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Maths-

General

Easy

Question

Let $stack r with rightwards arrow on top equals left parenthesis stack a with rightwards arrow on top cross times stack b with rightwards arrow on top right parenthesis s i n invisible function application x plus left parenthesis stack b with rightwards arrow on top cross times stack c with rightwards arrow on top right parenthesis c o s invisible function application y plus 2 left parenthesis stack c with rightwards arrow on top cross times stack a with rightwards arrow on top right parenthesis$ where $stack a with rightwards arrow on top stack c with rightwards arrow on top$ are three noncoplanar vectors. If $stack r with rightwards arrow on top$ is perpendicular to $stack a with rightwards arrow on top plus stack b with rightwards arrow on top plus stack v with rightwards arrow on top$ , then minimum value of $x squared plus y squared$

$π^{2}$
$\frac{π^{2}}{4}$
$\frac{5 π^{2}}{4}$
none of these

The correct answer is: $fraction numerator 5 pi to the power of 2 end exponent over denominator 4 end fraction$

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Two adjacent sides of a parallelogram $A B C D$ are given by $stack A B with minus on top equals 2 i with ‾ on top plus 10 j with ‾ on top plus 11 k with ‾ on top$ and $stack A D with minus on top equals negative i with ‾ on top plus 2 j with ‾ on top plus 2 k with ‾ on top$ . The side $A D$ is rotated by an acute angle $alpha$ in the plane of parallelogram so that $A D$ becomes AD’. If $A D$ makes a right angle with the side AB then the cosine of angle $alpha$ is given by

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If $a with ‾ on top equals 2 i with ‾ on top plus 3 j with ‾ on top plus k with ‾ on top comma a with ‾ on top cross times b with ‾ on top equals 7 i with ‾ on top minus 3 j with ‾ on top minus 5 k with ‾ on top comma a with ‾ on top times b with ‾ on top equals 1$ then $stack b with ‾ on top equals$

If $a with ‾ on top equals 2 i with ‾ on top plus 3 j with ‾ on top plus k with ‾ on top comma a with ‾ on top cross times b with ‾ on top equals 7 i with ‾ on top minus 3 j with ‾ on top minus 5 k with ‾ on top comma a with ‾ on top times b with ‾ on top equals 1$ then $stack b with ‾ on top equals$

parallel

If $Error converting from MathML to accessible text.$ are non coplanar vectors then the roots of equation $open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell left square bracket b with not stretchy bar on top cross times c with not stretchy bar on top c with not stretchy bar on top cross times a with not stretchy bar on top a with not stretchy bar on top cross times b with not stretchy bar on top end cell end table close square brackets x squared$ $Error converting from MathML to accessible text.$ $plus left square bracket b with not stretchy bar on top minus c with not stretchy bar on top c with not stretchy bar on top minus a with not stretchy bar on top a with not stretchy bar on top minus b with not stretchy bar on top right square bracket plus 1 equals 0$

If $Error converting from MathML to accessible text.$ are non coplanar vectors then the roots of equation $open square brackets table attributes columnalign left left left columnspacing 1em end attributes row cell left square bracket b with not stretchy bar on top cross times c with not stretchy bar on top c with not stretchy bar on top cross times a with not stretchy bar on top a with not stretchy bar on top cross times b with not stretchy bar on top end cell end table close square brackets x squared$ $Error converting from MathML to accessible text.$ $plus left square bracket b with not stretchy bar on top minus c with not stretchy bar on top c with not stretchy bar on top minus a with not stretchy bar on top a with not stretchy bar on top minus b with not stretchy bar on top right square bracket plus 1 equals 0$

If the squares of the lengths of the tangents from a point P to the circles $x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent$ , $x to the power of 2 end exponent plus y to the power of 2 end exponent equals b to the power of 2 end exponent$ and $x to the power of 2 end exponent plus y to the power of 2 end exponent equals c to the power of 2 end exponent$ are in A.P then $a to the power of 2 end exponent comma b to the power of 2 end exponent comma c to the power of 2 end exponent$ are in

If the squares of the lengths of the tangents from a point P to the circles $x to the power of 2 end exponent plus y to the power of 2 end exponent equals a to the power of 2 end exponent$ , $x to the power of 2 end exponent plus y to the power of 2 end exponent equals b to the power of 2 end exponent$ and $x to the power of 2 end exponent plus y to the power of 2 end exponent equals c to the power of 2 end exponent$ are in A.P then $a to the power of 2 end exponent comma b to the power of 2 end exponent comma c to the power of 2 end exponent$ are in

In the following redox reaction, $Cu left parenthesis OH right parenthesis subscript 2 left parenthesis straight s right parenthesis plus straight N subscript 2 straight H subscript 4 left parenthesis aq right parenthesis not stretchy rightwards arrow Cu left parenthesis straight s right parenthesis plus straight N subscript 2 left parenthesis straight g right parenthesis$ Number of moles of $Cu left parenthesis OH right parenthesis subscript 2$ reduced by 1 mole of $straight N subscript 2 straight H subscript 4$ is

In the following redox reaction, $Cu left parenthesis OH right parenthesis subscript 2 left parenthesis straight s right parenthesis plus straight N subscript 2 straight H subscript 4 left parenthesis aq right parenthesis not stretchy rightwards arrow Cu left parenthesis straight s right parenthesis plus straight N subscript 2 left parenthesis straight g right parenthesis$ Number of moles of $Cu left parenthesis OH right parenthesis subscript 2$ reduced by 1 mole of $straight N subscript 2 straight H subscript 4$ is

Chemistry-General

parallel

The vector $i with minus on top cross times left square bracket left parenthesis a with ‾ on top cross times b with ‾ on top right parenthesis cross times i with ‾ on top right parenthesis right square bracket plus j with ‾ on top cross times left square bracket left parenthesis a with ‾ on top cross times b with ‾ on top right parenthesis cross times j with ‾ on top right square bracket plus k with ‾ on top cross times left square bracket left parenthesis a with ‾ on top cross times b with ‾ on top right parenthesis cross times k with ‾ on top right square bracket$ equals

The vector $i with minus on top cross times left square bracket left parenthesis a with ‾ on top cross times b with ‾ on top right parenthesis cross times i with ‾ on top right parenthesis right square bracket plus j with ‾ on top cross times left square bracket left parenthesis a with ‾ on top cross times b with ‾ on top right parenthesis cross times j with ‾ on top right square bracket plus k with ‾ on top cross times left square bracket left parenthesis a with ‾ on top cross times b with ‾ on top right parenthesis cross times k with ‾ on top right square bracket$ equals

Let $a with ‾ on top equals i with ˆ on top plus j with ˆ on top plus k with rightwards arrow on top times b with rightwards arrow on top equals left square bracket x right square bracket i with ˆ on top plus left curly bracket s i n invisible function application left parenthesis x right parenthesis right curly bracket j comma c equals c subscript 1 i with ˆ on top plus c subscript 2 j with ˆ on top plus c subscript 3 k with ˆ on top$ be three non zero vectors,
such that $vertical line stack c with rightwards arrow on top vertical line equals 1$ angle between $stack a with ‾ on top$ and $stack b with ‾ on top$ is $fraction numerator pi over denominator 4 end fraction$ , $stack c with rightwards arrow on top$ is perpendicular to both $stack a with rightwards arrow on top$ and $stack b with ‾ on top$ . (where $left square bracket x right square bracket equals$ Greatest integer function and $left curly bracket x right curly bracket equals$ Fractional part of $x right parenthesis$ . If $f left parenthesis x right parenthesis equals left parenthesis stack a with rightwards arrow on top left parenthesis stack b with ‾ on top cross times stack c with ‾ on top right parenthesis right parenthesis to the power of 2 end exponent$ , then which of the following is correct for $f left parenthesis x right parenthesis$

Let $a with ‾ on top equals i with ˆ on top plus j with ˆ on top plus k with rightwards arrow on top times b with rightwards arrow on top equals left square bracket x right square bracket i with ˆ on top plus left curly bracket s i n invisible function application left parenthesis x right parenthesis right curly bracket j comma c equals c subscript 1 i with ˆ on top plus c subscript 2 j with ˆ on top plus c subscript 3 k with ˆ on top$ be three non zero vectors,
such that $vertical line stack c with rightwards arrow on top vertical line equals 1$ angle between $stack a with ‾ on top$ and $stack b with ‾ on top$ is $fraction numerator pi over denominator 4 end fraction$ , $stack c with rightwards arrow on top$ is perpendicular to both $stack a with rightwards arrow on top$ and $stack b with ‾ on top$ . (where $left square bracket x right square bracket equals$ Greatest integer function and $left curly bracket x right curly bracket equals$ Fractional part of $x right parenthesis$ . If $f left parenthesis x right parenthesis equals left parenthesis stack a with rightwards arrow on top left parenthesis stack b with ‾ on top cross times stack c with ‾ on top right parenthesis right parenthesis to the power of 2 end exponent$ , then which of the following is correct for $f left parenthesis x right parenthesis$

If $a subscript 1 end subscript$ and $a subscript 2 end subscript$ are two values of a for which the unit vector $a i with minus on top plus b j with minus on top plus 1 half k with minus on top$ is linearly dependent with $i with minus on top plus 2 j$ and $stack j with minus on top minus 2 stack k with minus on top$ then $fraction numerator 1 over denominator a subscript 1 end subscript end fraction plus fraction numerator 1 over denominator a subscript 2 end subscript end fraction$ is equal to

If $a subscript 1 end subscript$ and $a subscript 2 end subscript$ are two values of a for which the unit vector $a i with minus on top plus b j with minus on top plus 1 half k with minus on top$ is linearly dependent with $i with minus on top plus 2 j$ and $stack j with minus on top minus 2 stack k with minus on top$ then $fraction numerator 1 over denominator a subscript 1 end subscript end fraction plus fraction numerator 1 over denominator a subscript 2 end subscript end fraction$ is equal to

parallel

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