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Maths-

General

Easy

Question

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x t a n invisible function application 2 x minus 2 x t a n invisible function application x right parenthesis divided by left parenthesis left parenthesis 1 minus c o s invisible function application 2 x right parenthesis squared right parenthesis$

1
$1 / 2$
0
2

The correct answer is: $1 divided by 2$

Related Questions to study

The Lewis symbols of three unknown elements are as follows :
i)
ii)
iii)
Write the formula of the different compounds formed by their combination.

The Lewis symbols of three unknown elements are as follows :
i)
ii)
iii)
Write the formula of the different compounds formed by their combination.

Chemistry-General

How many $sigma$ and $pi$ -bonds are there in the molecule of tetracyanoethylene ?

How many $sigma$ and $pi$ -bonds are there in the molecule of tetracyanoethylene ?

Chemistry-General

If $stack a with ‾ on top comma stack b with ‾ on top$ are non -zero, non -parallel vectors then the value of $vertical line stack a with ‾ on top plus stack b with ‾ on top plus stack a with ‾ on top cross times stack b with rightwards arrow on top vertical line to the power of 2 end exponent plus vertical line stack a with rightwards arrow on top plus stack b with ‾ on top minus stack a with ‾ on top cross times stack b with ‾ on top vertical line to the power of 2 end exponent$ equals

If $stack a with ‾ on top comma stack b with ‾ on top$ are non -zero, non -parallel vectors then the value of $vertical line stack a with ‾ on top plus stack b with ‾ on top plus stack a with ‾ on top cross times stack b with rightwards arrow on top vertical line to the power of 2 end exponent plus vertical line stack a with rightwards arrow on top plus stack b with ‾ on top minus stack a with ‾ on top cross times stack b with ‾ on top vertical line to the power of 2 end exponent$ equals

parallel

The vectors $stack x with ‾ on top$ and $stack y with ‾ on top$ satisfy the equation $p stack x with rightwards arrow on top plus q stack y with ‾ on top equals stack a with rightwards arrow on top$ (where $p comma q$ are scalar constants and $stack a with rightwards arrow on top$ is a known vector). It is given that $stack x with ‾ on top times stack y with rightwards arrow on top greater or equal than fraction numerator vertical line stack a with rightwards arrow on top vertical line to the power of 2 end exponent over denominator 4 p q end fraction$ , then $∣ x over y ∣$ is equal to $left parenthesis p q greater than 0 right parenthesis$

The vectors $stack x with ‾ on top$ and $stack y with ‾ on top$ satisfy the equation $p stack x with rightwards arrow on top plus q stack y with ‾ on top equals stack a with rightwards arrow on top$ (where $p comma q$ are scalar constants and $stack a with rightwards arrow on top$ is a known vector). It is given that $stack x with ‾ on top times stack y with rightwards arrow on top greater or equal than fraction numerator vertical line stack a with rightwards arrow on top vertical line to the power of 2 end exponent over denominator 4 p q end fraction$ , then $∣ x over y ∣$ is equal to $left parenthesis p q greater than 0 right parenthesis$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis l o g subscript e invisible function application left parenthesis 1 plus x cubed right parenthesis divided by s i n cubed invisible function application x$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis l o g subscript e invisible function application left parenthesis 1 plus x cubed right parenthesis divided by s i n cubed invisible function application x$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis invisible function application left parenthesis ∛ x minus 1 right parenthesis divided by left parenthesis x minus 1 right parenthesis$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis invisible function application left parenthesis ∛ x minus 1 right parenthesis divided by left parenthesis x minus 1 right parenthesis$

parallel

A: For a reaction at equilibrium, the Gibb’s free energy of reaction is minimum at constant temperature and pressure.

A: For a reaction at equilibrium, the Gibb’s free energy of reaction is minimum at constant temperature and pressure.

Chemistry-General

$L subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 3 to the power of t minus 1 right parenthesis divided by left parenthesis √ left parenthesis 1 plus x right parenthesis minus 1 right parenthesis$

$L subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 3 to the power of t minus 1 right parenthesis divided by left parenthesis √ left parenthesis 1 plus x right parenthesis minus 1 right parenthesis$

$L t subscript left parenthesis x rightwards arrow 2 right parenthesis invisible function application left parenthesis left parenthesis 2 x squared minus 7 x minus 4 right parenthesis right parenthesis divided by left parenthesis left parenthesis 2 x minus 1 right parenthesis left parenthesis √ x minus 2 right parenthesis right parenthesis$

$L t subscript left parenthesis x rightwards arrow 2 right parenthesis invisible function application left parenthesis left parenthesis 2 x squared minus 7 x minus 4 right parenthesis right parenthesis divided by left parenthesis left parenthesis 2 x minus 1 right parenthesis left parenthesis √ x minus 2 right parenthesis right parenthesis$

parallel

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis c o s invisible function application a x minus c o s invisible function application b x right parenthesis divided by x squared$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis c o s invisible function application a x minus c o s invisible function application b x right parenthesis divided by x squared$

The value of $sin invisible function application open parentheses a over 2 close parentheses open curly brackets 1 half plus sum from k equals 1 to n of cos invisible function application left parenthesis k a right parenthesis close curly brackets$ is equal to

The value of $sin invisible function application open parentheses a over 2 close parentheses open curly brackets 1 half plus sum from k equals 1 to n of cos invisible function application left parenthesis k a right parenthesis close curly brackets$ is equal to

depending upon its numerical value 8. Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by

depending upon its numerical value 8. Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by

Chemistry-General

parallel

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis x s i n invisible function application a minus a s i n invisible function application x right parenthesis divided by left parenthesis x minus a right parenthesis$

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis x s i n invisible function application a minus a s i n invisible function application x right parenthesis divided by left parenthesis x minus a right parenthesis$

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis t a n invisible function application left parenthesis x minus a right parenthesis right parenthesis divided by left parenthesis x squared minus a squared right parenthesis left parenthesis a not equal to 0 right parenthesis$

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis t a n invisible function application left parenthesis x minus a right parenthesis right parenthesis divided by left parenthesis x squared minus a squared right parenthesis left parenthesis a not equal to 0 right parenthesis$

For the reaction $N subscript 2 end subscript O subscript 4 end subscript left parenthesis g right parenthesis 2 N O subscript 2 end subscript left parenthesis g right parenthesis$ , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant Kp will be

For the reaction $N subscript 2 end subscript O subscript 4 end subscript left parenthesis g right parenthesis 2 N O subscript 2 end subscript left parenthesis g right parenthesis$ , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant Kp will be

Chemistry-General

parallel

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