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


Taking L.H.S,
tan to the power of negative 1 end exponent open parentheses fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction close parentheses space minus space tan to the power of negative 1 end exponent open parentheses fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction close parentheses

We space know space that comma
tan to the power of negative 1 end exponent space left parenthesis straight A right parenthesis space minus space tan to the power of negative 1 end exponent space left parenthesis straight B right parenthesis space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight A plus straight B over denominator 1 plus AB end fraction close parentheses

Thus comma space straight L. straight H. straight S space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction end style minus begin display style fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction end style over denominator 1 plus open parentheses begin display style fraction numerator 6 straight x minus 8 straight x cubed over denominator 1 minus 12 straight x squared end fraction end style close parentheses open parentheses begin display style fraction numerator 4 straight x over denominator 1 minus 4 straight x squared end fraction end style close parentheses end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style fraction numerator left parenthesis 6 straight x minus 8 straight x cubed right parenthesis left parenthesis 1 minus 4 straight x squared right parenthesis minus 4 straight x left parenthesis 1 minus 12 straight x squared right parenthesis over denominator left parenthesis 1 minus 12 straight x squared right parenthesis left parenthesis 1 minus 4 straight x right parenthesis squared end fraction end style over denominator begin display style fraction numerator 4 straight x left parenthesis 6 straight x minus 8 straight x cubed right parenthesis over denominator left parenthesis 1 minus 12 straight x squared right parenthesis left parenthesis 1 minus 4 straight x squared right parenthesis end fraction end style end fraction close parentheses space

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator left parenthesis 6 straight x minus 24 straight x cubed minus 8 straight x cubed plus 32 straight x to the power of 5 minus 4 straight x plus 48 straight x cubed over denominator 1 minus 4 straight x squared minus 12 straight x squared plus 48 straight x to the power of 4 plus 24 straight x squared minus 32 straight x to the power of 4 end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 32 straight x to the power of 5 plus 16 straight x to the power of 6 plus 2 straight x over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x left parenthesis 16 straight x to the power of 4 plus 2 straight x over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses

equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x left parenthesis 16 straight x to the power of 4 plus 8 straight x squared plus 1 over denominator 16 straight x to the power of 4 plus 8 straight x squared plus 1 end fraction close parentheses

equals space tan to the power of negative 1 end exponent space 2 straight x

Thus comma space straight L. straight H. straight S space equals space straight R. straight H. straight S

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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.
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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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