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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

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

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

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4.
Using the properties of determinants, solve the following for x:

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

Let space increment space equals space 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

Applying space straight C subscript 2 rightwards arrow straight C subscript 2 minus straight C subscript 1 space and space straight C subscript 3 space rightwards arrow space straight C subscript 3 space minus straight C subscript 1 increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row cell straight x plus 6 end cell cell space minus 7 end cell cell space minus 4 end cell row cell straight x minus 1 end cell cell space 3 end cell cell space space space 7 end cell end table close vertical bar Applying space straight R subscript 2 space rightwards arrow straight R subscript 2 minus straight R subscript 1 space and space straight R subscript 3 space rightwards arrow straight R subscript 3 minus straight R subscript 1 space space space increment equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row 4 cell space minus 11 end cell cell space minus 1 end cell row cell negative 3 end cell cell space minus 1 end cell cell space space 10 end cell end table close vertical bar space Applying space straight R subscript 2 rightwards arrow straight R subscript 2 space plus space straight R subscript 3 increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space space 4 end cell cell space space minus 3 end cell row 1 cell space minus 12 end cell cell space space space 9 end cell row cell negative 3 end cell cell space minus 1 end cell cell space 10 end cell end table close vertical bar Applying space straight R subscript 3 space rightwards arrow straight R subscript 3 plus left parenthesis 3 right parenthesis straight R subscript 2 space increment space equals space open vertical bar table row cell straight x plus 2 end cell cell space 4 end cell cell space minus 3 end cell row 1 cell negative 12 end cell cell space space 9 end cell row 0 cell negative 37 end cell cell space 37 end cell end table close vertical bar Expanding space along space straight C subscript 1

increment space equals space left parenthesis straight x plus 2 right parenthesis space open vertical bar table row cell negative 12 end cell cell space space 9 end cell row cell negative 37 end cell cell space 37 end cell end table close vertical bar minus 1 open vertical bar table row 4 cell space minus 3 end cell row cell negative 37 end cell cell space space 37 end cell end table close vertical bar increment equals left parenthesis straight x plus 2 right parenthesis thin space left parenthesis negative 444 plus 333 right parenthesis minus 1 left parenthesis 148 minus 111 right parenthesis increment equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis negative 111 right parenthesis minus 1 left parenthesis 37 right parenthesis therefore space increment space equals space 0 space equals space minus 111 straight x minus 259 therefore space space straight x space equals space minus 259 over 111 equals negative 7 over 3

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Solve the following for x:

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

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