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

Download the PDF Sample Papers Free for off line practice and view the Solutions online.

CBSE Class 12 Mathematics Solved Question Paper 2015

Short Answer Type

Write the element $straight a subscript 12$ of the matrix $straight A space equals space open square brackets straight a subscript ij close square brackets subscript 2 cross times 2 end subscript comma$ whose elements $straight a subscript ij$ are given by $aij space equals space straight e to the power of 2 ix end exponent space sin space jx.$

1662 Views

2. $If space straight A space equals space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets comma space then space show space that space straight A squared minus 4 straight A minus 5 straight I space equals 0 comma space and space hence space find space straight A to the power of negative 1 end exponent space space space space space space space space space space space space space space space space space space space space space space space space$

straight A space equals space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets straight A squared equals open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets space open square brackets table row 1 2 2 row 2 1 2 row 2 2 1 end table close square brackets space space space equals space open square brackets table row cell 1 cross times 1 plus 2 cross times 2 plus 2 cross times 2 end cell cell space space 1 cross times 2 plus 2 cross times 1 plus 2 cross times 2 end cell cell space space space space space 1 cross times 2 plus 2 cross times 2 plus 2 cross times 1 end cell row cell 2 cross times 1 plus 1 cross times 2 plus 2 cross times 2 end cell cell space 2 cross times 2 plus 1 cross times 1 plus 2 cross times 2 end cell cell space space space space 2 cross times 2 plus 1 cross times 2 plus 2 cross times 1 end cell row cell 2 cross times 1 plus 2 cross times 2 plus 1 cross times 2 end cell cell space 2 cross times 2 plus 2 cross times 1 plus 1 cross times 2 end cell cell space space space space 2 cross times 2 plus 2 cross times 2 plus 1 cross times 1 end cell end table close square brackets space equals space open square brackets table row cell 1 plus 4 plus 4 end cell cell space space space 2 plus 2 plus 4 end cell cell space space 2 plus 4 plus 2 end cell row cell 2 plus 2 plus 4 end cell cell space space 4 plus 1 plus 4 end cell cell space space 4 plus 2 plus 2 end cell row cell 2 plus 4 plus 2 end cell cell space 4 plus 2 plus 2 end cell cell space 4 plus 4 plus 1 end cell end table close square brackets equals space open square brackets table row 9 8 8 row 8 9 8 row 8 8 9 end table close square brackets

Conisder space straight A squared minus 4 straight A minus 5 straight I space equals space open square brackets table row 9 8 8 row 8 9 8 row 8 8 9 end table close square brackets space minus space 4 open square brackets table row 1 cell space 2 end cell cell space 2 end cell row 2 1 2 row 2 cell space 2 end cell 1 end table close square brackets space minus space 5 open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets space equals open square brackets table row 9 cell space 8 end cell cell space 8 end cell row 8 cell space 9 end cell cell space 8 end cell row 8 cell space 8 end cell cell space 9 end cell end table close square brackets space minus space open square brackets table row 4 cell space 8 end cell cell space 8 end cell row 8 cell space 4 end cell cell space 8 end cell row 8 cell space 8 end cell cell space 4 end cell end table close square brackets space minus space open square brackets table row 5 cell space 0 end cell cell space 0 end cell row 0 cell space 5 end cell cell space 0 end cell row 0 0 cell space 5 end cell end table close square brackets space equals space open square brackets table row cell 9 minus 9 end cell cell space 8 minus 8 end cell cell space space 8 minus 8 end cell row cell 8 minus 8 end cell cell space 9 minus 9 end cell cell space 8 minus 8 end cell row cell 8 minus 8 end cell cell space 8 minus 8 end cell cell space 9 minus 9 end cell end table close square brackets equals space open square brackets table row 0 0 0 row 0 0 0 row 0 0 0 end table close square brackets

Now straight A squared minus 4 straight A minus 5 straight I space equals space 0 straight A squared minus 4 straight A space equals space 5 straight I straight A squared straight A to the power of negative 1 end exponent space minus 4 AA to the power of negative 1 end exponent space equals space 5 IA to the power of negative 1 end exponent left parenthesis Postmultiply space by space straight A to the power of negative 1 end exponent right parenthesis straight A minus 4 straight I space equals space 5 straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 2 end cell cell space 2 end cell row 2 1 cell space 2 end cell row 2 cell space 2 end cell cell space 1 end cell end table close square brackets space minus space open square brackets table row 4 cell space 0 end cell cell space 0 end cell row 0 cell space 4 end cell cell space 0 end cell row 0 cell space 0 end cell cell space 4 end cell end table close square brackets space equals space 5 straight A to the power of negative 1 end exponent open square brackets table row cell negative 3 end cell cell space space space 2 end cell cell space space space 2 end cell row 2 cell negative 3 end cell cell space space space 2 end cell row 2 cell space space 2 end cell cell negative 3 end cell end table close square brackets space equals space 5 straight A to the power of negative 1 end exponent

straight A to the power of negative 1 end exponent space equals space open square brackets table row cell fraction numerator negative 3 over denominator 5 end fraction end cell cell space 2 over 5 end cell cell space 2 over 5 end cell row cell 2 over 5 end cell cell fraction numerator negative 3 over denominator 5 end fraction end cell cell 2 over 5 end cell row cell 2 over 5 end cell cell 2 over 5 end cell cell fraction numerator negative 3 over denominator 5 end fraction end cell end table close square brackets

616 Views

If space straight A space equals space open vertical bar table row 2 cell space space 0 end cell cell negative 1 end cell row 5 cell space 1 end cell cell space 0 end cell row 0 cell space 1 end cell cell space 3 end cell end table close vertical bar comma space then space find thin space straight A to the power of negative 1 end exponent space using space elementary space row space operations.

Applying space straight R subscript 3 space rightwards arrow space straight R subscript 3 plus left parenthesis negative 1 right parenthesis straight R subscript 2 straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell negative 1 half end cell row 0 cell space 1 end cell cell space 5 over 2 end cell row 0 cell space 1 end cell cell 1 half end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell space space 0 end cell cell space 0 end cell row cell negative 5 over 2 end cell 1 cell space 0 end cell row cell 5 over 2 end cell cell negative 1 end cell cell space 1 end cell end table close square brackets Applying space straight R subscript 3 space rightwards arrow space left parenthesis 2 right parenthesis straight R subscript 3 straight A to the power of negative 1 end exponent open square brackets table row 1 cell space 0 end cell cell negative 1 half end cell row 0 1 cell space 5 over 2 end cell row 0 cell space 0 end cell 1 end table close square brackets space equals space open square brackets table row cell 1 half end cell 0 cell space space space 0 end cell row cell negative 5 over 2 end cell 1 cell space space 0 end cell row 5 cell negative 2 end cell cell space space 2 end cell end table close square brackets

460 Views

Using the properties of determinants, solve the following for x:

$open vertical bar table row cell straight x plus 2 end cell cell space space straight x plus 6 end cell cell space straight x minus 1 end cell row cell straight x plus 6 end cell cell straight x minus 1 end cell cell straight x plus 2 end cell row cell straight x minus 1 end cell cell straight x plus 2 end cell cell straight x plus 6 end cell end table close vertical bar space equals space 0$

751 Views

Solve the following for x:

$sin to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis minus 2 sin to the power of negative 1 end exponent straight x equals straight pi over 2$

536 Views

Show that:
$2 sin to the power of negative 1 end exponent open parentheses 3 over 5 close parentheses minus tan to the power of negative 1 end exponent open parentheses 17 over 31 close parentheses equals space straight pi over 4$

638 Views

Long Answer Type

Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b+ad) for (a, b), (c, d) $element of$ A. Then find
(i) The identify element of * in A.
(ii) Invertible elements of A, and write the inverse of elements (5, 3) and $open parentheses 1 half comma space 4 close parentheses.$

1348 Views

Let space straight f colon straight W rightwards arrow straight W space be space defined space as straight f left parenthesis straight n right parenthesis space equals space open curly brackets table row cell straight n minus 1 comma space space if space straight n space is space odd end cell row cell straight n plus 1 comma space if space straight n space is space even end cell end table close curly brackets

Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers.

579 Views

Find the absolute maximum and absolute minimum values of the function f given by
$straight f left parenthesis straight x right parenthesis space equals sin squared straight x minus cosx comma space straight x space element of space left parenthesis 0 comma space straight pi right parenthesis$

619 Views

Short Answer Type

10.

If $straight a with rightwards arrow on top space equals space 2 straight i with hat on top plus straight j with hat on top plus 3 straight k with hat on top space space space and space straight b with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 5 straight j with hat on top space minus space 2 straight k with hat on top comma space then space find space open vertical bar straight a with rightwards arrow on top cross times space straight b with rightwards arrow on top close vertical bar.$