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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

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

7.
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 $open parentheses 1 half comma space 4 close parentheses.$

Let A = Q x Q, where Q is the set of rational numbers.
Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for
(a, b), (c, d) ∈ A.
(i)
We need to find the identity element of the operation * in A.
Let (x, y) be the identity element in A.
Thus,
(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a, b) ∈ A
⇒(ax, b + ay) = (a, b)
⇒ ax = a and b + ay =b
⇒ y = 0 and x = 1
Therefore, (1, 0) ∈ A is the identity element in A with respect to the operation *.

(ii) We need to find the invertible elements of A.
Let (p, q) be the inverse of the element (a, b)
Thus,
$left parenthesis straight a comma space straight b right parenthesis asterisk times left parenthesis straight p comma space straight q right parenthesis space equals space left parenthesis 1 comma space 0 right parenthesis rightwards double arrow space left parenthesis ap comma space straight b plus aq right parenthesis space equals left parenthesis 1 comma space 0 right parenthesis rightwards double arrow space ap space equals space 1 space and space straight b plus aq space equals space 0 rightwards double arrow space straight p space equals 1 over straight a space space and space straight q equals negative straight b over straight a space space Thus space the space inverse space elements space of space left parenthesis straight a comma space straight b right parenthesis space is space open parentheses 1 over straight a comma space minus straight b over straight a close parentheses space space Now space let space us space find space the space inverse space of space left parenthesis 5 comma space 3 right parenthesis space and space open parentheses 1 half comma space 4 close parentheses space Hence comma space inverse space of space left parenthesis 5 comma space 3 right parenthesis space is space open parentheses 1 fifth comma space minus 3 over 5 close parentheses And space inverse space of space open parentheses 1 half comma space 4 close parentheses space is space open parentheses 2 comma space fraction numerator negative 4 over denominator begin display style 1 half end style end fraction close parentheses space equals space space left parenthesis 2 comma space minus 8 right parenthesis$

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