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    Mathematics

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    CBSE Class 12

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    CBSE Class 12 Mathematics Solved Question Paper 2016

    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?

    1308 Views
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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
    676 Views
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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.
    1266 Views
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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
    1206 Views
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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

    Given space that straight y space equals space sin to the power of negative 1 end exponent space open square brackets fraction numerator 6 straight x minus 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets if space straight y space equals space sin to the power of negative 1 end exponent straight x comma space then space dy over dx space equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x minus 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets rightwards double arrow space space straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 6 straight x over denominator 5 end fraction minus fraction numerator 4 square root of 1 space minus space 4 straight x squared end root over denominator 5 end fraction close square brackets rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets fraction numerator 2 straight x.3 over denominator 3 end fraction space minus space fraction numerator 4 square root of 1 space minus space left parenthesis 2 straight x right parenthesis squared end root over denominator 5 end fraction close square brackets rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets 2 straight x.3 over 5 minus 4 over 5 square root of 1 space minus space left parenthesis 2 straight x right parenthesis squared end root close square brackets rightwards double arrow space straight y space equals space sin to the power of negative 1 end exponent open square brackets 2 straight x space square root of 1 minus open parentheses 4 over 5 close parentheses squared end root minus space 4 over 5 square root of 1 space minus left parenthesis 2 straight x right parenthesis squared end root close square brackets we space know space that comma sin to the power of negative 1 end exponent straight p space minus space sin to the power of negative 1 end exponent straight q space equals space sin to the power of negative 1 end exponent space left parenthesis straight p square root of 1 space minus space straight q squared end root minus straight q-th root of 1 space minus space straight p squared end root right parenthesis Here comma space straight p space equals space 2 straight x space space and space straight q space equals 4 over 5 Differentiating space the space above space functions space with space respect space straight x comma space space we space have comma space dy over dx space equals space fraction numerator 1 over denominator square root of 1 minus left parenthesis 2 straight x right parenthesis squared end root end fraction space straight x space 2 minus 0 rightwards double arrow space space dy over dx space equals space fraction numerator 2 over denominator square root of 1 minus 4 straight x squared end root end fraction space
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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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