Subject

Mathematics

Class

CBSE Class 12

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1. If space straight A space equals space open parentheses table row cell cos space straight alpha end cell cell sin space straight alpha end cell row cell negative sin space straight alpha end cell cell cos space straight alpha end cell end table close parentheses comma space find space straight alpha space satisfying space 0 space less than space straight alpha space less than space straight pi over 2 space when space straight A space plus space straight A to the power of straight T space space equals space square root of 2 straight I subscript 2 semicolon space Where space straight A to the power of straight T space is space
transpose space of space straight A.
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2.

If A is a 3 x 3 matrix |3A| = k|A|, then write the value of k.

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3.

For what values of k, the system of linear equations

x + y + z = 2
2x + y - z = 3
3x + 2y + kz = 4

has a unique solution?

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4. Solve space for space straight x space colon space tan to the power of negative 1 end exponent space left parenthesis straight x space plus 1 right parenthesis space plus space tan to the power of negative 1 end exponent straight x space plus tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent 3 straight x.
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5. Prove space that space tan to the power of negative 1 end exponent space open parentheses fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction close parentheses minus tan to the power of negative 1 end exponent space open parentheses fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction close parentheses space equals space tan to the power of negative 1 end exponent space 2 straight x semicolon space vertical line 2 straight x vertical line less than fraction numerator 1 over denominator square root of 3 end fraction
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6.

A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However, typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?

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7. If space space straight f open parentheses straight x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin open parentheses straight a plus 1 close parentheses plus 2 sinx over denominator straight x end fraction comma space straight x less than 0 end cell row cell 2 space space space space space space space space space space space space space space space space space space space space space space comma space x equals 0 end cell row cell fraction numerator square root of 1 plus b x end root minus 1 over denominator straight x end fraction space space space space space space comma space straight x greater than 0 end cell end table close
is continuous at x = 0, then find the values of a and b.


Given space that space straight f space is space continous space at space straight x space equals space 0

If space space straight f open parentheses straight x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin open parentheses straight a plus 1 close parentheses plus 2 sinx over denominator straight x end fraction comma space straight x space less than space 0 end cell row cell 2 space space space space space space space space space space space space space space space space space space space space space space comma space x equals 0 end cell row cell fraction numerator square root of 1 plus b x end root minus 1 over denominator straight x end fraction space space space space space space comma space straight x space greater than space 0 end cell end table close

Since space straight f left parenthesis straight x right parenthesis space is space continous space at space straight x equals 0 comma space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis space equals limit as straight x rightwards arrow 0 of straight f space left parenthesis 0 right parenthesis
Thus space straight R. straight H. straight L space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis

space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis 0 plus straight h right parenthesis

equals space limit as straight h rightwards arrow 0 of space space fraction numerator square root of 1 plus bh end root minus 1 over denominator straight h end fraction

equals limit as straight h rightwards arrow 0 of space fraction numerator square root of 1 plus bh end root minus 1 over denominator straight h end fraction space straight x space fraction numerator square root of 1 plus bh end root plus 1 over denominator square root of 1 plus bh end root plus 1 end fraction

equals space limit as straight h rightwards arrow 0 of space fraction numerator 1 plus bh minus 1 over denominator straight h left parenthesis square root of 1 plus bh end root plus 1 right parenthesis end fraction

equals space limit as straight h rightwards arrow 0 of space fraction numerator straight b over denominator square root of 1 plus bh end root plus 1 end fraction

equals straight b over 2

Given space that space straight f space left parenthesis straight x right parenthesis space equals space 2

rightwards double arrow space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis space space equals space straight f left parenthesis 0 right parenthesis

rightwards double arrow space straight b over 2 space space equals space 2 space

rightwards double arrow space straight b equals space 4

Similarly comma space

straight L. straight H. straight L space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis straight x right parenthesis

space equals space limit as straight x rightwards arrow 0 of straight f space left parenthesis 0 minus straight h right parenthesis

equals space limit as straight h rightwards arrow 0 of fraction numerator sin space left parenthesis straight a plus 1 right parenthesis left parenthesis 0 minus straight h right parenthesis plus 2 sin left parenthesis 0 minus straight h right parenthesis over denominator 0 minus straight h end fraction

equals space limit as straight h rightwards arrow 0 of fraction numerator negative sin left parenthesis straight a plus 1 right parenthesis straight h minus 2 space sin space straight h over denominator negative straight h end fraction

equals space limit as straight h rightwards arrow 0 of fraction numerator negative sin space left parenthesis straight a plus 1 right parenthesis straight h over denominator negative straight h end fraction space plus space limit as straight h rightwards arrow 0 of fraction numerator negative 2 sin space straight h over denominator negative straight h end fraction

equals space limit as straight h rightwards arrow 0 of fraction numerator sin left parenthesis straight a plus 1 right parenthesis straight h over denominator straight h end fraction fraction numerator left parenthesis straight a plus 1 right parenthesis over denominator left parenthesis straight a plus 1 right parenthesis end fraction space plus space 2 stack space lim with straight h rightwards arrow 0 below fraction numerator sin space straight h over denominator straight h end fraction

equals space straight a plus 1 plus 2 space space space space space space open square brackets therefore space lim space fraction numerator sin space straight theta over denominator straight theta end fraction equals 1 space close square brackets

Given space that space straight f space left parenthesis straight x right parenthesis equals 2

rightwards double arrow stack space lim with straight x space rightwards arrow 0 below space straight f space left parenthesis straight x right parenthesis space equals space straight f space left parenthesis 0 right parenthesis
rightwards double arrow space straight a plus 1 plus space 2 space space equals space 2
rightwards double arrow space straight a space equals negative 1
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8. if space cos space left parenthesis straight a plus straight y right parenthesis space equals space cos space straight y space then space prove space that space dy over dx space equals fraction numerator cos squared left parenthesis straight a plus straight y right parenthesis over denominator sin space straight a end fraction
Hence space show space that space sin space straight a space fraction numerator straight d squared straight y over denominator dx squared end fraction plus sin space 2 space left parenthesis straight a plus straight y right parenthesis dy over dx equals 0
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9. Find space dy over dx space if space straight y equals sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 minus 4 straight x squared end root over denominator 5 end fraction close square brackets
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10.

Find the equation of tangents to the curve y= x3+2x-4, which are perpendicular to line x+14y+3 =0.

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