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Question

If $alpha comma beta comma gamma$ are the roots of x³ – 3x + 2 = 0, then the value of the determinant $open vertical bar table row alpha beta gamma row beta gamma alpha row gamma alpha beta end table close vertical bar$ is equal to

–3
2
1
none of these

The correct answer is: none of these

Related Questions to study

If $straight capital delta$ ABC is a scalene triangle, then the value of $open vertical bar table row cell sin invisible function application A end cell cell sin invisible function application B end cell cell sin invisible function application C end cell row cell cos invisible function application A end cell cell cos invisible function application B end cell cell cos invisible function application C end cell row 1 1 1 end table close vertical bar$ is

If $straight capital delta$ ABC is a scalene triangle, then the value of $open vertical bar table row cell sin invisible function application A end cell cell sin invisible function application B end cell cell sin invisible function application C end cell row cell cos invisible function application A end cell cell cos invisible function application B end cell cell cos invisible function application C end cell row 1 1 1 end table close vertical bar$ is

Consider the system of equations-x – 2y + 3z = –1–x + y – 2z = k x – 3y + 4z = 1
STATEMENT-1: The system of equations has no solution for k $not equal to$ 3
STATEMENT-2: The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3

Consider the system of equations-x – 2y + 3z = –1–x + y – 2z = k x – 3y + 4z = 1
STATEMENT-1: The system of equations has no solution for k $not equal to$ 3
STATEMENT-2: The determinant $open vertical bar table row 1 3 cell negative 1 end cell row cell negative 1 end cell cell negative 2 end cell k row 1 4 1 end table close vertical bar$ $not equal to$ 0, for k $not equal to$ 3

Suppose, x > 0, y > 0, z > 0 and $capital delta$ (a, b, c) = $open vertical bar table row cell x log invisible function application 2 end cell 3 cell 15 plus log invisible function application left parenthesis a to the power of x end exponent right parenthesis end cell row cell y log invisible function application 3 end cell 5 cell 25 plus log invisible function application left parenthesis b to the power of y end exponent right parenthesis end cell row cell z log invisible function application 5 end cell 7 cell 35 plus log invisible function application left parenthesis c to the power of z end exponent right parenthesis end cell end table close vertical bar$
Statement - 1 : $capital delta$ (8, 27, 125) = 0
Statement - 2 : $capital delta$ $open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator 1 over denominator 3 end fraction comma fraction numerator 1 over denominator 5 end fraction close parentheses$ = 0

Suppose, x > 0, y > 0, z > 0 and $capital delta$ (a, b, c) = $open vertical bar table row cell x log invisible function application 2 end cell 3 cell 15 plus log invisible function application left parenthesis a to the power of x end exponent right parenthesis end cell row cell y log invisible function application 3 end cell 5 cell 25 plus log invisible function application left parenthesis b to the power of y end exponent right parenthesis end cell row cell z log invisible function application 5 end cell 7 cell 35 plus log invisible function application left parenthesis c to the power of z end exponent right parenthesis end cell end table close vertical bar$
Statement - 1 : $capital delta$ (8, 27, 125) = 0
Statement - 2 : $capital delta$ $open parentheses fraction numerator 1 over denominator 2 end fraction comma fraction numerator 1 over denominator 3 end fraction comma fraction numerator 1 over denominator 5 end fraction close parentheses$ = 0

parallel

If $fraction numerator cos space x over denominator cos space y end fraction equals 2$ and $cos space left parenthesis x minus y right parenthesis equals fraction numerator square root of 3 over denominator 2 end fraction$ then y=

If $fraction numerator cos space x over denominator cos space y end fraction equals 2$ and $cos space left parenthesis x minus y right parenthesis equals fraction numerator square root of 3 over denominator 2 end fraction$ then y=

If $tan space left parenthesis pi divided by 4 plus theta right parenthesis plus tan space left parenthesis pi divided by 4 minus theta right parenthesis equals a$ then $t a n cubed space left parenthesis pi divided by 4 plus theta right parenthesis plus t a n cubed space left parenthesis pi divided by 4 minus theta right parenthesis equals$

If $tan space left parenthesis pi divided by 4 plus theta right parenthesis plus tan space left parenthesis pi divided by 4 minus theta right parenthesis equals a$ then $t a n cubed space left parenthesis pi divided by 4 plus theta right parenthesis plus t a n cubed space left parenthesis pi divided by 4 minus theta right parenthesis equals$

$text If end text cos left parenthesis theta minus alpha right parenthesis equals a times cos space left parenthesis theta minus beta right parenthesis equals b$ then $s i n squared space left parenthesis alpha minus beta right parenthesis plus 2 a b c o s space left parenthesis alpha minus beta right parenthesis equals$

$text If end text cos left parenthesis theta minus alpha right parenthesis equals a times cos space left parenthesis theta minus beta right parenthesis equals b$ then $s i n squared space left parenthesis alpha minus beta right parenthesis plus 2 a b c o s space left parenthesis alpha minus beta right parenthesis equals$

parallel

if $fraction numerator cos space x minus cos space alpha over denominator cos space x minus cos space beta end fraction equals fraction numerator sin squared begin display style space end style alpha cos space beta over denominator sin squared begin display style space end style beta cos space alpha end fraction text then end text cos text end text x equals$

if $fraction numerator cos space x minus cos space alpha over denominator cos space x minus cos space beta end fraction equals fraction numerator sin squared begin display style space end style alpha cos space beta over denominator sin squared begin display style space end style beta cos space alpha end fraction text then end text cos text end text x equals$

Assertion: $open vertical bar table row cell cos invisible function application left parenthesis theta plus alpha right parenthesis end cell cell cos invisible function application left parenthesis theta plus beta right parenthesis end cell cell cos invisible function application left parenthesis theta plus gamma right parenthesis end cell row cell sin invisible function application left parenthesis theta plus alpha right parenthesis end cell cell sin invisible function application left parenthesis theta plus beta right parenthesis end cell cell sin invisible function application left parenthesis theta plus gamma right parenthesis end cell row cell sin invisible function application left parenthesis beta minus gamma right parenthesis end cell cell sin invisible function application left parenthesis gamma minus alpha right parenthesis end cell cell sin invisible function application left parenthesis alpha minus beta right parenthesis end cell end table close vertical bar$ is independent of $theta$
Reason: If f( $theta$ ) = c, then f( $theta$ ) is independent of $theta$ .

Assertion: $open vertical bar table row cell cos invisible function application left parenthesis theta plus alpha right parenthesis end cell cell cos invisible function application left parenthesis theta plus beta right parenthesis end cell cell cos invisible function application left parenthesis theta plus gamma right parenthesis end cell row cell sin invisible function application left parenthesis theta plus alpha right parenthesis end cell cell sin invisible function application left parenthesis theta plus beta right parenthesis end cell cell sin invisible function application left parenthesis theta plus gamma right parenthesis end cell row cell sin invisible function application left parenthesis beta minus gamma right parenthesis end cell cell sin invisible function application left parenthesis gamma minus alpha right parenthesis end cell cell sin invisible function application left parenthesis alpha minus beta right parenthesis end cell end table close vertical bar$ is independent of $theta$
Reason: If f( $theta$ ) = c, then f( $theta$ ) is independent of $theta$ .

Statement-1 : The function f(x) = |x³| is differentiable at x = 0
Statement-2 : at x = 0, $f to the power of straight prime$ (x) = 0

Statement-1 : The function f(x) = |x³| is differentiable at x = 0
Statement-2 : at x = 0, $f to the power of straight prime$ (x) = 0

parallel

Statement-1 : The function y = sin–1 (cos x) is not differentiable at $x equals n pi comma n element of Z$ is particular at x = $pi$

Statement-2 : $fraction numerator d y over denominator d x end fraction$ = $fraction numerator negative sin invisible function application x over denominator vertical line sin invisible function application x vertical line end fraction$ so the function is not differentiable at the points where sin x = 0.

Statement-1 : The function y = sin–1 (cos x) is not differentiable at $x equals n pi comma n element of Z$ is particular at x = $pi$

Statement-2 : $fraction numerator d y over denominator d x end fraction$ = $fraction numerator negative sin invisible function application x over denominator vertical line sin invisible function application x vertical line end fraction$ so the function is not differentiable at the points where sin x = 0.

Statement-1 : f $left parenthesis x right parenthesis equals x to the power of n end exponent s i n invisible function application open parentheses fraction numerator 1 over denominator x end fraction close parentheses semicolon x not equal to 0 equals 0 semicolon x equals 0$ is differentiable for all real values of x (n $greater or equal than$ 2)
Statement-2 : For n $greater or equal than$ 2, Right derivative = Left derivative (for all real values of x)

Statement-1 : f $left parenthesis x right parenthesis equals x to the power of n end exponent s i n invisible function application open parentheses fraction numerator 1 over denominator x end fraction close parentheses semicolon x not equal to 0 equals 0 semicolon x equals 0$ is differentiable for all real values of x (n $greater or equal than$ 2)
Statement-2 : For n $greater or equal than$ 2, Right derivative = Left derivative (for all real values of x)

Statement-1 : f(x) = cos²x + cos³ $open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses$ – cos x cos³ $open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses$ Then f‘(x) = 0
Statement-2 : Derivative of constant function is zero

Statement-1 : f(x) = cos²x + cos³ $open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses$ – cos x cos³ $open parentheses x plus fraction numerator pi over denominator 3 end fraction close parentheses$ Then f‘(x) = 0
Statement-2 : Derivative of constant function is zero

parallel

Let f and g be real valued functions defined on interval (–1, 1) such that $g to the power of ′′ left parenthesis x right parenthesis text is continuous, end text g left parenthesis 0 right parenthesis not equal to 0. g to the power of straight prime left parenthesis 0 right parenthesis equals 0 comma g to the power of ′′ left parenthesis 0 right parenthesis not equal to 0 comma straight & f left parenthesis x right parenthesis equals g$ (x)sin x
Statement-1 : $stack l i m with x rightwards arrow 0 below$ [g(x) cot x –g(0) cosec x] = $f to the power of ′′$ (0)
Statement-2 : $f to the power of straight prime$ (0) = g (0)

Let f and g be real valued functions defined on interval (–1, 1) such that $g to the power of ′′ left parenthesis x right parenthesis text is continuous, end text g left parenthesis 0 right parenthesis not equal to 0. g to the power of straight prime left parenthesis 0 right parenthesis equals 0 comma g to the power of ′′ left parenthesis 0 right parenthesis not equal to 0 comma straight & f left parenthesis x right parenthesis equals g$ (x)sin x
Statement-1 : $stack l i m with x rightwards arrow 0 below$ [g(x) cot x –g(0) cosec x] = $f to the power of ′′$ (0)
Statement-2 : $f to the power of straight prime$ (0) = g (0)

Statement-1 : If f(x) = $fraction numerator left parenthesis e to the power of k x end exponent minus 1 right parenthesis sin invisible function application blank k x over denominator 4 x to the power of 2 end exponent end fraction$ (x $not equal to$ 0) and f(0) = 9 is continuous at x = 0 then k = ± 6.
Statement-2 : For continuous function $stack l i m with x rightwards arrow 0 below$ f(x) = f(0)

Statement-1 : If f(x) = $fraction numerator left parenthesis e to the power of k x end exponent minus 1 right parenthesis sin invisible function application blank k x over denominator 4 x to the power of 2 end exponent end fraction$ (x $not equal to$ 0) and f(0) = 9 is continuous at x = 0 then k = ± 6.
Statement-2 : For continuous function $stack l i m with x rightwards arrow 0 below$ f(x) = f(0)

Statement-I : Let f(x) = $fraction numerator 1 minus tan invisible function application x over denominator 4 x minus pi end fraction$ , x $not equal to$ $fraction numerator pi over denominator 4 end fraction$ , x $element of$ $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ . If f(x) is continuous in $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ , Then f $open parentheses fraction numerator pi over denominator 4 end fraction close parentheses$ = $negative fraction numerator 1 over denominator 2 end fraction$ .
Statement-II : f(x) is continuous at x = a if $stack l i m with x rightwards arrow a below$ f(x) = f(a)

Statement-I : Let f(x) = $fraction numerator 1 minus tan invisible function application x over denominator 4 x minus pi end fraction$ , x $not equal to$ $fraction numerator pi over denominator 4 end fraction$ , x $element of$ $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ . If f(x) is continuous in $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ , Then f $open parentheses fraction numerator pi over denominator 4 end fraction close parentheses$ = $negative fraction numerator 1 over denominator 2 end fraction$ .
Statement-II : f(x) is continuous at x = a if $stack l i m with x rightwards arrow a below$ f(x) = f(a)

parallel

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