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

If $x subscript 1 end subscript blank a n d blank x subscript 2 end subscript$ are the roots of the equation $e to the power of 2 end exponent blank. x to the power of ln invisible function application x end exponent equals x to the power of 3 end exponent blank w i t h blank x subscript 1 end subscript greater than x subscript 2 end subscript comma$ then

$x_{1} = 2 x_{2}$
$x_{1} = x_{2}^{2}$
$2 x_{1} = x_{2}^{2}$
$x_{1}^{2} = x_{2}^{3}$

The correct answer is: $x subscript 1 end subscript equals x subscript 2 end subscript superscript 2 end superscript$

$e to the power of 2 end exponent bullet x to the power of ln invisible function application x end exponent equals x to the power of 3 end exponent$
Taking log on both sides, we get $ln invisible function application open parentheses e to the power of 2 end exponent bullet x to the power of ln invisible function application x end exponent close parentheses equals ln invisible function application open parentheses x to the power of 3 end exponent close parentheses$
$rightwards double arrow open parentheses ln invisible function application x close parentheses to the power of 2 end exponent minus 3 ln invisible function application x plus 2 equals 0$
$rightwards double arrow open parentheses ln invisible function application x minus 2 close parentheses open parentheses ln invisible function application x minus 1 close parentheses equals 0$
If $ln invisible function application x equals 2 rightwards double arrow x equals e to the power of 2 end exponent$
If $ln invisible function application x equals 1 rightwards double arrow x equals e$
Since $x subscript 1 end subscript greater than x subscript 2 end subscript comma blank w e blank g e t blank x subscript 1 end subscript equals e to the power of 2 end exponent blank a n d blank x subscript 2 end subscript equals e$
$rightwards double arrow x subscript 2 end subscript superscript 2 end superscript equals x subscript 1 end subscript$

Related Questions to study

$sec to the power of 2 end exponent invisible function application theta equals fraction numerator 4 x y over denominator open parentheses x plus y close parentheses to the power of 2 end exponent end fraction$ is true if and only if

$sec to the power of 2 end exponent invisible function application theta equals fraction numerator 4 x y over denominator open parentheses x plus y close parentheses to the power of 2 end exponent end fraction$ is true if and only if

Product of roots of the equation $fraction numerator log subscript 8 end subscript invisible function application open parentheses 8 divided by x to the power of 2 end exponent close parentheses over denominator open parentheses log subscript 8 end subscript invisible function application x close parentheses to the power of 2 end exponent end fraction equals 3$ is

Product of roots of the equation $fraction numerator log subscript 8 end subscript invisible function application open parentheses 8 divided by x to the power of 2 end exponent close parentheses over denominator open parentheses log subscript 8 end subscript invisible function application x close parentheses to the power of 2 end exponent end fraction equals 3$ is

The value of the expression $fraction numerator 2 open parentheses sin invisible function application 1 degree plus sin invisible function application 2 degree plus sin invisible function application 3 degree plus horizontal ellipsis plus sin invisible function application 89 degree close parentheses over denominator 2 open parentheses cos invisible function application 1 degree plus cos invisible function application 2 degree plus horizontal ellipsis plus cos invisible function application 44 degree close parentheses plus 1 end fraction$ equals

The value of the expression $fraction numerator 2 open parentheses sin invisible function application 1 degree plus sin invisible function application 2 degree plus sin invisible function application 3 degree plus horizontal ellipsis plus sin invisible function application 89 degree close parentheses over denominator 2 open parentheses cos invisible function application 1 degree plus cos invisible function application 2 degree plus horizontal ellipsis plus cos invisible function application 44 degree close parentheses plus 1 end fraction$ equals

parallel

In $blank increment A B C comma$ if $blank A equals 30 degree comma blank b equals 2 comma blank c equals square root of 3 plus 1 comma blank$ then $fraction numerator C minus B over denominator 2 end fraction blank$ is equal to

In $blank increment A B C comma$ if $blank A equals 30 degree comma blank b equals 2 comma blank c equals square root of 3 plus 1 comma blank$ then $fraction numerator C minus B over denominator 2 end fraction blank$ is equal to

$I f blank x comma blank y comma blank z blank a r e blank i n blank A. P comma blank t h e n fraction numerator sin invisible function application x minus sin invisible function application z over denominator cos invisible function application z minus cos invisible function application x end fraction blank i s blank e q u a l blank t o$

$I f blank x comma blank y comma blank z blank a r e blank i n blank A. P comma blank t h e n fraction numerator sin invisible function application x minus sin invisible function application z over denominator cos invisible function application z minus cos invisible function application x end fraction blank i s blank e q u a l blank t o$

The general value of $x$ satisfying the equation $2 cot to the power of 2 end exponent invisible function application x plus 2 square root of 3 cot invisible function application x plus 4 blank c o s e c blank x plus 8 equals 0$ is

The general value of $x$ satisfying the equation $2 cot to the power of 2 end exponent invisible function application x plus 2 square root of 3 cot invisible function application x plus 4 blank c o s e c blank x plus 8 equals 0$ is

parallel

If $open parentheses 21.4 close parentheses to the power of a end exponent equals open parentheses 0.00214 close parentheses to the power of b end exponent equals 100 comma$ then the value of $fraction numerator 1 over denominator b end fraction minus fraction numerator 1 over denominator b end fraction$ is

If $open parentheses 21.4 close parentheses to the power of a end exponent equals open parentheses 0.00214 close parentheses to the power of b end exponent equals 100 comma$ then the value of $fraction numerator 1 over denominator b end fraction minus fraction numerator 1 over denominator b end fraction$ is

In triangle $A B C comma fraction numerator a over denominator b end fraction equals fraction numerator 2 over denominator 3 end fraction blank$ and $sec to the power of 2 end exponent invisible function application A equals fraction numerator 8 over denominator 5 end fraction.$ Then the number of triangles satisfying these conditions is

In triangle $A B C comma fraction numerator a over denominator b end fraction equals fraction numerator 2 over denominator 3 end fraction blank$ and $sec to the power of 2 end exponent invisible function application A equals fraction numerator 8 over denominator 5 end fraction.$ Then the number of triangles satisfying these conditions is

The greatest value of $sin to the power of 4 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta$ is

The greatest value of $sin to the power of 4 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta$ is

parallel

If in a $increment A B C comma cos invisible function application 3 A plus cos invisible function application 3 B plus cos invisible function application 3 C equals 1 comma blank$ then one angle must be exactly equal to

If in a $increment A B C comma cos invisible function application 3 A plus cos invisible function application 3 B plus cos invisible function application 3 C equals 1 comma blank$ then one angle must be exactly equal to

The solution of $4 sin to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x plus c o s e c to the power of 2 end exponent x plus cot to the power of 2 end exponent invisible function application x minus 6 equals 0$ is

The solution of $4 sin to the power of 2 end exponent invisible function application x plus tan to the power of 2 end exponent invisible function application x plus c o s e c to the power of 2 end exponent x plus cot to the power of 2 end exponent invisible function application x minus 6 equals 0$ is

In triangle $A B C comma blank$ if $blank A minus B equals 120 degree blank$ and $blank R equals 8 r$ where $R blank$ and $blank r$ have their usual meaning, then $cos invisible function application C$ equals

In triangle $A B C comma blank$ if $blank A minus B equals 120 degree blank$ and $blank R equals 8 r$ where $R blank$ and $blank r$ have their usual meaning, then $cos invisible function application C$ equals

parallel

In the given figure, $A B$ is the diameter of the circle, centered at $´ O to the power of ´ end exponent.$ If $angle C O A equals 60 degree comma blank A B equals 2 r comma blank A C equals d blank$ and $blank C D equals l comma$ then $l$ is equal to

In the given figure, $A B$ is the diameter of the circle, centered at $´ O to the power of ´ end exponent.$ If $angle C O A equals 60 degree comma blank A B equals 2 r comma blank A C equals d blank$ and $blank C D equals l comma$ then $l$ is equal to
Maths-General

$I f tan invisible function application x equals n tan invisible function application y comma blank n element of R to the power of plus end exponent comma$ then the maximum value of $s e c to the power of 2 end exponent open parentheses x minus y close parentheses$ is equal to

$I f tan invisible function application x equals n tan invisible function application y comma blank n element of R to the power of plus end exponent comma$ then the maximum value of $s e c to the power of 2 end exponent open parentheses x minus y close parentheses$ is equal to

If the equation $2 to the power of x end exponent plus 4 to the power of y end exponent equals 2 to the power of y end exponent plus 4 to the power of x end exponent$ is solved for $y$ in terms of $x comma blank$ where $x less than 0 comma$ then the sum of the solutions is

If the equation $2 to the power of x end exponent plus 4 to the power of y end exponent equals 2 to the power of y end exponent plus 4 to the power of x end exponent$ is solved for $y$ in terms of $x comma blank$ where $x less than 0 comma$ then the sum of the solutions 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.