracket x y z right square bracket comma B equals open square brackets table attributes columnspacing 1em end attributes row cell a straight h invisible function application g end cell row cell h b f end cell row cell g f c end cell end table close square brackets$ and $C equals open square brackets table attributes columnalign left columnspacing 1em end attributes row x row y row z end table close square brackets$ Then, $B to the power of straight prime left parenthesis A B right parenthesis equals 0$ if

The function $f colon R not stretchy rightwards arrow R$ given by $f left parenthesis x right parenthesis equals x cubed minus 1$ is

The function $f colon R not stretchy rightwards arrow R$ given by $f left parenthesis x right parenthesis equals x cubed minus 1$ is

parallel

$f left parenthesis x right parenthesis equals x plus square root of x squared end root$ is a function from R to R, then $f left parenthesis x right parenthesis$ is

$f left parenthesis x right parenthesis equals x plus square root of x squared end root$ is a function from R to R, then $f left parenthesis x right parenthesis$ is

Which of the following functions is one-to -one?

Which of the following functions is one-to -one?

A mapping $f colon N not stretchy rightwards arrow N$ , where N is the set of natural numbers is defined as $f left parenthesis n right parenthesis equals open curly brackets table attributes columnspacing 1em end attributes row cell n squared comma text for end text n text odd end text end cell row cell 2 n plus 1 comma text for end text n text even end text end cell end table close$ For $n element of N$ Then, $f$ is

A mapping $f colon N not stretchy rightwards arrow N$ , where N is the set of natural numbers is defined as $f left parenthesis n right parenthesis equals open curly brackets table attributes columnspacing 1em end attributes row cell n squared comma text for end text n text odd end text end cell row cell 2 n plus 1 comma text for end text n text even end text end cell end table close$ For $n element of N$ Then, $f$ is

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.