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.


Let ABC be the right angle triangle with base b and hypotenuse h. 
Given that b+ h = k
Let A be the area of the right triangle.

straight A equals 1 half cross times straight b cross times square root of straight h squared minus straight b squared end root
rightwards double arrow space space straight A squared space equals space 1 fourth straight b squared left parenthesis straight h squared minus straight b squared right parenthesis
rightwards double arrow space straight A squared equals space space straight b squared over 4 open parentheses open parentheses straight k minus straight b close parentheses squared minus straight b squared close parentheses space space space space space space open square brackets because space straight h space equals space straight k minus straight b close square brackets
rightwards double arrow straight A squared space equals space straight b squared over 4 open parentheses straight k squared plus straight b squared minus 2 kb minus straight b squared close parentheses
rightwards double arrow straight A squared space equals space straight b squared over 4 left parenthesis straight k squared minus 2 kb right parenthesis
rightwards double arrow straight A squared equals space fraction numerator straight b squared straight k squared minus 2 kb cubed over denominator 4 end fraction
Differentiating the above function with respect to be, we have

2 straight A dA over db equals fraction numerator 2 bk squared minus 6 kb squared over denominator 4 end fraction space... left parenthesis 1 right parenthesis
rightwards double arrow space space dA over db equals fraction numerator bk squared minus 3 kb squared over denominator 2 straight A end fraction
For the area to be maximum, we have

dA over db equals 0
rightwards double arrow space bk squared minus 3 kb squared space equals 0
rightwards double arrow space bk space equals space 3 straight b squared
rightwards double arrow space straight b space equals space straight k over 3
Again differentiating the function in equation (1), with respect to b, we have
2 open parentheses dA over db close parentheses squared plus 2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction equals fraction numerator 2 straight k squared minus 12 kb over denominator 4 end fraction space... left parenthesis 2 right parenthesis
Now substituting dA over db equals 0 space and space straight b space equals space straight k over 3 in equation (2), we have
2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction space equals space fraction numerator 2 straight k squared minus 12 straight k open parentheses begin display style straight k over 3 end style close parentheses over denominator 4 end fraction
space space space rightwards double arrow 2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction space equals space fraction numerator 6 straight k squared minus 12 straight k squared over denominator 12 end fraction
space space rightwards double arrow space 2 straight a fraction numerator straight d squared straight A over denominator db squared end fraction equals negative straight k squared over 2
space space space rightwards double arrow space fraction numerator straight d squared straight A over denominator db squared end fraction equals negative fraction numerator straight k squared over denominator 4 straight A end fraction less than 0
Thus area is maximum at straight b equals straight k over 3.
Now, straight h equals straight k minus straight k over 3 equals fraction numerator 2 straight k over denominator 3 end fraction
Let straight theta be the angle between the base of the triangle and the hypotenuse of the right angle. 

Thus comma space cosθ equals straight b over straight h equals fraction numerator begin display style straight k over 3 end style over denominator begin display style fraction numerator 2 straight k over denominator 3 end fraction end style end fraction equals 1 half
rightwards double arrow space straight theta equals space cos to the power of negative 1 end exponent open parentheses 1 half close parentheses equals space straight pi over 3

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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

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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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