Subject

Mathematics

Class

CBSE Class 12

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

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 Multiple Choice QuestionsShort Answer Type

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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


Consider the determinant
increment space equals open vertical bar table row cell 1 plus straight a end cell 1 1 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
Taking abc common outside, we have

increment equals abc space open vertical bar table row cell 1 over straight a plus 1 end cell cell 1 over straight b end cell cell space 1 over straight c end cell row cell 1 over straight a end cell cell 1 over straight b plus 1 end cell cell 1 over straight c end cell row cell 1 over straight a end cell cell 1 over straight b end cell cell space 1 over straight c plus 1 end cell end table close vertical bar
Applying the transformation, straight C subscript 1 rightwards arrow straight C subscript 1 plus straight C subscript 2 plus straight C subscript 3 comma

increment equals space abc open vertical bar table row cell 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c space space space end cell cell 1 over straight b end cell cell 1 over straight c end cell row cell 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c end cell cell space 1 over straight b plus 1 end cell cell 1 over straight c end cell row cell 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c end cell cell 1 over straight b end cell cell space space 1 over straight c plus 1 end cell end table close vertical bar

rightwards double arrow increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses space open vertical bar table row 1 cell 1 over straight b end cell cell 1 over straight c end cell row cell 1 space end cell cell 1 over straight b plus 1 end cell cell 1 over straight c end cell row 1 cell 1 over straight b end cell cell 1 over straight c plus 1 end cell end table close vertical bar
Apply the transformations, straight R subscript 2 rightwards arrow straight R subscript 1 space and space straight R subscript 3 rightwards arrow straight R subscript 3 minus straight R subscript 1

increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses space open vertical bar table row 1 cell space 1 over straight b end cell cell 1 over straight c end cell row 1 1 0 row 1 0 1 end table close vertical bar
Expanding along straight C subscript 1, we have
increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses cross times 1 cross times open vertical bar table row 1 0 row 0 1 end table close vertical bar
rightwards double arrow space increment equals abc open parentheses 1 plus 1 over straight a plus 1 over straight b plus 1 over straight c close parentheses equals abc plus ab plus bc plus ca

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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

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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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