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CBSE Class 12 Mathematics Solved Question Paper 2013

Short Answer Type

Write the principal value of $tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis plus cos to the power of negative 1 end exponent open parentheses negative 1 half close parentheses.$

1188 Views

Write the value of $open parentheses 2 tan to the power of negative 1 end exponent 1 fifth close parentheses$

264 Views

Find the value of a if $open square brackets table row cell straight a minus straight b end cell cell space space space 2 straight a plus straight c end cell row cell 2 straight a minus straight b end cell cell space 3 straight c plus straight d end cell end table close square brackets space equals space open square brackets table row cell negative 1 end cell cell space space space 5 end cell row 0 cell space 13 end cell end table close square brackets$

171 Views

If $open vertical bar table row cell straight x plus 1 end cell cell space space straight x minus 1 end cell row cell straight x minus 3 end cell cell space straight x plus 2 end cell end table close vertical bar space equals space open vertical bar table row 4 cell space space minus 1 end cell row 1 cell space space space space 3 end cell end table close vertical bar comma$ then write the value of x.

193 Views

If $open square brackets table row cell space 9 end cell cell space minus 1 end cell cell space space space 4 end cell row cell negative 2 end cell cell space space space 1 end cell cell space space space 3 end cell end table close square brackets space equals space straight A plus open square brackets table row 1 cell space space 2 end cell cell negative 1 end cell row 0 cell space 4 end cell 9 end table close square brackets comma$ then find the matrix A.

207 Views

Find the value of the following:
$tan 1 half open square brackets sin to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction plus cos to the power of negative 1 end exponent fraction numerator 1 minus straight y squared over denominator 1 plus straight y squared end fraction close square brackets comma space open vertical bar straight x close vertical bar space less than 1 comma space space straight y greater than 0 space and space xy less than 1$

218 Views

Prove that $tan to the power of negative 1 end exponent open parentheses 1 half close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 fifth close parentheses plus tan to the power of negative 1 end exponent open parentheses 1 over 8 close parentheses space equals straight pi over 4$

179 Views

Using properties of determinants prove the following:
$open vertical bar table row 1 cell space space straight x end cell cell space space straight x squared end cell row cell straight x squared end cell cell space 1 end cell straight x row straight x cell space straight x end cell 1 end table close vertical bar space equals space left parenthesis 1 minus straight x cubed right parenthesis squared$

247 Views

Show that the function $straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 3 close vertical bar comma space straight x element of bold R bold comma$ is continuous but not differentiable at x=3.

281 Views

Long Answer Type

10.
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

Let the award money given for honesty, regularity and hard work be Rs. x, Rs. y and Rs. z respectively.
Since total cash award is Rs. 6,000.
$therefore space straight x plus straight y plus straight z space equals space 6 comma 000 space... left parenthesis 1 right parenthesis$
Three times the award money for hard work and honesty amounts to Rs.11,000.
$therefore space space straight x plus 3 straight z space equals space 11 comma 000 rightwards double arrow space space straight x plus 0 cross times straight y plus 3 straight z space equals space 11 comma 000 space... left parenthesis 2 right parenthesis$
Award money for honesty and hard work is double that given for regularity.
$therefore space straight x plus straight z space equals space 2 straight y rightwards double arrow space space straight x minus 2 straight y plus straight z space equals space 0 space space space... left parenthesis 3 right parenthesis$
The above system of equations can be written in matrix form AX = B as:

$open square brackets table row 1 cell space space space 1 end cell cell space space space 1 end cell row 1 cell space space space 0 end cell cell space space space 3 end cell row 1 cell space minus 2 end cell cell space space space 1 end cell end table close square brackets space open square brackets table row x row y row z end table close square brackets space equals space open square brackets table row 6000 row 11000 row 0 end table close square brackets$
Here,

$straight A space equals space open square brackets table row 1 cell space space space 1 end cell cell space space 1 end cell row 1 cell space space space 0 end cell cell space space 3 end cell row 1 cell space minus 2 end cell cell space space 1 end cell end table close square brackets comma space straight X space equals space open square brackets table row straight x row straight y row straight z end table close square brackets space space and space straight B space equals space open square brackets table row 6000 row 11000 row 0 end table close square brackets open vertical bar straight A close vertical bar space equals space 1 left parenthesis 0 plus 6 right parenthesis minus 1 left parenthesis 1 minus 3 right parenthesis plus 1 left parenthesis negative 2 minus 0 right parenthesis space equals space 6 not equal to 0$
Thus, A is non-singular. Hence, it is invertible.

$Adj space straight A space equals space open square brackets table row 6 cell space minus 3 end cell cell space space space space space 3 end cell row 2 cell space space space 0 end cell cell space space minus 2 end cell row cell negative 2 end cell cell space space 3 end cell cell space minus 1 end cell end table close square brackets$
$therefore space space straight A to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator open vertical bar straight A close vertical bar end fraction left parenthesis adj space straight A right parenthesis space equals space 1 over 6 open square brackets table row 6 cell space minus 3 end cell cell space space 3 end cell row 2 cell space space 0 end cell cell negative 2 end cell row cell negative 2 end cell cell space 3 end cell cell negative 1 end cell end table close square brackets straight X space equals space straight A to the power of negative 1 end exponent straight B space equals space 1 over 6 open square brackets table row 6 cell space space minus 3 end cell cell space space space 3 space end cell row 2 cell space space space 0 end cell cell negative 2 end cell row cell negative 2 end cell cell space space 3 end cell cell negative 1 end cell end table close square brackets open square brackets table row 6000 row 11000 row 0 end table close square brackets equals 1 over 6 open square brackets table row cell 36000 minus 33000 plus 0 end cell row cell 12000 plus 0 minus 0 end cell row cell negative 12000 plus 33000 minus 0 end cell end table close square brackets equals 1 over 6 open square brackets table row 3000 row 12000 row 21000 end table close square brackets rightwards double arrow open square brackets table row straight x row straight y row straight z end table close square brackets space equals open square brackets table row 500 row 2000 row 35000 end table close square brackets$

Hence, x = 500, y = 2000, and z = 3500.
Thus, award money given for honesty, regularity and hardwork is Rs. 500, Rs.2000 and Rs. 3500 respectively.
The school can include awards for obedience.