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Maths-

General

Easy

Question

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=

$\sqrt{3} + 4$
$\sqrt{3} + 1$
$\sqrt{3} - 4$
$\sqrt{3} - 1$

The correct answer is: $square root of 3 minus 4$

$table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell space of 1em cos invisible function application left parenthesis x minus y right parenthesis equals fraction numerator square root of 3 over denominator 2 end fraction not stretchy rightwards double arrow x minus y equals pi over 6 end cell row cell not stretchy rightwards double arrow x equals pi over 6 plus y therefore cos invisible function application x equals 2 cos invisible function application y end cell row cell not stretchy rightwards double arrow fraction numerator square root of 3 over denominator 2 end fraction cos invisible function application y minus 1 half sin invisible function application y equals 2 cos invisible function application y end cell row cell not stretchy rightwards double arrow open parentheses fraction numerator square root of 3 minus 4 over denominator 2 end fraction close parentheses cos invisible function application y equals 1 half sin invisible function application y not stretchy rightwards double arrow tan invisible function application y equals square root of 3 minus 4 end cell end table$

Related Questions to study

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 $s i n squared space left parenthesis alpha minus beta right parenthesis plus 2 a b c o s space left parenthesis alpha minus beta right parenthesis equals$

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

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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

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.

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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

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)

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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)

Statement 1 : f(x) = xⁿ sin $open parentheses fraction numerator 1 over denominator x end fraction close parentheses$ 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) = xⁿ sin $open parentheses fraction numerator 1 over denominator x end fraction close parentheses$ 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).

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If H₂O₂ is mixed with Fe²⁺, which reaction is more likely:

If H₂O₂ is mixed with Fe²⁺, which reaction is more likely:

Chemistry-General

For the following cell reaction $P b open parentheses s close parentheses plus H g subscript 2 end subscript S O subscript 4 end subscript open parentheses s close parentheses P b S O subscript 4 end subscript open parentheses s close parentheses plus 2 H g open parentheses l close parentheses E subscript text cell end text end subscript superscript ring operator end superscript equals 0.92 V comma$ $K subscript S p end subscript open parentheses P b S O subscript 4 end subscript close parentheses equals 2 cross times 10 to the power of negative 8 end exponent$ , $K subscript S p end subscript open parentheses H g S O subscript 4 end subscript close parentheses equals 1 cross times 10 to the power of negative 6 end exponent$ Hence, E_cell is:

For the following cell reaction $P b open parentheses s close parentheses plus H g subscript 2 end subscript S O subscript 4 end subscript open parentheses s close parentheses P b S O subscript 4 end subscript open parentheses s close parentheses plus 2 H g open parentheses l close parentheses E subscript text cell end text end subscript superscript ring operator end superscript equals 0.92 V comma$ $K subscript S p end subscript open parentheses P b S O subscript 4 end subscript close parentheses equals 2 cross times 10 to the power of negative 8 end exponent$ , $K subscript S p end subscript open parentheses H g S O subscript 4 end subscript close parentheses equals 1 cross times 10 to the power of negative 6 end exponent$ Hence, E_cell is:

Chemistry-General

Extraction of zinc from zinc blende is achieved by-

Extraction of zinc from zinc blende is achieved by-

Chemistry-General

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