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

1.

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.

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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
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3. 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


open vertical bar straight A close vertical bar space equals space open vertical bar table row 2 cell space space space 0 end cell cell space minus 1 end cell row 5 cell space space space 1 end cell cell space space space 0 end cell row 0 cell space space space 1 end cell cell space space space 3 end cell end table close vertical bar
space space equals space 2 left parenthesis 3 minus 0 right parenthesis minus 0 left parenthesis 15 minus 0 right parenthesis minus 1 left parenthesis 5 minus 0 right parenthesis
space space equals 6 minus 0 minus 5
space space equals space 1
space space not equal to 0
space
Hence space straight A to the power of negative 1 end exponent space exists.
straight A to the power of negative 1 end exponent straight A space equals space 1
straight A to the power of negative 1 end exponent open square brackets table row 2 cell space space 0 end cell cell space minus 1 end cell row cell space 5 space end cell cell space 1 end cell cell space space space 0 end cell row 0 cell space 1 end cell cell space space space 3 end cell end table close square brackets space equals space open square brackets table row 1 cell space 0 end cell cell space 0 end cell row 0 cell space 1 end cell cell space 0 end cell row 0 cell space 0 end cell cell space 1 end cell end table close square brackets
Applying space straight R subscript 1 space rightwards arrow space open parentheses 1 half close parentheses straight R subscript 1
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 5 1 cell space space 0 end cell row 0 1 cell space 3 end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell space 0 end cell cell space space space 0 end cell row 0 cell space 1 end cell cell space space space 0 end cell row 0 cell space 0 end cell cell space space space 1 end cell end table close square brackets
Applying space straight R subscript 2 space rightwards arrow space straight R subscript 2 space plus left parenthesis negative 5 right parenthesis straight R subscript 1
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 space 5 over 2 end cell row 0 1 cell space 3 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 space space 0 end cell row cell negative 5 over 2 end cell cell space 1 end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space space 1 end cell end table close square brackets
Applying space straight R subscript 3 space rightwards arrow 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 space 0 end cell cell space space minus 1 half end cell row 0 cell space space 1 end cell cell space space space space 5 over 2 end cell row 0 cell space space 0 end cell cell space space space space 1 half end cell end table close square brackets space equals space open square brackets table row cell 1 half end cell cell 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 1 end table close square brackets
Applying space straight R subscript 3 space rightwards arrow 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 space minus 1 half end cell row 0 cell space 1 end cell cell space space 5 over 2 end cell row 0 cell space 0 end cell cell space 1 end cell end table close square brackets space equals open square brackets table row cell 1 half end cell cell space 0 end cell 0 row cell negative 5 over 2 end cell 1 0 row 5 cell negative 2 end cell cell space 2 end cell end table close square brackets
Applying space straight R subscript 1 space rightwards arrow space straight R subscript 1 space plus space open parentheses 1 half close parentheses straight R subscript 3
space space space space space space space space space space space space straight R subscript 2 space rightwards arrow space straight R subscript 2 space plus space open parentheses negative 5 over 2 close parentheses straight R subscript 3
space space straight A to the power of negative 1 end exponent open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets space equals space open square brackets table row 3 cell space minus 1 end cell cell space space space 1 end cell row cell negative 15 end cell cell space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space space 2 end cell end table close square brackets
straight A to the power of negative 1 end exponent space equals space open square brackets table row 3 cell space minus 1 end cell cell space space space space 1 end cell row cell negative 15 end cell cell space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 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

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5.

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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6.

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

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 Multiple Choice QuestionsLong 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.

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8. 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. 
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9.

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

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 Multiple Choice QuestionsShort 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.

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