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CBSE Class 12 Mathematics Solved Question Paper 2016

Short Answer Type

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

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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. $Solve space for space straight x space colon space tan to the power of negative 1 end exponent space left parenthesis straight x space plus 1 right parenthesis space plus space tan to the power of negative 1 end exponent straight x space plus tan to the power of negative 1 end exponent space left parenthesis straight x plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent 3 straight x.$

Given space that comma tan to the power of negative 1 end exponent space left parenthesis straight x minus 1 right parenthesis space plus space tan to the power of negative 1 end exponent space straight x space plus space 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 rightwards double arrow space space tan to the power of negative 1 end exponent space left parenthesis straight x minus 1 right parenthesis space plus space 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 space 3 straight x space minus tan to the power of negative 1 end exponent space straight x space.... space left parenthesis straight i right parenthesis we space know space that comma space tan to the power of negative 1 end exponent space straight A space plus space tan to the power of negative 1 end exponent space straight B 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 and comma space tan to the power of negative 1 end exponent straight A space minus space tan to the power of negative 1 end exponent straight B space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator straight A minus straight B over denominator 1 plus AB end fraction close parentheses Thus comma space tan to the power of negative 1 end exponent left parenthesis straight x minus 1 right parenthesis space plus space 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 space equals space open parentheses fraction numerator straight x minus 1 plus straight x plus 1 over denominator 1 minus left parenthesis straight x minus 1 right parenthesis left parenthesis straight x plus 1 right parenthesis end fraction close parentheses equals space space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 minus left parenthesis straight x to the power of 2 minus end exponent 1 right parenthesis end fraction close parentheses equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction close parentheses space space.. space left parenthesis ii right parenthesis Similarly comma space tan to the power of negative 1 end exponent 3 straight x space minus space tan to the power of negative 1 end exponent space straight x space space equals space space tan to the power of negative 1 end exponent space open parentheses fraction numerator 3 straight x minus straight x over denominator 1 plus 3 straight x left parenthesis straight x right parenthesis end fraction close parentheses equals space space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction close parentheses space.. space left parenthesis iii right parenthesis From space space equ. space left parenthesis straight i right parenthesis space comma space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis comma space we space have comma tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction close parentheses space space equals space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction close parentheses space rightwards double arrow space fraction numerator 2 straight x over denominator 2 minus straight x squared end fraction space equals fraction numerator 2 straight x over denominator 1 plus 3 straight x squared end fraction cross space multiply space the space above space equ. rightwards double arrow space 2 minus straight x squared space equals space 1 plus 3 straight x squared rightwards double arrow space 4 straight x squared minus 1 rightwards double arrow space 4 straight x squared space equals 1 rightwards double arrow space straight x space equals space plus for minus of space 1 half

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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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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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