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Question

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

$a \in (1, 4)$
$a \in (4, \infty)$
$a \in (2, 3)$
$a \in (2, \infty)$

The correct answer is: $a element of open parentheses 4 comma blank infinity close parentheses$

We must have $log subscript 1 divided by 3 end subscript invisible function application open parentheses log subscript 3 end subscript invisible function application open parentheses sin invisible function application x plus a close parentheses close parentheses less than 0 blank for all x element of R$
$rightwards double arrow log subscript 3 end subscript invisible function application left parenthesis sin invisible function application x plus a right parenthesis greater than 1 blank for all x element of R$
$rightwards double arrow sin invisible function application x plus a greater than 3 blank for all x element of R$
$rightwards double arrow a greater than 3 minus sin invisible function application x blank for all x element of R$
$rightwards double arrow a greater than 4$

Related Questions to study

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 $mu equals 10$

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

parallel

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

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

parallel

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

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

parallel

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

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

parallel

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

Let $stack a with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack b with rightwards arrow on top short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow a n d short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack c with rightwards arrow on top$ be three non-coplanar vectors and $stack d with rightwards arrow on top$ be a non-zero vector which is perpendicular to $open parentheses stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack c with rightwards arrow on top close parentheses$ and is represented as $stack d with rightwards arrow on top equals x open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses plus y open parentheses stack b with rightwards arrow on top cross times stack c with rightwards arrow on top close parentheses plus z open parentheses stack c with rightwards arrow on top cross times stack a with rightwards arrow on top close parentheses$ . Then

Let $stack a with rightwards arrow on top comma short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack b with rightwards arrow on top short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow a n d short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow short rightwards arrow over leftwards arrow stack c with rightwards arrow on top$ be three non-coplanar vectors and $stack d with rightwards arrow on top$ be a non-zero vector which is perpendicular to $open parentheses stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack c with rightwards arrow on top close parentheses$ and is represented as $stack d with rightwards arrow on top equals x open parentheses stack a with rightwards arrow on top cross times stack b with rightwards arrow on top close parentheses plus y open parentheses stack b with rightwards arrow on top cross times stack c with rightwards arrow on top close parentheses plus z open parentheses stack c with rightwards arrow on top cross times stack a with rightwards arrow on top close parentheses$ . Then

parallel

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