Solve the differential equation  from Class 12 CBSE Previous Y

Subject

Mathematics

Class

CBSE Class 12

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

11.

Use Properties of determinants, prove that:
open vertical bar table row cell 1 plus straight a end cell cell space 1 end cell cell space 1 end cell row 1 cell 1 plus straight b end cell 1 row 1 1 cell 1 plus straight c end cell end table close vertical bar space equals space abc plus bc plus ca plus ab

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 Multiple Choice QuestionsLong Answer Type

12.

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award x each, y each and z each for the three respective values to 3, 2 and 1 students respectively with a total award money of 1,600. School B wants to spend 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount for one prize on each value is 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

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13.

If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is 60 degree.

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 Multiple Choice QuestionsShort Answer Type

14. If space straight f left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x straight t space sint space dt comma space write space the space value space of space straight f apostrophe left parenthesis straight x right parenthesis
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15.

Find the value of 'p' for which the vectors 3 straight i with hat on top plus 2 straight j with hat on top plus 9 straight k with hat on top space and space straight i with hat on top minus 2 straight p straight j with hat on top plus 3 straight k with hat on top are parallel.

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16.

If the cartesian equations of a line are fraction numerator 3 minus straight x over denominator 5 end fraction equals fraction numerator straight y plus 4 over denominator 7 end fraction equals fraction numerator 2 straight z minus 6 over denominator 4 end fraction comma write the vector equation for the line. 

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17.

If integral subscript 0 superscript straight a fraction numerator 1 over denominator 4 plus straight x squared end fraction dx equals straight pi over 8, find the value of a.

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18. If space straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space are space perpendicular space vectors comma space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space 13 space and space open vertical bar straight a with rightwards arrow on top close vertical bar space equals 5 space and space find space the space value space of space open vertical bar straight b with rightwards arrow on top close vertical bar.
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19.

Solve the differential equation left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent


Given differential equation is:
left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent
rightwards double arrow dy over dx plus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space equals space fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction
This is a linear differential equation of the form
dy over dx plus Py space equals space straight Q
where space straight P space equals space fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space and space straight Q space equals fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction
Therefore comma space
straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent
Thus space the space solution space is comma
straight y left parenthesis straight I. straight F right parenthesis space equals space integral straight Q thin space left parenthesis straight I. straight F right parenthesis space dx
rightwards double arrow straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals integral fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent dx
Substitute space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space straight t semicolon
straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space cross times fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dt
Thus comma space
straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space integral tdt
rightwards double arrow space space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals straight t squared over 2 plus straight C
rightwards double arrow space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses squared over 2 plus straight C 

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20.

Show that the four points A, B, C and D with position vectors
4 straight i with hat on top plus 5 straight j with hat on top plus straight k with hat on top comma negative straight j with hat on top minus straight k with hat on top comma space 3 straight i with hat on top plus 9 straight j with hat on top plus 4 straight k with hat on top space and space 4 left parenthesis negative straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis respectively are coplanar.

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