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 value of $c$ in Lagrange’s theorem for the function $f open parentheses x close parentheses equals log invisible function application sin invisible function application x$ in the interval $left square bracket divided by 6 comma blank 5 divided by 6 right square bracket$

$\frac{}{4}$
$\frac{}{2}$
$\frac{2}{3}$
None of these

The correct answer is: $fraction numerator blank over denominator 2 end fraction$

$f open parentheses fraction numerator 5 pi over denominator 6 end fraction close parentheses equals log invisible function application sin invisible function application open parentheses fraction numerator 5 pi over denominator 6 end fraction close parentheses equals log invisible function application sin invisible function application fraction numerator pi over denominator 6 end fraction equals log invisible function application fraction numerator 1 over denominator 2 end fraction equals negative log invisible function application 2 comma$
$f open parentheses fraction numerator pi over denominator 6 end fraction close parentheses equals log invisible function application sin invisible function application fraction numerator pi over denominator 6 end fraction equals negative log invisible function application 2$
$f to the power of ´ end exponent open parentheses c close parentheses equals fraction numerator 1 over denominator sin invisible function application x end fraction cos invisible function application x equals cot invisible function application x$
By Lagrange’s mean value theorem,
$fraction numerator f left parenthesis 5 pi divided by 6 right parenthesis blank – blank f left parenthesis pi divided by 6 right parenthesis over denominator left parenthesis 5 pi divided by 6 right parenthesis blank – left parenthesis pi divided by 6 right parenthesis end fraction equals cot invisible function application c$
$rightwards double arrow c o t blank c equals 0 blank rightwards double arrow blank c equals fraction numerator pi over denominator 2 end fraction$
Thus, $c equals pi over 2 element of left parenthesis straight pi over 6 comma fraction numerator 5 straight pi over denominator 6 end fraction right parenthesis$

Related Questions to study

Let $f left parenthesis x right parenthesis$ be a function such that $f to the power of ´ end exponent open parentheses x close parentheses equals log subscript 1 divided by 3 end subscript invisible function application left square bracket log subscript 3 end subscript invisible function application left parenthesis sin invisible function application x plus a right parenthesis right square bracket$ . If $f left parenthesis x right parenthesis$ is decreasing for all real values of $x$ , then

Let $f left parenthesis x right parenthesis$ be a function such that $f to the power of ´ end exponent open parentheses x close parentheses equals log subscript 1 divided by 3 end subscript invisible function application left square bracket log subscript 3 end subscript invisible function application left parenthesis sin invisible function application x plus a right parenthesis right square bracket$ . If $f left parenthesis x right parenthesis$ is decreasing for all real values of $x$ , then

The volume of the greatest cylinder which can be inscribed in a cone of height 30 cm and semi-vertical angle $30 degree$ is

The volume of the greatest cylinder which can be inscribed in a cone of height 30 cm and semi-vertical angle $30 degree$ is

Consider the system of equation $x plus y plus z equals 6 comma blank x plus 2 y plus 3 z equals 10$ and $x plus 2 y plus lambda z equals mu$
Statement 1 If the system has infinite number of solutions, then $mu equals 10$
Statement 2: The determinant $open vertical bar table row 1 1 6 row 1 2 10 row 1 2 mu end table close vertical bar equals 0$ for $mu equals 10$

Consider the system of equation $x plus y plus z equals 6 comma blank x plus 2 y plus 3 z equals 10$ and $x plus 2 y plus lambda z equals mu$
Statement 1 If the system has infinite number of solutions, then $mu equals 10$
Statement 2: The determinant $open vertical bar table row 1 1 6 row 1 2 10 row 1 2 mu end table close vertical bar equals 0$ for $mu equals 10$

parallel

If $p q r not equal to 0$ and the system of equations $blank open parentheses p plus a close parentheses x plus b y plus c z equals 0 a x plus open parentheses q plus b close parentheses y plus c z equals 0$ $a x plus b y plus open parentheses r plus c close parentheses z equals 0$ Has a non-trivial solution, then value of $fraction numerator a over denominator p end fraction plus fraction numerator b over denominator q end fraction plus fraction numerator c over denominator r end fraction$ is

If $p q r not equal to 0$ and the system of equations $blank open parentheses p plus a close parentheses x plus b y plus c z equals 0 a x plus open parentheses q plus b close parentheses y plus c z equals 0$ $a x plus b y plus open parentheses r plus c close parentheses z equals 0$ Has a non-trivial solution, then value of $fraction numerator a over denominator p end fraction plus fraction numerator b over denominator q end fraction plus fraction numerator c over denominator r end fraction$ is

If $a comma blank b comma blank c$ are non-zeros, then the system of equations $open parentheses alpha plus a close parentheses x plus alpha y plus alpha z equals 0 comma blank alpha x plus open parentheses alpha plus b close parentheses y plus alpha z equals 0 comma alpha x plus alpha y plus open parentheses alpha plus c close parentheses z equals 0$ has a non-trivial solution if

If $a comma blank b comma blank c$ are non-zeros, then the system of equations $open parentheses alpha plus a close parentheses x plus alpha y plus alpha z equals 0 comma blank alpha x plus open parentheses alpha plus b close parentheses y plus alpha z equals 0 comma alpha x plus alpha y plus open parentheses alpha plus c close parentheses z equals 0$ has a non-trivial solution if

$increment subscript 1 end subscript equals open vertical bar table row cell y to the power of 5 end exponent z to the power of 6 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent z to the power of 6 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent y to the power of 5 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis y to the power of 6 end exponent minus z to the power of 6 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis z to the power of 6 end exponent minus x to the power of 6 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis x to the power of 6 end exponent minus y to the power of 6 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell end table close vertical bar$ and $increment subscript 2 end subscript equals open vertical bar table row x cell y to the power of 2 end exponent end cell cell z to the power of 3 end exponent end cell row cell x to the power of 4 end exponent end cell cell y to the power of 5 end exponent end cell cell z to the power of 6 end exponent end cell row cell x to the power of 7 end exponent end cell cell y to the power of 8 end exponent end cell cell z to the power of 9 end exponent end cell end table close vertical bar$ . Then $increment subscript 1 end subscript increment subscript 2 end subscript$ is equal to

$increment subscript 1 end subscript equals open vertical bar table row cell y to the power of 5 end exponent z to the power of 6 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent z to the power of 6 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x to the power of 4 end exponent y to the power of 5 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis y to the power of 6 end exponent minus z to the power of 6 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis z to the power of 6 end exponent minus x to the power of 6 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis x to the power of 6 end exponent minus y to the power of 6 end exponent right parenthesis end cell row cell y to the power of 2 end exponent z to the power of 3 end exponent left parenthesis z to the power of 3 end exponent minus y to the power of 3 end exponent right parenthesis end cell cell x z to the power of 3 end exponent left parenthesis x to the power of 3 end exponent minus z to the power of 3 end exponent right parenthesis end cell cell x y to the power of 2 end exponent left parenthesis y to the power of 3 end exponent minus x to the power of 3 end exponent right parenthesis end cell end table close vertical bar$ and $increment subscript 2 end subscript equals open vertical bar table row x cell y to the power of 2 end exponent end cell cell z to the power of 3 end exponent end cell row cell x to the power of 4 end exponent end cell cell y to the power of 5 end exponent end cell cell z to the power of 6 end exponent end cell row cell x to the power of 7 end exponent end cell cell y to the power of 8 end exponent end cell cell z to the power of 9 end exponent end cell end table close vertical bar$ . Then $increment subscript 1 end subscript increment subscript 2 end subscript$ is equal to

parallel

The system of equations $a x minus y minus z equals alpha minus 1$ $x minus alpha y minus z equals alpha minus 1$ $x minus y minus alpha z equals alpha minus 1$ Has no solution if $alpha$ is

The system of equations $a x minus y minus z equals alpha minus 1$ $x minus alpha y minus z equals alpha minus 1$ $x minus y minus alpha z equals alpha minus 1$ Has no solution if $alpha$ is

The value of the determinant $open vertical bar table row 1 1 1 row cell blank to the power of m end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 1 end subscript end cell row cell blank to the power of m end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 2 end subscript end cell end table close vertical bar$ is equal to

The value of the determinant $open vertical bar table row 1 1 1 row cell blank to the power of m end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 1 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 1 end subscript end cell row cell blank to the power of m end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 1 end exponent C subscript 2 end subscript end cell cell blank to the power of m plus 2 end exponent C subscript 2 end subscript end cell end table close vertical bar$ is equal to

If $alpha comma blank beta comma blank gamma$ are the angles of a triangle and the system of equations $cos invisible function application open parentheses alpha minus beta close parentheses x plus cos invisible function application open parentheses beta minus gamma close parentheses y plus cos invisible function application open parentheses gamma minus alpha close parentheses z equals 0 cos invisible function application open parentheses alpha plus beta close parentheses x plus cos invisible function application open parentheses beta plus gamma close parentheses y plus cos invisible function application open parentheses gamma plus alpha close parentheses z equals 0 sin invisible function application open parentheses alpha plus beta close parentheses x plus sin invisible function application open parentheses beta plus gamma close parentheses y plus sin invisible function application open parentheses gamma plus alpha close parentheses z equals 0$ Has non-trivial solutions, then triangle is necessarily

If $alpha comma blank beta comma blank gamma$ are the angles of a triangle and the system of equations $cos invisible function application open parentheses alpha minus beta close parentheses x plus cos invisible function application open parentheses beta minus gamma close parentheses y plus cos invisible function application open parentheses gamma minus alpha close parentheses z equals 0 cos invisible function application open parentheses alpha plus beta close parentheses x plus cos invisible function application open parentheses beta plus gamma close parentheses y plus cos invisible function application open parentheses gamma plus alpha close parentheses z equals 0 sin invisible function application open parentheses alpha plus beta close parentheses x plus sin invisible function application open parentheses beta plus gamma close parentheses y plus sin invisible function application open parentheses gamma plus alpha close parentheses z equals 0$ Has non-trivial solutions, then triangle is necessarily

parallel

If $x not equal to y not equal to z$ and $open vertical bar table row x cell x to the power of 2 end exponent end cell cell 1 plus x to the power of 3 end exponent end cell row y cell y to the power of 2 end exponent end cell cell 1 plus y to the power of 3 end exponent end cell row z cell z to the power of 2 end exponent end cell cell 1 plus z to the power of 3 end exponent end cell end table close vertical bar equals 0$ , then the value of $x y z$ is

If $x not equal to y not equal to z$ and $open vertical bar table row x cell x to the power of 2 end exponent end cell cell 1 plus x to the power of 3 end exponent end cell row y cell y to the power of 2 end exponent end cell cell 1 plus y to the power of 3 end exponent end cell row z cell z to the power of 2 end exponent end cell cell 1 plus z to the power of 3 end exponent end cell end table close vertical bar equals 0$ , then the value of $x y z$ is

If $f open parentheses x close parentheses equals a plus b x plus c x to the power of 2 end exponent$ and $alpha comma blank beta comma blank gamma$ are the roots of the equation $x to the power of 3 end exponent equals 1$ , then $open vertical bar table row a b c row b c a row c a b end table close vertical bar$ is equal to

If $f open parentheses x close parentheses equals a plus b x plus c x to the power of 2 end exponent$ and $alpha comma blank beta comma blank gamma$ are the roots of the equation $x to the power of 3 end exponent equals 1$ , then $open vertical bar table row a b c row b c a row c a b end table close vertical bar$ is equal to

If $A equals open square brackets table row alpha 2 row 2 alpha end table close square brackets$ and $open vertical bar A to the power of 3 end exponent close vertical bar equals 125 comma blank$ then the value of $alpha$ is

If $A equals open square brackets table row alpha 2 row 2 alpha end table close square brackets$ and $open vertical bar A to the power of 3 end exponent close vertical bar equals 125 comma blank$ then the value of $alpha$ is

parallel

If $omega blank open parentheses not equal to 1 close parentheses$ is a cube root of unity, then value of the determinant $open vertical bar table row 1 cell 1 plus i plus omega to the power of 2 end exponent end cell cell omega to the power of 2 end exponent end cell row cell 1 minus i end cell cell negative 1 end cell cell omega to the power of 2 end exponent minus 1 end cell row cell negative i end cell cell negative i plus omega minus 1 end cell cell negative 1 end cell end table close vertical bar$ is

If $omega blank open parentheses not equal to 1 close parentheses$ is a cube root of unity, then value of the determinant $open vertical bar table row 1 cell 1 plus i plus omega to the power of 2 end exponent end cell cell omega to the power of 2 end exponent end cell row cell 1 minus i end cell cell negative 1 end cell cell omega to the power of 2 end exponent minus 1 end cell row cell negative i end cell cell negative i plus omega minus 1 end cell cell negative 1 end cell end table close vertical bar$ is

If $A subscript 1 end subscript comma blank B subscript 1 end subscript comma blank C subscript 1 end subscript comma blank horizontal ellipsis$ are, respectively, the cofactors of the elements $a subscript 1 end subscript comma blank b subscript 1 end subscript comma blank c subscript 1 end subscript comma blank horizontal ellipsis$ of the determinant $increment equals open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar comma blank increment not equal to 0$ , then the value of $open vertical bar table row cell B subscript 2 end subscript end cell cell C subscript 2 end subscript end cell row cell B subscript 3 end subscript end cell cell C subscript 3 end subscript end cell end table close vertical bar$ is equal to

If $A subscript 1 end subscript comma blank B subscript 1 end subscript comma blank C subscript 1 end subscript comma blank horizontal ellipsis$ are, respectively, the cofactors of the elements $a subscript 1 end subscript comma blank b subscript 1 end subscript comma blank c subscript 1 end subscript comma blank horizontal ellipsis$ of the determinant $increment equals open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar comma blank increment not equal to 0$ , then the value of $open vertical bar table row cell B subscript 2 end subscript end cell cell C subscript 2 end subscript end cell row cell B subscript 3 end subscript end cell cell C subscript 3 end subscript end cell end table close vertical bar$ is equal to

Let $stack a with rightwards arrow on top$ be a unit vector perpendicular to unit vectors $stack b with rightwards arrow on top a n d stack c with rightwards arrow on top$ inclined at an angle a, then $stack b with rightwards arrow on top cross times stack c with rightwards arrow on top$ is

Let $stack a with rightwards arrow on top$ be a unit vector perpendicular to unit vectors $stack b with rightwards arrow on top a n d stack c with rightwards arrow on top$ inclined at an angle a, then $stack b with rightwards arrow on top cross times stack c with rightwards arrow on top$ 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.