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Let $f open parentheses x close parentheses equals open curly brackets table row cell x plus 2 comma blank minus 1 less or equal than x less than 0 end cell row cell 1 comma blank x equals 0 blank end cell row cell fraction numerator x over denominator 2 end fraction comma blank 0 less than x less or equal than 1 end cell end table close$ Then on $left square bracket negative 1 comma blank 1 right square bracket$ , this function has

A minimum
A maximum
Either a maximum or a minimum
Neither a maximum nor a minimum

The correct answer is: Neither a maximum nor a minimum

$f open parentheses 0 close parentheses greater than f left parenthesis 0 to the power of plus end exponent right parenthesis$ and $f open parentheses 0 close parentheses less than f left parenthesis 0 to the power of minus end exponent right parenthesis$ , hence $x equals 0$ is neither a maximum nor a minimum

Related Questions to study

If $ϕ left parenthesis x right parenthesis$ is a polynomial function and $ϕ to the power of ´ end exponent open parentheses x close parentheses greater than ϕ open parentheses x close parentheses comma blank for all blank x greater or equal than 1$ and $ϕ open parentheses 1 close parentheses equals 0$ , then

If $ϕ left parenthesis x right parenthesis$ is a polynomial function and $ϕ to the power of ´ end exponent open parentheses x close parentheses greater than ϕ open parentheses x close parentheses comma blank for all blank x greater or equal than 1$ and $ϕ open parentheses 1 close parentheses equals 0$ , then

If $f open parentheses x close parentheses equals x plus sin invisible function application x semicolon g open parentheses x close parentheses equals e to the power of negative x end exponent semicolon u equals square root of c plus 1 end root minus square root of c semicolon v equals square root of c minus square root of c minus 1 end root semicolon left parenthesis c greater than 1 right parenthesis$ , then

If $f open parentheses x close parentheses equals x plus sin invisible function application x semicolon g open parentheses x close parentheses equals e to the power of negative x end exponent semicolon u equals square root of c plus 1 end root minus square root of c semicolon v equals square root of c minus square root of c minus 1 end root semicolon left parenthesis c greater than 1 right parenthesis$ , then

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 $left square bracket divided by 6 comma blank 5 divided by 6 right square bracket$

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 $left square bracket divided by 6 comma blank 5 divided by 6 right square bracket$

parallel

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

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