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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.

4254 Views

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

920 Views

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

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.

1117 Views

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

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?

Let charges for typing one English page be Rs. x.

Let charges for typing one Hindi page be Rs.y.

Thus from the given statements, we have,

10x + 3y = 145

3x + 10y = 180

Thus the above system can be written as,

$open square brackets table row 10 3 row 3 10 end table close square brackets open square brackets table row straight x row straight y end table close square brackets space space equals space open square brackets table row 145 row 180 end table close square brackets rightwards double arrow space AX space equals space straight B comma space where comma space straight A space equals space open square brackets table row 10 3 row 3 10 end table close square brackets comma space straight X space space equals space open square brackets table row straight x row straight y end table close square brackets space and space open square brackets table row 145 row 180 end table close square brackets Multiply space straight A to the power of negative 1 end exponent space on space both space the space sides comma space we space have comma straight A to the power of negative 1 end exponent space straight x space AX space equals space straight A to the power of negative 1 end exponent straight B rightwards double arrow space IX space space equals space straight A to the power of negative 1 end exponent straight B rightwards double arrow space straight X space equals space straight A to the power of negative 1 end exponent straight B Thus comma space we space need space to space find space the space inverse space of space the space matrix space straight A. We space know space that comma space if space space straight P space equals space open square brackets table row straight a straight b row straight c straight d end table close square brackets space then space straight p to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator ad minus bc end fraction space open square brackets table row straight d cell negative straight b end cell row cell negative straight c end cell straight a end table close square brackets Thus comma space straight A to the power of negative 1 end exponent space equals space fraction numerator 1 over denominator 10 space straight x space 10 minus 3 space straight x space 3 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets equals fraction numerator 1 over denominator 100 minus 9 end fraction open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets equals 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets therefore space space straight X space equals space 1 over 91 open square brackets table row 10 cell negative 3 end cell row cell negative 3 end cell 10 end table close square brackets space space open square brackets table row 145 row 180 end table close square brackets equals 1 over 91 open square brackets table row cell 10 space straight x space 145 space minus end cell cell 3 straight x space 180 end cell row cell negative 3 space straight x space 145 space plus end cell cell 10 straight x 180 end cell end table close square brackets space equals 1 over 91 open square brackets table row 910 row 1365 end table close square brackets equals open square brackets table row 10 row 15 end table close square brackets rightwards double arrow space open square brackets table row straight x row straight y end table close square brackets space equals space open square brackets table row 10 row 15 end table close square brackets therefore comma straight x space equals 10 space and space straight y space equals space 15 Amount space taken space from space shyam space equals 2 space straight x space 5 space equals Rs. space 10 Actual space rate space equals space 15 space straight x space 5 space equals Rs. space 75 Difference space amount space equals space 75 minus 10 space equals space 65 Rs. space 65 space was space less space charged space from space the space poor space boy space Shyam.$
Humanity is reflected in this problem.